-Quadratic Equations- Solve each by factoring, write each equation in standard form first.
Answers
Answer
The solutions to the quadratic equations are
[tex]\begin{gathered} a^2-4a-45 \\ \text{Solution: }a=-5\text{ or }9 \\ \\ 5y^2+4y=0 \\ \text{Solution: }y=0\text{ or }-\frac{4}{5} \end{gathered}[/tex]SOLUTION
Problem Statement
The question gives us 2 quadratic equations and we are required to solve them by factoring, first writing them in their standard forms.
The quadratic equations given are:
[tex]\begin{gathered} a^2-4a-45=0 \\ 5y^2+4y=0 \end{gathered}[/tex]Method
To solve the questions, we need to follow these steps:
(We will represent the independent variable as x for this explanation. We know they are "a" and "y" in the questions given)
The steps outlined below are known as the method of Completing the Square.
Step 1: Find the square of the half of the coefficient of x.
Step 2: Add and subtract the result from step 1.
Step 3: Re-write the Equation. This will be the standard form of the equation
Step 4. Solve for x
We will apply these steps to solve both questions.
Implementation
Question 1:
[tex]\begin{gathered} a^2-4a-45=0 \\ \text{Step 1: Find the square of the half of the coefficient of }a \\ (-\frac{4}{2})^2=(-2)^2=4 \\ \\ \text{Step 2: Add and subtract 4 to the equation} \\ a^2-4a-45+4-4=0 \\ \\ \text{Step 3: Rewrite the Equation} \\ a^2-4a+4-45-4=0 \\ (a^2-4a+4)-49=0 \\ (a^2-4a+4)=(a-2)^2 \\ \therefore(a-2)^2-49=0 \\ \text{ In standard form, we have:} \\ (a-2)^2=49 \\ \\ \text{Step 4: Solve for }a \\ (a-2)^2=49 \\ \text{ Find the square root of both sides} \\ \sqrt[]{(a-2)^2}=\pm\sqrt[]{49} \\ a-2=\pm7 \\ \text{Add 2 to both sides} \\ \therefore a=2\pm7 \\ \\ \therefore a=-5\text{ or }9 \end{gathered}[/tex]Question 2:
[tex]\begin{gathered} 5y^2+4y=0 \\ \text{ Before we begin solving, we should factorize out 5} \\ 5(y^2+\frac{4}{5}y)=0 \\ \\ \text{Step 1: Find the square of the coefficient of the half of y} \\ (\frac{4}{5}\times\frac{1}{2})^2=(\frac{2}{5})^2=\frac{4}{25} \\ \\ \text{Step 2: Add and subtract }\frac{4}{25}\text{ to the equation} \\ \\ 5(y^2+\frac{4}{5}y+\frac{4}{25}-\frac{4}{25})=0 \\ \\ \\ \text{Step 3: Rewrite the Equation} \\ 5((y^2+\frac{4}{5}y+\frac{4}{25})-\frac{4}{25})=0 \\ 5(y^2+\frac{4}{5}y+\frac{4}{25})-5(\frac{4}{25})=0 \\ 5(y^2+\frac{4}{5}y+\frac{4}{25})-\frac{4}{5}=0 \\ \\ (y^2+\frac{4}{5}y+\frac{4}{25})=(y+\frac{2}{5})^2 \\ \\ \therefore5(y+\frac{2}{5})^2-\frac{4}{5}=0 \\ \\ \text{ In standard form, the Equation becomes} \\ 5(y+\frac{2}{5})^2=\frac{4}{5} \\ \\ \\ \text{Step 4: Solve for }y \\ 5(y+\frac{2}{5})^2=\frac{4}{5} \\ \text{ Divide both sides by 5} \\ \frac{5}{5}(y+\frac{2}{5})^2=\frac{4}{5}\times\frac{1}{5} \\ (y+\frac{2}{5})^2=\frac{4}{25} \\ \\ \text{ Find the square root of both sides} \\ \sqrt[]{(y+\frac{2}{5})^2}=\pm\sqrt[]{\frac{4}{25}} \\ \\ y+\frac{2}{5}=\pm\frac{2}{5} \\ \\ \text{Subtract }\frac{2}{5}\text{ from both sides} \\ \\ y=-\frac{2}{5}\pm\frac{2}{5} \\ \\ \therefore y=0\text{ or }-\frac{4}{5} \end{gathered}[/tex]Final Answer
The solutions to the quadratic equations are
[tex]\begin{gathered} a^2-4a-45 \\ \text{Solution: }a=-5\text{ or }9 \\ \\ 5y^2+4y=0 \\ \text{Solution: }y=0\text{ or }-\frac{4}{5} \end{gathered}[/tex]Related Questions
