Using the Pythagorean theorem, the distance between two points (x1, y1) and (x2, y2) is gotten as follows:
[tex]\begin{gathered} d^2=(x_2-x_1)^2+(y_2-y_1)^2 \\ \text{Thus:} \\ d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2} \end{gathered}[/tex]Since we have the two coordinates: (2, 8) and (-8, 2)
where:
(x1, y1)= (2, 8)
(x2, y2) = (-8, 2)
Therefore, the distance between them is:
[tex]\begin{gathered} d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2} \\ d=\sqrt[]{((-8)_{}-2_{})^2+(2-8)^2} \\ d=\sqrt[]{((-10_{})^2+(-6)^2} \\ d=\sqrt[]{100+36} \\ d=\sqrt[]{136} \\ d=11.66 \\ d=11.7\text{ (to one decimal place)} \end{gathered}[/tex]Therefore, the distance between the two p is: 11.7
Correct option is: Option D
system of equationsb+c= -55b-c= 17
Let's solve the system of equations:
b + c = - 55
b - c = 17
Step 1: Let's isolate b on the first equation:
b + c = - 55
b = - 55 - c
Step 2: Let's solve for c on the second equation, substituting b:
b - c = 17
-55 - c - c = 17
-55 - 2c = 17
Adding 55 at both sides:
-2c - 55 + 55 = 17 + 55
-2c = 72
Dividing by - 2 at both sides:
-2c/-2 = 72/-2
c = -36
Step 3: Let's solve for b on the first equation, susbtituting c:
b + c = - 55
b + (-36) = - 55
b - 36 = - 55
Adding 36 at both sides:
b - 36 + 36 = - 55 + 36
I think you are ready to finish and calculate the value for b.
Given f(x) and g(x) = f(k⋅x), use the graph to determine the value of k.A.) - 2B.) -1/2C.) 1/2D.) 2
In order to solve this problem we have to remember that the equation of any line takes the form
[tex]y(x)=mx+b[/tex]Therefore,
[tex]y(kx)=\text{mkx}+b[/tex]In other words, multiplying k by x is just multiplying the slope m by a factor of k.
The slope of g(x) is
[tex]m=2[/tex]and the slope of f(x) is
[tex]m=1[/tex]We see than the slope of g(x) is 2 times the slope of f(x); therefore, k = 2 which is choice D.
hello and thank you for helping me and this is a trigonometry question bit for the question has give exact value and it won't accept decimals as an answer and thank you for your time.
1) In this question let's calculate the sin(θ) and cos(θ)
Given that
[tex]\begin{gathered} \text{If }\sin (\theta)=\frac{5\pi}{4} \\ \sin (\theta)\text{ }\Rightarrow\sin (\frac{5\pi}{4})\text{ }=-\frac{\sqrt[]{2}}{2} \\ \cos (\frac{5\pi}{4})=-\frac{\sqrt[]{2}}{2} \end{gathered}[/tex]2) In this question, we're calculating the value of the sine and the cosine in radians.
We must remember that 5π/4 ⇒ to 225º, and that it's in the Quadrant III
If we subtract
225 -180 =45 So the sine of 5π/4 is -√2/2 and the cosine (5π/4 ) = -√2/2
2.3) The sign of the Quadrant
Since 225º is in Quadrant III both results are negative ones.
What transaction occurs when an investor decides to liquidate assets?
A. buy
B. hold
C. sell
D. speculate
Answer:
What transaction occurs when an investor decides to liquidate assets?
A. buy
B. hold
(C. sell)
D. speculate
Step-by-step explanation:
I got a 5/5 on the test and i got the answer from a quizlet (:
Sell is the answer
The correct option is (C).
Given,
In the question:
What transaction occurs when an investor decides to liquidate assets?
Now, According to the question:
when an investor decides to liquidate assets.
when an investor decides to liquidate assets means he or she want to sell the property in the open market, in other words liquidate assets means
converting non- liquid assets into liquid assets.
In investing, liquidation occurs when an investor closes their position in an asset. Liquidating an asset is usually carried out when an investor or portfolio manager needs cash to re-allocate funds or rebalance a portfolio. An asset that is not performing well may also be partially or fully liquidated.
According to the statement
Therefore, Sell is the answer
The correct option is (C).
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Inside. Make sure you don’t enter any spaces in your answers. This answer needs to be rounded to the nearest hundredth.
ANSWER
c = 14.14
EXPLANATION
To find the length of side AB, which is the hypotenuse of the right triangle ABC, we have to apply the Pythagorean Theorem,
[tex]AB^2=AC^2+BC^2[/tex]Replace the known values and solve for c,
[tex]c^2=10^2+10^2=100+100=200\Rightarrow c=\sqrt{200}\approx14.14[/tex]Hence, the value of c is 14.14, rounded to the nearest hundredth.
