Can you please help me with 44Please use all 3 forms of the expression such as : down/up. As _,_ And limits
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Answer:
[tex]\begin{gathered} x-\text{intercept}=-3\text{ and 1} \\ y-\text{intercept}=\text{ -9} \end{gathered}[/tex][tex]\begin{gathered} x=1\text{ multiplicity 3} \\ x=-3\text{ multiplicity 2} \\ \lim _{x\rightarrow\infty}(x-1)^3(x+3)^2=\infty \\ \lim _{x\rightarrow-\infty}(x-1)^3(x+3)^2=-\infty \end{gathered}[/tex]Step-by-step explanation:
To find the x-intercepts factor the function to the simplest form:
[tex]h(x)=(x-1)^3(x+3)^2[/tex]As we can see the zeros to the function would be 1 and -3, then its:
[tex]\begin{gathered} x-\text{intercept}=-3\text{ and 1} \\ y-\text{intercept}=\text{ -9} \end{gathered}[/tex]Zero has a "multiplicity", which refers to the number of times that its associated factor appears in the polynomial. Therefore, this function has multiplicity:
[tex]\begin{gathered} x=1\text{ multiplicity 3} \\ x=-3\text{ multiplicity 2} \end{gathered}[/tex]For the end behavior:
down/up
As x approaches infinity f(x) approaches infinity
As x approaches -infinity f(x) approaches -infinity
[tex]\begin{gathered} \lim _{x\rightarrow\infty}(x-1)^3(x+3)^2=\infty \\ \lim _{x\rightarrow-\infty}(x-1)^3(x+3)^2=-\infty \end{gathered}[/tex]
Related Questions
Which of the following ordered pairs is a solution to the graph of the system of inequalities? Select all that apply(5.2)(-3,-4)(0.-3)(0.1)(-4,1)
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For this type of question, we should draw a graph and find the area of the common solutions
[tex]\begin{gathered} \because-2x-3\leq y \\ \therefore y\ge-2x-3 \end{gathered}[/tex][tex]\begin{gathered} \because y-1<\frac{1}{2}x \\ \therefore y-1+1<\frac{1}{2}x+1 \\ \therefore y<\frac{1}{2}x+1 \end{gathered}[/tex]Now we can draw the graphs of them
The red line represents the first inequality
The blue line represents the second inequality
The area of the two colors represents the area of the solutions,
Let us check the given points which one lies in this area
Point (5, -2) lies on the area of the solutions
∴ (5, -2) is a solution
Point (-3, -4) lies in the blue area only
∴ (-3, -4) not a solution
Point (0, -3) lies in the red line and the red line is solid, which means any point on it will be on the area of the solutions
∴ (0, -3) is a solution
Point (0, 1) lies in the blue line and the blue line is dashed, which means any point that lies on it not belong to the area of the solutions
∴ (0, 1) is not a solution
Point (-4, 1) lies on the area of the solutions
∴ (-4, 1) is a solution
The solutions are (5, -2), (0, -3), and (-4, 1)
Rewrite the following equation in slope-intercept form. x - 7y = 20 Write your answer using integers, proper fractions, and improper fractions in simplest form.
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In this case, we'll have to carry out several steps to find the solution.
Step 01:
x - 7y = 20
slope-intercept form = ?