See the attached for the math problem
Answers
1. If the cake rises by ¹/₃ as it bakes, the number of cups of cake batter needed for the four cakes is 140.
2. If ¹/₄ in. is used between layers and ¹/₂ in. is used on the top and sides, the number of cups of icing needed for the four cakes is 47.
How are the numbers determined?The number of cups of cake batter and icing can be determined using the mathematical operations of multiplication, addition, division, and subtraction.
First, the volumes of each cake and its batter are calculated using the given dimensions and the rise.
Using the division operation, the number of cups of cake batter for each cake is determined and multiplied by four.
We understand that the normal volume of the cake will increase with the icing, helping us to calculate the increased volume after the icing.
The difference between the two volumes becomes the volume of the icing required, which is divided by 14.4 in³ to get the number of cups of icing required.
a) Cups of Cake Butter:The volume of each cake = Length x Width x Height
Length = 14 inches
Width = 12 inches
Height = 4 inches (2 x 2)
= 12 x 14 x 4
= 672 in³.
Rise of the cake as it bakes = ¹/₃
The normal volume before rising = 1
Risen volume = 1¹/₃
1¹/₃ = 672 in³
The normal volume of cake batter before the ¹/₃ rise = 504 in³ (672/1¹/₃).
1 cup = 14.4 in³, the total cups for each cake = 35 cups (504 in³/14.4 in³).
The total cups of cake batter for the 4 cakes = 140 cups (35 x 4).
b) Cups of Icing:The total quantity of icing = 1¹/₄ (¹/₄ + ¹/₂ + ¹/₂).
The new volume after the icing = 840 in³ (672 x 1¹/₄)
The difference in volume after the icing = 168 in³ in (840 in³ - 672 in³)
If 1 cup = 14.4 in³, the cups of icing for each cake = 11.67 cups (168 in³/14.4 in³).
The total cups of icing for the 4 cakes = 47 cups (11.67 x 4).
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Question Completion:Four double-layered cakes, each 12.0 in. x 14.0 in., have been ordered for a special event. Each layer is 2.0 in. high.
a) If the cake rises by ¹/₃ as it bakes, how many cups of cake batter are needed? (1 cup = 14.4 in³. Hint: The ¹/₃ rise should be treated as a constant.)
b) How many cups of icing are needed if ¹/₄ in. is used between layers and ¹/₂ in. is used on the top and sides? (Assume icing is not layered on top of the icing. 1 cup = 14.4 in³.)
Can someone help me with this geometry question?A.Triangular prismB.Hexagonal prismC.Triangular pyramidD.Hexagonal pyramid
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B. Hexagonal Prism
1) One prism is defined, in terms of naming it by the base.
2) Counting the edges of the base in this net surface, we can tell that this is a Hexagonal Prism for the base is a hexagon.
If p(x) is a polynomial function where p(x) = 3(x + 1)(x - 2)(2x-5)a. What are the x-intercepts of the graph of p(x)?b. What is the end behavior (as x→ ∞, f(x)→?? and as x→ -∞, f (x)→ ??) of p(x))?c. Find an equation for a polynomial q(x) that has x-intercepts at -2, 3⁄4, and 7.
Answers
Hello there. To solve this question, we have to remember some properties about polynomial functions.
Given the polynomial function
[tex]p(x)=3(x+1)(x-2)(2x-5)[/tex]We want to determine:
a) What are the x-intercepts of the graph of p(x)?