Find the x-coordinate for the point of intersection by using the equations method of solving. Show all the work. f (x)=2x+6g(x)= -3x+1
y = 2x + 6
y = -3x + 1
Equality
2x + 6 = -3x + 1 6 is in the left side and is positive so we substract 6 in both sides
2x +6 - 6 = -3x + 1 - 6
Simplify
2x = -3x - 5
Add -3x in both sides
2x + 3x = -3x + 3x - 5
Simplify
5x = -5
2x + 3x = 1 - 6
5x = -5
x = -5/5
This is the x-coordinate
x = -1
-1/2 (2/5y - 2) (1/10y-4)
we multiply the first parenthesis by its coefficient
[tex]\begin{gathered} ((-\frac{1}{2}\times\frac{2}{5}y)+(-\frac{1}{2}\times-2))(\frac{1}{10}y-4) \\ \\ (-\frac{2}{10}y+\frac{2}{2})(\frac{1}{10}y-4) \\ \\ (-\frac{1}{5}y+1)(\frac{1}{10}y-4) \end{gathered}[/tex]now multiply each value and add the solutions
[tex]\begin{gathered} (-\frac{1}{5}y\times\frac{1}{10}y)+(-\frac{1}{5}y\times-4)+(1\times\frac{1}{10}y)+(1\times-4) \\ \\ (-\frac{1}{50}y^2)+(\frac{4}{5}y)+(\frac{1}{10}y)+(-4) \\ \\ -\frac{1}{50}y^2+(\frac{4}{5}y+\frac{1}{10}y)-4 \\ \\ -\frac{1}{50}y^2+\frac{9}{10}y-4 \end{gathered}[/tex]If the area of the rectangle to be drawn is 12 square units, where should points C and D be located, if they lie vertically below A and B, to make this rectangle?
Answer:
C(2,-2), D(-1,-2)
Explanation:
The area of a rectangle is calculated using the formula:
[tex]A=L\times W[/tex]• From the graph, AB = 3 units.
,• Given that the area = 12 square units
[tex]\begin{gathered} 12=3\times L \\ L=\frac{12}{3}=4 \end{gathered}[/tex]This means that the distance from B to C and A to D must be 4 units each.
Count 4 units vertically downwards from A and B.
The coordinates of C and D are:
• C(2,-2)
,• D(-1,-2)
The first option is correct.
The function f(x) = 6x represents the number of lightbulbs f(x) that are needed for x chandeliers. How many lightbulbs are needed for 7 chandeliers? Show your work
There are a total of 42 lightbulbs needed for 7 chandeliers
How to determine the number of lightbulbs needed?From the question, the equation of the function is given as
f(x) = 6x
Where
x represents the number of chandeliersf(x) represents the number of lightbulbs
For 7 chandeliers, we have
x = 7
Substitute x = 7 in f(x) = 6x
So, we have
f(7) = 6 x 7
Evaluate the product
f(7) = 42
Hence, the number of lightbulbs needed for 7 chandeliers is 42
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The number of lightbulbs needed for 7 chandeliers would be; 42
What is an equation?An equation is an expression that shows the relationship between two or more numbers and variables.
From the given problem, the equation of the function is;
f(x) = 6x
Where
x be the number of chandeliers and f(x) represents the number of lightbulbs.
For 7 chandeliers, x = 7
Now Substitute x = 7 in f(x) = 6x
Therefore, f(7) = 6 x 7
Evaluate the product;
f(7) = 42
Hence, the number of lightbulbs needed for 7 chandeliers would be; 42
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Solve for a side in right triangles. AC = ?. Round to the nearest hundredth
The length of segment AC is 2.96 units
How to determine the side length AC?From the question, the given parameters are
Line segment AB = 7 units
Angle A = 65 degrees
The line segment AC can be calculated using the following cosine ratio
cos(Angle) = Adjacent/Hypotenuse
Where
Adjacent = Side length AC
Hypotenuse = Side length AB
So, we have
cos(65) = AC/AB
This gives
cos(65) = AC/7
Make AC the subject
AC =7 * cos(65)
Evaluate
AC = 2.96
Hence, the side length AC has a value of 2.96 units
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(c) Given that q= 8d^2, find the other two real roots.