Step 02:
Slope-intercept form of the line
y = mx + b
x - 7y = 20
x = 20 + 7y
x - 20 = 7y
7y = x - 20
[tex]y\text{ = }\frac{x}{7}\text{ - }\frac{20}{7}[/tex]The answer is:
y = x/7 - 20/7
Linear Programming WorksheetGraph each feasible region. maximize or minimize each objective
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Given:
x+2y = 8
x=2, y=0
Substitute x=2 then find value of x as,
2+2y=8
2y=6
y=3
(x,y) = (0,3)
Now, substitute y=0 then find value of y as,
x+2(0)=8
x=8
(x,y) = (8,0)
It is given that P = x+3y
(x,y) = (0,3) then P= 0+3x3
P=9
The maximum valu P=9 and vertiex (0,3)
(x,y) = (8,0) then P=8+0= 8
The mininmum val
In windy cold weather, the increased rate of heat loss makes the temperature feel colder than the actual temperature. To describe an equivalent temperature that more closely matches how it “feels,” weather reports often give a windchill index, WCI. The WCI is a function of both the temperature F(in degrees Fahrenheit) and the wind speed v (in miles per hour). For wind speeds v between 4 and 45 miles per hour, the WCI is given by the formula(FORMULA SHOWN IN PHOTO)A) What is the WCI for a temperature of 10 F in a wind of 20 miles per hour?B) A weather forecaster claims that a wind of 36 miles per hour has resulted in a WCI of -50 F. What is the actual temperature to the nearest degree?
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Let's remember what the variables mean:
F= temperature (in Fahrenheit),
v= wind speed.
A) The formula "works" when the wind speed is between 4 and 45 miles per hour. The question asks for a wind speed of 20 miles per hour. Then, we can apply the formula. Here,
[tex]\begin{cases}F=10 \\ v=20\end{cases}[/tex]Then,
[tex]\begin{gathered} WCI(10,20)=91.4-\frac{(10.45+6.69\cdot\sqrt[]{20}-0.447\cdot20)(91.4-10)}{22}\approx\ldots \\ \ldots91.4-116.2857=-24.8857 \end{gathered}[/tex]Approximating, the answer is
[tex]-25F[/tex]B) This question is just about to find F in the provided equation after replacing the given v and WCI. Let's do that:
[tex]\begin{gathered} -50=91.4-\frac{(10.45+6.69\cdot\sqrt[]{36}-0.447\cdot36)(91.4-F)}{22}, \\ -141.4=-\frac{(10.45+6.69\cdot\sqrt[]{36}-0.447\cdot36)(91.4-F)}{22}, \\ -3110.8=-(10.45+6.69\cdot\sqrt[]{36}-0.447\cdot36)(91.4-F), \\ 3110.8=(10.45+6.69\cdot\sqrt[]{36}-0.447\cdot36)(91.4-F), \\ \frac{3110.8}{10.45+6.69\cdot\sqrt[]{36}-0.447\cdot36}=91.4-F, \\ F=91.4-\frac{3110.8}{10.45+6.69\cdot\sqrt[]{36}-0.447\cdot36}\approx1.2 \end{gathered}[/tex]Then, the actual temperature is
[tex]1F[/tex]What makes a function a function?
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In a relationship between two variables x and y, the data set is a function, if every element of the domain corresponds to exactly one element of the range
that means
one element of x corresponds to exactly one element of y
In any function, there is an input value (independent variable or x variable) and there is an output value (dependent variable or y variable)
What makes a function a function? ------> one element of the input (variable x) corresponds to exactly one element of the output (variable y)
A right triangle is shown in the graph.
right triangle on coordinate plane with hypotenuse labeled t and one endpoint of hypotenuse at r comma s and the other endpoint at x comma y, vertical line from point x comma y and horizontal line from r comma s that meet at right angle of triangle, horizontal dotted line from point r comma s to point s on y axis, horizontal dotted line from point x comma y to point y on y axis, vertical dotted line from point r comma s to point r on x axis, and vertical dotted line from right angle to point x on x axis
Part A: Use the Pythagorean Theorem to derive the standard equation of the circle with center at (r, s) and a point on the circle at (x, y). Show all necessary math work. (3 points)
Part B: If (r, s) = (7, –4) and t = 10, determine the domain and range of the circle. (4 points)
Part C: Is the point (9, 1) inside the border of the circle if (r, s) = (7, –4) and t = 10? Explain using mathematical evidence. (3 points
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Part a: The standard equation of circle: (x - r)² + (y - s)² = t².
Part b: Domain = {17, -3} and Range = {-14, 6}.
Part c: Point (9, 1) lies inside the circle.