For this, we have to determine the roots of the polynomial function p(x). In this case, we have to determine for which values of x we have
[tex]p(x)=0[/tex]Since p(x) is written in canonical form, we find that
[tex]p(x)=3(x+1)(x-2)(2x-5)=0[/tex]A product is equal to zero if at least one of its factors is equal to zero, hence
[tex]x+1=0\text{ or }x-2=0\text{ or }2x-5=0[/tex]Solving the equations, we find that
[tex]x=-1\text{ or }x=2\text{ or }x=\dfrac{5}{2}[/tex]Are the solutions of the polynomial equation and therefore the x-intercepts of p(x).
b) What is the end-behavior of p(x) as x goes to +∞ or x goes to -∞?
For this, we have to take the limit of the function.
In general, for polynomial functions, those limits are either equal to ∞ or -∞, depending on the degree of the polynomial and the leading coefficient.
For example, a second degree polynomial function with positive leading coefficient is a parabola concave up and both limits for the function as x goes to ∞ or x goes to -∞ is equal to ∞.
On the other hand, an odd degree function usually has an odd number of factors (the number of x-intercepts in the complex plane) hence the limits might be different.
In this case, we have a third degree polynomial equation and we find that, as the leading coefficient is positive and all the other factors are monoic, that
[tex]\begin{gathered} \lim_{x\to\infty}p(x)=\infty \\ \\ \lim_{x\to-\infty}p(x)=-\infty \end{gathered}[/tex]That is, it gets larger and larger when x is increasing arbitrarily, while it get smaller and smaller as x is decreasing.
c) To find the equation for a polynomial q(x) that has x-intercepts at -2, 3/4 and 7.
The canonical form of a polynomial of degree n with x-intercepts at x1, x2, ..., xn and leading coefficient equals a is written as
[tex]f(x)=a\cdot(x-x_1)(x-x_2)\cdots(x-x_n)[/tex]So in this case, there are infinitely many polynomials satisfying this condition. Choosing a = 1, we find that q(x) is equal to
[tex]\begin{gathered} q(x)=(x-(-2))\cdot\left(x-\dfrac{3}{4}\right)\cdot(x-7) \\ \\ \boxed{q(x)=(x+2)\cdot\left(x-\dfrac{3}{4}\right)\cdot(x-7)} \end{gathered}[/tex]These are the answers to this question.
what is the image of -3 -7 after a reflection over the x-axis
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Given the point (-3, -7)
We need to find the image after a reflection over the x-axis
The rule of reflection over the x-axis is:
[tex](x,y)\rightarrow(x,-y)[/tex]So, the image of the given point will be:
[tex](-3,-7)\rightarrow(-3,7)[/tex]so, the answer is option D. (-3, 7)
A rocket is shot off from a launcher. The accompanying table represents the height of the rocket at given times, where x is time, in seconds, and y is height, in feet. Write a quadratic regression equation for this set of data, rounding all coefficients to the nearest tenth. Using this equation, find the height, to the nearest foot, at a time of 3.8 seconds.
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Given
The data can be modeled using a quadratic regression equation.
Using the general form of a quadratic equation:
[tex]y=ax^2\text{ + bx + c}[/tex]We should use a regression calculator to obtain the required coefficients. The graph of the equation is shown below:
The coefficients of the equation is:
[tex]\begin{gathered} a\text{ = -17.5 (nearest tenth)} \\ b\text{ = }249.0\text{ (nearest tenth)} \\ c\text{ = }-0.5 \end{gathered}[/tex]Hence, the regression equation is:
[tex]y=-17.5x^2\text{ + 249.0x -0.5}[/tex]We can find the height (y) at a time of 3.8 seconds by substitution:
[tex]\begin{gathered} y=-17.5(3.8)^2\text{ + 249}(3.8)\text{ - 0.5} \\ =\text{ }693 \end{gathered}[/tex]Hence, the height at time 3.8 seconds is 693 ft
the variable w varies inversely as the cube of v. if k is the constant of variation, which equation represents this situation?a: qv^=kb: q^3 v= kc: q/v^3=kd: q^3/v=k picture listed below
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Solution
Given that:
[tex]\begin{gathered} q\propto\frac{1}{v^3} \\ \\ \Rightarrow q=\frac{k}{v^3} \\ \\ \Rightarrow k=qv^3 \end{gathered}[/tex]Option A.