Polynomials
Given the equation:
[tex]x^5-3x^4+mx^3+nx^2+px+q=0[/tex]Where all the coefficients are real numbers, and it has 3 real roots of the form:
[tex]x_1=\log _2a,x_2=\log _2b,x_3=\log _2c[/tex]It has two imaginary roots of the form: di and -di. Recall both roots must be conjugated.
a) Knowing the sum of the roots must be equal to the inverse negative of the coefficient of the fourth-degree term:
[tex]\begin{gathered} \log _2a+\log _2b+\log _2c+di-di=3 \\ \text{Simplifying:} \\ \log _2a+\log _2b+\log _2c=3 \\ \text{Apply log property:} \\ \log _2(abc)=3 \\ abc=2^3 \\ abc=8 \end{gathered}[/tex]b) It's additionally given the values of a, b, and c are consecutive terms of a geometric sequence. Assume that sequence has first term a1 and common ratio r, thus:
[tex]a=a_1,b=a_1\cdot r,c=a_1\cdot r^2[/tex]Using the relationship found in a):
[tex]\begin{gathered} a_1\cdot a_1\cdot r\cdot a_1\cdot r^2=8 \\ \text{Simplifying:} \\ (a_1\cdot r)^3=8 \\ a_1\cdot r=2 \end{gathered}[/tex]As said above, the real roots are:
[tex]x_1=\log _2a,x_2=\log _2b,x_3=\log _2c[/tex]Since b = a1*r, then b = 2, thus:
[tex]x_2=\log _22=1[/tex]One of the real roots has been found to be 1. We still don't know the others.
c) We know the product of the roots of a polynomial equals the inverse negative of the independent term, thus:
[tex]\log _2a_1\cdot2\cdot\log _2(a_1\cdot r^2)\cdot(di)\cdot(-di)=-q[/tex]Since q = 8 d^2:
[tex]\begin{gathered} \log _2a_1\cdot2\cdot\log _2(a_1\cdot r^2)\cdot(di)\cdot(-di)=-8d^2 \\ \text{Operate:} \\ 2\log _2a_1\cdot\log _2(a_1\cdot r^2)\cdot(-d^2i^2)=-8d^2 \\ \log _2a_1\cdot\log _2(a_1\cdot r^2)=-8 \end{gathered}[/tex]From the relationships obtained in a) and b):
[tex]a_1=\frac{2}{r}[/tex]Substituting:
[tex]\begin{gathered} \log _2(\frac{2}{r})\cdot\log _2(2r)=-8 \\ By\text{ property of logs:} \\ (\log _22-\log _2r)\cdot(\log _22+\log _2r)=-8 \end{gathered}[/tex]Simplifying:
[tex]\begin{gathered} (1-\log _2r)\cdot(1+\log _2r)=-8 \\ (1-\log ^2_2r)=-8 \\ \text{Solving:} \\ \log ^2_2r=9 \end{gathered}[/tex]We'll take the positive root only:
[tex]\begin{gathered} \log _2r=3 \\ r=8 \end{gathered}[/tex]Thus:
[tex]a_1=\frac{2}{8}=\frac{1}{4}[/tex]The other roots are:
[tex]\begin{gathered} x_1=\log _2\frac{1}{4}=-2 \\ x_3=\log _216=4 \end{gathered}[/tex]Real roots: -2, 1, 4
H = -16t^2 + 36t + 56 Where H is the height of the ball after t seconds have passed.
we have the equation
H = -16t^2 + 36t + 56
This equation represents a vertical parabola open downward, which means, the vertex is a maximum
The time t when the ball reaches its maximum value corresponds to the x-coordinate of the vertex
so
Convert the given equation into vertex form
H=a(t-h)^2+k
where
(h,k) is the vertex
step 1
Complete the square
H = -16t^2 + 36t + 56
Factor -16
H=-16(t^2-36/16t)+56
H=-16(t^2-36/16t+81/64)+56+81/4
Rewrite as perfect squares
H=-16(t-9/8)^2+76.25
the vertex is (9/8,76.25)
therefore
the time is 9/8 sec or 1.125 seconds when the ball reaches its maximumSubtract and simplify the answer. 8/9 - 1/3
Solution
We want to simplify
[tex]\frac{8}{9}-\frac{1}{3}[/tex]Now
[tex]\begin{gathered} \frac{8}{9}-\frac{1}{3}=\frac{8}{9}-\frac{1\times3}{3\times3} \\ \frac{8}{9}-\frac{1}{3}=\frac{8}{9}-\frac{3}{9} \\ \frac{8}{9}-\frac{1}{3}=\frac{8-3}{9} \\ \frac{8}{9}-\frac{1}{3}=\frac{5}{9} \end{gathered}[/tex]Therefore, the answer is
[tex]\frac{5}{9}[/tex]in a sale normal prices are reduced by 15%. The sale price of a CD player is £102. work out the normal price of the CD player
The normal price for the CD player is $117.30
How to calculate the value?Since the normal prices are reduced by 15%, the percentage for the normal price will be:
= 100% + 15%
= 115%
Also, the sale price of a CD player is £102.