What is termed as the Pythagorean Theorem?The Pythagorean theorem, or Pythagorean theorem, explains the relation between the three sides of such a right-angled triangle. The the hypotenuse's square is equal to the total of the squares of the remaining two sides of a triangle, according to Pythagoras' theorem.For the given question,
The right triangle are given with two of ts vertices as (r, s) and (x, y).
The distance between these two points is 't'.
Part a: The standard equation of the circle.
Centre of circle = (r,s) and
Point on the circle = (x, y)
Using Pythagorean Theorem,
(x - r)² + (y - s)² = t²
Thus, the standard equation of the circle is (x - r)² + (y - s)² = t²
Where, t is the radius of the circle.
Part b: Domain and range.
(r, s) = (7, –4) and t = 10,
For x values in the domain r ± t and y values in the range s ± t, the circle would be defined.
Domain = 7 ± 10 = {17, -3}
Range = -4 ± 10 = {-14, 6}
Part c: Point (9, 1) lies inside or not.
(r, s) = (7, –4) and t = 10
Point (9, 1) = (x, y)
Put the values;
(x - r)² + (y - s)² ≤ t²
(9 - 7)² + (1 + 4)² ≤ 10²
2² + 5² ≤ 10²
4 + 25 ≤ 100
29 ≤ 100
Thus, the points (9, 1) lies inside the circle.
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A straight line l1 with equation 5x - 7 = 0 cuts the x axis at point A. Straight line l2 is perpendicular to straight line l1 and passes through point A. What is the coordinates of point A and the equation of the straight line l2?
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The coordinates of point A are (7/5, 0), and the perpendicular line that also passes through that point is:
y = 0.
How to get the perpendicular line?Here we want to get a line perpendicular to:
5x - 7 = 0
Solving this for x, we get:
5x = 7
x = 7/5.
This is a vertical line, so the perpendicular line will be a horizontal line, which is of the form:
y = a.
We know that the line:
x = 7/5.
Cuts the x-axis at point A.
Remember that the x-axis as coordinates (x, 0).
So the coordinates of point A are (7/5, 0).
Now, the perpendicular line:
y = a
Needs to pass through the point (7/5, 0), so the value of a must be zero, then the line is:
y = 0.
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Find the missing factor. x2 - 11x + 18 = (x - 2)( .) Enter the correct answer. 000 DONE Clear all DOO
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we have the second degree polynomial
[tex]x^2-11x+18[/tex]we must find two numbers a,b such that
[tex]\begin{gathered} x^2-11x+18=(x+a)(x+b)\text{ and} \\ a+b=11 \\ ab=18 \end{gathered}[/tex]We can see that, a=-2 and b=-9 fulfill the above conditions. Therefore, we have
[tex]x^2-11x+18=(x-2)(x-9)\text{ }[/tex]A company needs to take 10 sample sensor readings if the sensor collects data at 1/3 of a sample per second how long will it take the company to take all 10 samples
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Given:
Sample space = 10
Rate = 1/3 per second
Li’s family is saving money for their summer vacation. Their vacation savings account currently has a balance of $2,764. The family would like to have at least $5,000.Which inequality can be used to determine the amount of money the family still needs to save?
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EXPLANATION
Savings account balance = $2,764
Desired amount = $5,000
Let's call x to the amount of money the family needs.
The inequality that could be used to determine the amount of money the family needs is the following:
2,764 + x ≥ 5,000
What is lim (2x² - x + 3)/(3x² + 5) as x approaches + ∞?