Describe how the graph of the function g(x)=1/4|x|-2 can be obtained from the basic graph. Then graph the function.Start with the graph of h(x)=|x|. Then [__] it vertically by a factor of [__]. Finally, shift it [___] units.
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Start the graph of h(x) = |x|, then stretch it vertically by a factor of 1/4 . Finally, shift it down by 2 units
The original graph can be seen above
what it becomes can be seen below
Hence the final answer is option B
Rewrite the expression by factoring out (y+4).3(y + 4)+7y(y+4)
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Given-:
Function is:
[tex]3(y+4)+7y(y+4)[/tex]Find-:
Expression by factoring
Explanation-:
Factoring the function is:
[tex]\begin{gathered} =3(y+4)+7y(y+4) \\ \\ =(y+4)(3+7y) \end{gathered}[/tex]Factoring is:
[tex]=(y+4)(3+7y)[/tex]in two or more complete sentences, compare the slopes of the two functions. in your comparison, include which function has the greatest slope.
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Slope of g (x) : ZERO. The slope of any horizontal line is zero, 0.
Slope of f (x) :
let's take the points ( -4, 7) and (-2, 5) from the table
[tex]m=\frac{y_2-y_1}{x_2-x_1}=\frac{5-7}{-2-(-4)}=\frac{-2}{-2+4}=\frac{-2}{2}=-1[/tex]Answer: The slope of g (x ) is zero since it is a horizontal line while the slope of f (x) is -1. The slope of g(x) i greater than the slope of f(x).
what is the slope of a line perpendicular to this linewhat is the slope of a line parallel to this line
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Answer:
• Slope perpendicular to the line: 8/5
,• Slope parallel to the line: –5/8
Explanation
Given
[tex]5x+8y=7[/tex]To know the result, it is better if we work with the slope-intercept form:
[tex]y=mx+b[/tex]Then, to get this kind of form we have to isolate y from the given equation:
[tex]8y=7-5x[/tex][tex]y=\frac{7-5x}{8}[/tex][tex]y=-\frac{5}{8}x+\frac{7}{8}[/tex]Thus, in this case, m = –5/8 and b = 7/8.
Perpendicular lines have negative reciprocal lines:
[tex]m_2=-\frac{1}{m_1}[/tex]where m₁ is the slope of line 1 and m₂ is the line perpendicular to line 1.
Then, replacing the values:
[tex]m_2=-\frac{1}{-\frac{5}{8}}[/tex][tex]m_2=\frac{8}{5}[/tex]Finally, the slopes of parallel lines are the same, meaning:
[tex]m_2=m_1[/tex]where m₁ is the slope of line 1 and m₂ is the line parallel to line 1.
fine the slope of every line that is parallel to the line on the graph
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Every parallel line would have the same slope because the slope formula is Δy/Δx and the difference would be the same, so the slope for the line with the given points would be -1/6, or roughly 0.167.
What is parallel lines?Parallel lines in geometry are coplanar, straight lines that don't cross at any point. In the same three-dimensional space, parallel planes are any planes that never cross. Curves with a fixed minimum distance between them and no contact or intersection are said to be parallel.
What is slope?A line's steepness can be determined by looking at its slope. In mathematics, slope is determined by dividing the change in y by the change in x. Determine the coordinates of two points along the line that you choose. Find the difference between these two points' y-coordinates (rise). Find the difference between these two points' x-coordinates (run). Divide the difference in x-coordinates (rise/run or slope) by the difference in y-coordinates.