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Given:
lim (2x² - x + 3)/(3x² + 5)
We are to
A soup can has a radius of 4.3 cm and a height of 11.6 cm. What is the volume of the soup can to the nearest tenth of a cubic centimeter?A. 1816.8B. 49.9C. 168.4D. 673.8
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hello
to solve this problem, we need to identify the shape of the soup can first since soup is a liquid and carries the shape of whatever container its in.
volume of a cylinder is given as
[tex]\begin{gathered} V=\pi r^2h \\ \pi=3.142 \\ r=\text{radius} \\ h=\text{height} \end{gathered}[/tex][tex]\begin{gathered} v=\text{ ?} \\ r=4.3\operatorname{cm} \\ h=11.6\operatorname{cm} \\ \pi=3.142 \\ v=\pi r^2h \\ v=3.142\times4.3^2\times11.6 \\ v=673.9\operatorname{cm}^3 \end{gathered}[/tex]from the calculations above, the volume of the soup is equal to 673.9cm^3 which corresponds with option D
Which statement explains whether x=5 is the solution to 5x + 2 = 27? a. Yes, because 5x means x=5.b. No, because 5x doesn't mean x=5.c. No, because when x is replaced by 5 the equation is false. d. Yes, because when x is replaced by 5 the equation is true.
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Given
x = 5
5x + 2 = 27
Procedure
d. Yes, because when x is replaced by 5 the equation is true.
I’ve already done this problem, but I’m being told it’s wrong and I need to simplify but I don’t know how to do it with this question.
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Solve equation for x. 5x^2 - 4x =6
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Answer:
Step-by-step explanation:
use the quadratic formula
5x^2-4x-6
4+-[tex]\sqrt{16+120}[/tex] all over 10
4+-[tex]2\sqrt{34}[/tex]/10
2+-[tex]\sqrt{34}[/tex]/5
D
If 25% of your math class received an A, how many students were in your math class if 9students earned an A for the semester?
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Answer:36
Step-by-step explanation:
Assume the total number of students to be x
According to the question
25% of x =9
⇒x=900/25=36
What is 9207 /10 equivalent to?
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Answer:
9207/10 is equivalent to 920.7
Solve for x using the quadratic formula.3x^2 +10x+8=3
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The quadartic equation is 3x^2+10x+8=3.
Simplify the quadratic equation to obtain the equation in standard form ax^2+bx+c=0.
[tex]\begin{gathered} 3x^2+10x+8=3 \\ 3x^2+10x+5=0 \end{gathered}[/tex]The coefficent of x^2 is a=3, coefficient of x is b=10 and constant term is c=5.
The quadartic formula for the values of x is,
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]Substitute the values in the formula to obtain the value of x.
[tex]\begin{gathered} x=\frac{-10\pm\sqrt[]{(10)^2-4\cdot3\cdot5}}{2\cdot3} \\ =\frac{-10\pm\sqrt[]{100-60}}{6} \\ =\frac{-10\pm\sqrt[]{40}}{6} \\ =\frac{-10\pm2\sqrt[]{10}}{6} \\ =\frac{-5\pm\sqrt[]{10}}{3} \end{gathered}[/tex]The value of x is,
[tex]\frac{-5\pm\sqrt[]{10}}{3}[/tex]Hello, can you please help me solve this question ASAP!!!