Here the coordinates are (-6,0) and (0,-1)
ΔX = 0 – -6 = 6
ΔY = -1 – 0 = -1
Slope (m) =ΔY/ΔX
=-1/6
= -0.16666666666667
≈-0.167
The slope for the line with the given points would be -1/6, or roughly 0.167, because the slope formula is Δy/Δx and the difference would be the same for every parallel line.
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The function gives the cost to manufacture x items. C(x) = 15,000 + 8x - x2 -; X = 20,000 20,000 Find the average cost per unit of manufacturing h more items (i.e., the average rate of change of the total cost) at a production level of x, where x is as indicated a smaller values of h to check your estimates. Round your answers to five decimal places.) h 10 1 Cave 5.99950 5.9995 x Estimate the instantaneous rate of change of the total cost at the given production level x, specifying the units of measurement. c' (20,000) = 6 $/item A Need Help? Read It Watch It
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We can replace x=20000 in the function so:
[tex]c(20000)=15000+8(20000)-\frac{20000^2^{}}{20000}[/tex]and we simplify:
[tex]c(20000)=15500[/tex]now h=1 is the cost of one more item so we evaluate for 20001
[tex]\begin{gathered} c(20001)=15000+8(20001)-\frac{20001^2}{20000} \\ c(20001)=195010 \end{gathered}[/tex]So for h=1 will be :
[tex]C=0.599950[/tex]what is the substitution for f7=3(x)^2+2(x)-9
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Given a function f(x), whenever you want to evaluate the function, you simply change the variable for the value you where you want to evaluate the function at, and then perform the mathematical operations the function tells you to do.
In our case f(x) = 3x^2 + 2x -9
If we evaluate f(x) at x=7, then
[tex]f(7)=3(7)^2+2(7)\text{ -9 = 3 }\cdot\text{ 49 + 2}\cdot\text{ 7 - 9 = 152}[/tex]So f(7) = 152.
A tennis racket cost 12$ more than a hockey stick , if the price of the two is 31$ find the cost of a tennis racket
Answers
So $31 - 12 = 19
19 / 2 = 9.5
So 9.5 is the cost of the hockey stick, now lets find the cost of the racket
$9.5 + 12 = 21.5
Therefore the cost of the tennis racket is $21.5
Karen has 5 more quarters than dimes. She has $3.70. How many quarters and dimes she have?
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A dime is 10% of a dollar = 10/100 x 100 cent = 10 cents
A quater is 25% of a dollar = 25/100 x 100 cent = 25 cents
Since Karen has 5 more quaters than dimes
let quaters = q
let dimes = d
Then Karen has 5q : d = $ 3.70
$ 3.70 = 3.70 x 100 cents = 370 cents
The following are the distances (in miles) to the nearest airport for 13 families.10, 13, 15, 15, 20, 26, 27, 28, 30, 32, 34, 37, 39Notice that the numbers are ordered from least to greatest.Give the five-number summary and the interquartile range for the data set.
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We have the next given set for distances (in miles) to the nearest for 13airport families:
10, 13, 15, 15, 20, 26, 27, 28, 30, 32, 34, 37, 39
The minimum is the least number value. Then:
Minimum =10
In this case, we have 13 data, so :
- The middle number is the median:
10, 13, 15, 15, 20, 26, 27, 28, 30, 32, 34, 37, 39
Now, the lower quartile is given by the next equation:
[tex]=(n+1)\ast\frac{1}{4}[/tex]Replacing:
[tex]\begin{gathered} =(13+1)\ast\frac{1}{4} \\ =14\ast\frac{1}{4} \\ =3.5=4 \end{gathered}[/tex]The lower quartile is in the fourth position:
Lower quartile = 15
The upper quartile is given by the next equation:
[tex]\begin{gathered} =(n+1)\ast\frac{3}{4} \\ =(13+1)\ast\frac{3}{4} \\ =10.5=11 \end{gathered}[/tex]The upper quartile is located in the 11th position:
Upper quartile = 34
The interquartile range is given by:
IQR=upper quartile - lower quartile
IQR=34-15
The interquartile range =19
Scatter PlotWhich statement best describes the association betweenvariable X and variable Y?.moderate negative association. Perfect negative association. Moderate positive association. Perfect positive association
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It's moderate negative association
How many radians are equal to 360 degrees 2 2pi 1 Pi
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Answer:
2pi
Explanation:
By definition, 360 degrees are equal to 2π radians.