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SOLUTION:
Step 1:
In this question, we have that:
Step 2:
Part A:
We are meant to show that the equation:
[tex]5sinx=1+2cos^2x[/tex]can be written in the form
[tex]2sin^2\text{x + 5 sin x - 3=0}[/tex]Proof:
[tex]\begin{gathered} \text{5 sin x = 1 + 2 cos }^2x\text{ } \\ \text{But cos}^2x+sin^2x\text{ = 1} \\ \text{Then,} \\ \cos ^2x=1-sin^2x\text{ } \\ \text{Hence,} \\ 5sinx=1+2(1-sin^2x_{}) \\ 5sinx=1+2-2sin^2x \\ 5sinx=3-2sin^2x \end{gathered}[/tex]Re-arranging, we have that:
[tex]2sin^2x\text{ + 5 sin x - 3 = 0 }[/tex]Part B:
b) Hence, solve for x in the interval:
[tex]0\text{ }\leq\text{ x }\leq\text{ 2}\pi[/tex]top question says: Triangle ABC can be taken to triangle A'B'C' using rigid motions and a dilation. help me pls
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If triangle ABC can be taken to triangle A'B'C', it means that they are similar triangles. If tow triangles are similar, it means that the ratio of their corresponding sides are equal. Thus, we have
A'B'/AB = B'C'/BC = A'C'/AC
Thus, looking at the options, the true equations are
A) A'C'/B'A' = AC/BA
D) CA/C'A' = CB/C'B'
E) A'B'/AB = C'B'/CB
If we look at these options the ratios are always the same
Use disks and washers to find the volume of the solid the results when the area of the region y=x^3 y = 0, and x = 2 is revolved about the line x= 2
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Solution
The functions that define the region in consideration are given below:
[tex]\begin{gathered} y=x^3 \\ y=0 \\ x=2 \end{gathered}[/tex]The Washer Method:
- Plotting these functions would help us visualize the question better. This is done below:
- The question would like us to revolve around the region about line x = 2. The region is bounded by the Blue, Red, and Green line. This requires that we use the formula given below:
[tex]\begin{gathered} V=\int ^b_a{f(y)\mathrm{dy}} \\ \text{where,} \\ a\text{ and }b\text{ are the bounds of the integration along the y-axis} \end{gathered}[/tex]-
We can represent the region bounded by the function by rearranging the functions as follows:
[tex]undefined[/tex]If the price of gas was on average $2.85 per gallon, and thus was $1.36 cheaper than a year before, what is the percent of decrease in price?
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The price of gas = $2.85 per gallon
It was $1.36 cheaper than a year before.
So, the price before = 2.85 + 1.36 = $4.21
So, the percent of decrease = 1.36/4.21 = 0.323 = 32.3%
PLEASE HURRY!!!!!
Here is a hanger diagram. With a hanger diagram, when the diagram is balanced, there is equal weight on each side. Write an equation to represent the hanger. (Do NOT put spaces in your answer.)
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Answer:
x = 8
x = 3
x = 2
x = 1
x = 1
x = 2
x + x+ x+ x+ x+ 2 = 17
try any number
differentiate t^4 In(8cost)
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⇒It is way more appropriate if I use the product rule. That states that:
⇒f(x)g(x)=f'(x)g(x)+f(x)g'(x)
[tex]t^{4} In(8cos(t))\\=4t^{3}In(8cos(t))+t^{4} \frac{1}{8cos(t)} *(0cos(t)+8*(-sin(t))*1)\\=4t^{3}In(8cos(t))+\frac{t^{4}-8sin(t)}{8cos(t)}[/tex]
Note:
Given F(x)=In(x)
⇒[tex]F'(x)=\frac{1}{x}[/tex]
Goodluck
Answer:
t^3 (4 ln(cos8t) - t tant)
Step-by-step explanation:
Using the Product Rule:
dy/dt = t^4 * d(ln(8cost) / dt + ln(8cost) * d(t^4)/dt
= t^4 * 1/ (8cost) * (-8sint) + 4t^3 ln(8cost)
= -8t^4 sint / 8 cost + 4t^3 ln(8cost)
= -t^4 tan t + 4t^3 ln(8cost)
= t^3 (4 ln(cos8t) - t tant)
what is 2 to the 6 power
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What is the remainder when 5x3 + 2x2 - 7 is divided by x + 9?-93,7503,800-3,490
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Explanation
Given the expression
[tex]5x^3+2x^2-7[/tex]The remainder when it is divided by x+9 can be seen below;
[tex]r=5(-9)^3+2(-9)^2-7=-3645+162-7=-3490[/tex]Answer: -3490