This follows from the fact that the circumference of a circle is 2π times the radius. Therefore, if radius = 1, then
circumference = 2π
Since the circumference is the distance around a circle, and degrees are the "angular distance " around the circle, these two quantities can be related.
So if you think of the circle in terms of the circumference, a circle measures 2π. If you think in terms of degrees, a circle measures 360 degrees.
Therefore, we say
360 degrees = 2π (radians)
What is the mean for the data shown in the dot plot?
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We will determine the mean as follows:
[tex]x=\frac{1(4)+4(5)+3(6)+2(7)+1(10)}{11}\Rightarrow x=6[/tex]So, the mean will be 6.
Mr. McFall uses 2% cups of peanuts for every 1/2 cup of chocolate chips in a mixture. Enter the number of cups of peanuts for every 1 cup of chocolate chips. Remember to reduce.
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To solve this problem I'll use proportions.
2 1/8 cups ------------------------ 1/2 cup of chocolate.
x ----------------------- 1 cup od chocolate chips
x = (1*2 1/8) / 1/2
x = 17/8 / 2
x = 4 % cups of peanuts
Graph the following inequality.Note: To graph the inequality:Select the type of line below (solid or dashed).Plot two points on the line.Click on the side that should be shaded.
Answers
Given:
[tex]-4x-y>2[/tex]Consider the line,
[tex]-4x-y=2[/tex]Find the points on the line,
[tex]\begin{gathered} x=0 \\ -4(0)-y=2\Rightarrow y=-2 \\ x=1 \\ -4(1)-y=2\Rightarrow y=-6 \\ x=-1 \\ -4(-1)-y=2\Rightarrow y=2 \end{gathered}[/tex]The graph of the line is,
Now, find the region for inequality.
Consider any point from the right and the left side of the line and check which side satisfies the inequality.
[tex]\begin{gathered} R=(2,0) \\ -4x-y>2 \\ -4(2)-0=-8\text{ >2 is not true.} \end{gathered}[/tex]And,
[tex]\begin{gathered} L=(-2,0) \\ -4x-y>2 \\ -4(-2)-0=8>2\text{ is true} \end{gathered}[/tex]Therefore, the graph of the inequality is,
Note that inequality does not contain boundary points.
A number divisible by 2, 5 and 10 if the last digit is _______.
A. An even number
B. O
C. 0 or 5
D. An odd number
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Answer :- B) 0
Only a number ending with the digit 0 is divisible by 2,5 and 10
Example :-
20 ÷ 2 = 10
20 ÷ 5 = 4
20 ÷ 10 = 2
Here, 20 is the number that ends with 0.
Solve for t. If there are multiple solutions, enter them as a
Answers
we have the equation
[tex]\frac{12}{t}+\frac{18}{(t-2)}=\frac{9}{2}[/tex]Solve for t
step 1
Multiply both sides by 2t(t-2) to remove fractions
[tex]\frac{12\cdot2t(t-2)}{t}+\frac{18\cdot2t(t-2)}{(t-2)}=\frac{9\cdot2t(t-2)}{2}[/tex]simplify
[tex]12\cdot2(t-2)+18\cdot2t=9\cdot t(t-2)[/tex][tex]24t-48+36t=9t^2-18t[/tex][tex]\begin{gathered} 60t-48=9t^2-18t \\ 9t^2-18t-60t+48=0 \\ 9t^2-78t+48=0 \end{gathered}[/tex]Solve the quadratic equation
Using the formula
a=9
b=-78
c=48
substitute
[tex]t=\frac{-(-78)\pm\sqrt[]{-78^2-4(9)(48)}}{2(9)}[/tex][tex]t=\frac{78\pm66}{18}[/tex]The solutions for t are
t=8 and t=2/3
therefore
the answer is
t=2/3,8
Question 8 of 20If f(x) = 2x²+2 and g(x)=x2-1, find (f- g)(x).O A.3x²+3O B.x2²+1 O C.3x2² +1O D.x² +3
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Solution
Step 1
Write the two functions:
[tex]f\mleft(x\mright)=2x²+2\text{ }and\text{ }g\left(x\right)=x2-1[/tex]Step 2
(f- g)(x) = f(g) - g(x)
[tex]\begin{gathered} \left(f-g\right)\left(x\right)=2x²+2\text{ - \lparen}x^2-1) \\ \\ (f-g)(x)=2x^2+\text{ 2 - x}^2\text{ + 1} \\ \\ (f-g)(x)=x^2+3 \end{gathered}[/tex]Final answer
D. x² +3
Trapezoid W'X'Y'Z' is the image of trapezoid W XYZ under a dilation through point C What scale factor was used in the dilation?