2) Katie and Jacob are enlarging pictures in a school yearbook on the copy machine. The ratio of the width to the length of the enlarged photo will be the same as the ratio of the width to the length of the original photo. 25 points One of the photographs that they want to enlarge is a 3" x 4"photo. katie says that she can enlarge the photo to a 9" x 12", but Jacob disagrees. He says it will be 11" x 12". Who is correct? Explain your reasoning in words. * Enlarged Photo Original Photo 3 inches 4 inches
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The original picture Katie and Jacob want to enlarge is 3 by 4 photographs
This means that the initial length of the photograph is 3 and the intial width of the photographs is 4
If both of them want to enlarge the photograph, then the scaling factor must be the same for both the width and length
Katie enlarge the photo to a 9 x 12
The ratio of the original photograph is 3 to 4
That is, 3 : 4
Katie enlarge the photo to a 9 x 12
Ratio of the enlarged photo by katie is 9 to 12
That is, 9 : 12
Equate the two ratio together
3/4 = 9/12
Introduce cross multiplication
We have,
3 x 12 = 4 x 9
36 = 36
Therefore, the ratio which katie enlarged the photo results to a proportion
For Jacob
Jacob enlarged the photo to 11 x 12
Equating the two ratios
3/4 = 11/12
3 x 12 = 4 x 11
36 = 44
This does not give us a proportion
Therefore, Katie is correct while Jacob is wrong
need help asap look at attachment
Answers
Answer: Width =14, Length = 18
Step-by-step explanation:
L = W + 4
2W + 2L = 64
W+ L = 32
2W+ 4 = 32
2W = 28
W = 14
Kiran is solving 2x-3/x-1=2/x(x-1) for x, and he uses these steps.He checks his answer and finds that it isn’t a solution to the original equation, so he writes “no solutions.” Unfortunately, Kiran made a mistake while solving. Find his error and calculate the actual solution(s).
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Solution:
Given:
[tex]\begin{gathered} To\text{ solve,} \\ \frac{2x-3}{x-1}=\frac{2}{x(x-1)} \end{gathered}[/tex]Kiran multiplied the left-hand side of the equation by (x-1) and multiplied the right-hand side of the equation by x(x-1).
That was where he made the mistake. He ought to have multiplied both sides with the same quantity (Lowest Common Denominator) so as not to change the actual value of the question.
Multiplying both sides by the same quantity does not change the real magnitude of the question.
The actual solution goes thus,
[tex]\begin{gathered} \frac{2x-3}{x-1}=\frac{2}{x(x-1)} \\ \text{Multiplying both sides of the equation by the LCD,} \\ \text{The LCD is x(x-1)} \\ x(x-1)(\frac{2x-3}{x-1})=x(x-1)(\frac{2}{x(x-1)}) \\ x(2x-3)=2 \\ \text{Expanding the bracket,} \\ 2x^2-3x=2 \\ \text{Collecting all the terms to one side to make it a quadratic equation,} \\ 2x^2-3x-2=0 \end{gathered}[/tex]Solving the quadratic equation;
[tex]\begin{gathered} 2x^2-3x-2=0 \\ 2x^2-4x+x-2=0 \\ \text{Factorizing the equation,} \\ 2x(x-2)+1(x-2)=0 \\ (2x+1)(x-2)=0 \\ 2x+1=0 \\ 2x=0-1 \\ 2x=-1 \\ \text{Dividing both sides by 2,} \\ x=-\frac{1}{2} \\ \\ \\ OR \\ x-2=0 \\ x=0+2 \\ x=2 \end{gathered}[/tex]Therefore, the actual solutions to the expression are;
[tex]\begin{gathered} x=-\frac{1}{2} \\ \\ OR \\ \\ x=2 \end{gathered}[/tex]help meeeeeeeeee pleaseee !!!!!
Answers
The function 2x + 3x^2 represents the result of adding the two provided functions, f(x) and g(x).
Composite performance.An operation known as "function composition" takes two functions, f and g, and produces a new function, h, that is equal to both g and f and has the property that h(x) = g.
Given the f(x) = 2x and g(x) = 3x^2 functions
The sum of the two functions must be calculated as illustrated;
f(x) + g = (f+g)(x)
Put the provided functions in place of (f+g)(x) to have:
(f+g)(x) = 2x + 3x^2
Standard version of the expression is (f+g)(x) = 2x + 3x^2
Consequently, the sum of the functions f(x) and g(x) is2x + 3x^2
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