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The scale factor is basically by what we need to multiply the original to get the dilated one.
Simple.
We can see that the original one is Trapezoid WXYZ and the dilated one is W'X'Y'Z'.
THe dilated trapezoid is definitely bigger than original. So the scale factor should be larger than 1.
One side of original is "6" and the corresponding side of dilated trapezoid is "14".
So, what we have to do to "6", to get "14"??
This is the scale factor!
To get 14, we have to multiply 6 with, suppose, "x", so:
[tex]\begin{gathered} 6x=14 \\ x=\frac{14}{6} \\ x=\frac{7}{3} \end{gathered}[/tex]Hence, SF is 7/3
The first three terms of a sequence are given. Round to the nearest thousandth (if necessary). 6 36 1, 5' 25 . Find the 10th term
Answers
The first three terms of a sequence are given. Round to the nearest thousandth (if necessary)
1, 6/5, 36/25, Find the 10th term
__________________________________________________________________
1, 6/5, 36/25
(6/5)^(n-1)
n= 1
(6/5)^(1-1) = (6/5)^0 = 1
n= 2
(6/5)^(2-1) = (6/5)^1 = 6/5
n= 3
(6/5)^(3-1) = (6/5)^2 = 36/25
_______________________
n= 10
(6/5)^(10-1) = (6/5)^9 = 5. 1598
_______________________
Answer
Round to the nearest thousandth
The 10th term is 5.160
Please help me don't understand
Answers
Answer:
x=13
Step-by-step explanation:
50+3x=89
89-50=3x
39=3x
13=x
A school debate team has 4 girls and 6 boys. A total of 4 of the team members will be chosen top participate in the district debate. What is the probaility that 2 girland 2 boys will be selected?
Answers
The probability that 2 girls and 2 boys are selected is given by:
[tex]P(\text{ 2 boys, 2 girls })=\frac{4C2\times6C2}{10C4}=\frac{6\times15}{210}=\frac{3}{7}[/tex]Therefore, the correct choice is option D) 3/7.
R'Find the composition of transformationsthat map APQR to APQ'R'.QT RRotate [?] degrees aboutthe origin and then translate(unit(s) [ ].90180
Answers
For a transformation like the one shown in the graph, note that the points have switched as follows;
[tex]P(x,y)\Rightarrow P^{\prime}(-y,x)[/tex]This is a 90 degree anticlockwise rotation about the origin.
That means point P would now be translated as follows;
[tex]\begin{gathered} P(1,1)\Rightarrow P^{\prime}(-1,1) \\ Q(2,1)\Rightarrow Q^{\prime}(-1,2) \\ R(2,3)\Rightarrow R^{\prime}(-3,2) \end{gathered}[/tex]After this rotation, the coordinates have now moved from their position one unit upwards. That now makes the final coordinates;
[tex]undefined[/tex]I need geometry help please.
Answers
Lets remember the property "alternative-interior angles" when we have parallels lines:
Given two paralles lines, the following is true:
In our question, we have:
We know that the angles J + P + K=180
So, the triangles have all the angles the same, so they are similar by angle-angle-angle, and they have one side congruent, AC=CD, so the triangles are congruents.