Therefore, the normal price will be:
= Percentage for normal price × Price
= 115% × $102
= 1.15 × $102
= $117.30
The price is $117.30.
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Give the equation of the line parallel to a line through (-3, 4) and (-5, -6) that passes through the origin. y = 5x y = 5x + 1 y=-1/5x + 1 y = -1/5x y
To solve for the equation of the line parallel :
[tex]\begin{gathered} (-3,4)\Longrightarrow(x_1,y_1) \\ (-5,-6)\Longrightarrow(x_2,\text{y}_2) \end{gathered}[/tex]For parallel line equation:
Slope-intercept form: y=mx+b, where m is the slope and b is the y-intercept
First let's find the slope of the line.
To find the slope using two points, divide the difference of the y-coordinates by the difference of the x-coordinates.
[tex]\begin{gathered} \text{slope =}\frac{y_2-y_1}{x_2-x_1} \\ \text{slope}=\frac{-6-4}{-5--3} \\ \text{slope=}\frac{-10}{-5+3}=\frac{-10}{-2} \\ \text{slope =5} \end{gathered}[/tex]Slope= 5
[tex]\begin{gathered} y=mx+c \\ y=5x+c \\ \text{where c = y-intercept} \end{gathered}[/tex]The y-intercept is (0, b). The equation passes through the origin, so the y-intercept is 0.
[tex]\begin{gathered} y=5x+0 \\ y=5x \end{gathered}[/tex]Hence the
Solve the following inequality. Graph the solution set and then write it in interval notation .
Given:
-2x ≥ 6
Solve for x
Divide both sides by -2
-2x/-2 ≤ 6/-2
x ≤ -3
Graph:
Interval notation (-∞, -3 ]
Martin earns $7.50 per hour proofreading ads per hour proofreading ads at a local newspaper. His weekly wage can. e found by multiplying his salary times the number of hours h he works.1. Write an equation.2. Find f(15)3. Find f (25)
If Martin earns 7.50 per hour (that is h), then the equation for his weekly wage can be expressed as;
[tex]\begin{gathered} (A)f(h)=7.5h \\ (B)f(15)=7.5(15) \\ f(15)=112.5 \\ (C)f(25)=7.5(25) \\ f(25)=187.5 \end{gathered}[/tex]Therefore, answer number A shows the equation for his salary
Answer number 2 shows his salary at 15 hours ($112.5)
Answer number 3 shows his salary at 25 hours ($187.5)
For f(x)=x^2 and g(x)=x^2+9, find the following composite functions and state the domain of each.
(a) f.g (b) g.f (c) f.f (d) g.g
The composite functions in this problem are given as follows:
a) (f ∘ g)(x) = x^4 + 18x² + 81.
b) (g ∘ f)(x) = x^4 + 9.
c) (f ∘ f)(x) = x^4.
d) (g ∘ g)(x) = x^4 + 18x² + 90.
All these functions have a domain of all real values.
Composite functionsFor composite functions, the outer function is applied as the input to the inner function.
In the context of this problem, the functions are given as follows:
f(x) = x².g(x) = x² + 9.For item a, the composite function is given as follows:
(f ∘ g)(x) = f(x² + 9) = (x² + 9)² = x^4 + 18x² + 81.
For item b, the composite function is given as follows:
(g ∘ f)(x) = g(x²) = (x²)² + 9 = x^4 + 9.
For item c, the composite function is given as follows:
(f ∘ f)(x) = f(x²) = (x²)² = x^4.
For item d, the composite function is given as follows:
(g ∘ g)(x) = g(x² + 9) = (x² + 9)² + 9 = x^4 + 18x² + 90.
None of these functions have any restriction on the domain such as fractions or even roots, hence all of them have all real values as the domain.
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Select all the situations in which a proportional relationship is described.
Jackson saves $10 in the first month and $30 in the next 3 months.
Mia saves $8 in the first 2 months and $4 in the next month.
Piyoli spends $2 in the first 2 days of the week and $5 in the next 5 days.
Robert spends $2 in the first 3 days of the week and $5 in the next 4 days.
Answer:
Jackson saves $10 in the first month and $30 in the next 3 months.
Mia saves $8 in the first 2 months and $4 in the next month.
Piyoli spends $2 in the first 2 days of the week and $5 in the next 5 days.
Step-by-step explanation:
A proportional relationship is one that has a constant of proportionality.
In this case, the correct options are Mia, Piyoli, and Robert.
In the xy-plane, line n passes through point (0,0) and has a slope of 4. If line n also passes through point (3,a), what is the value of a?
a store donated 2 and 1/4 cases of cranes to a daycare center each case holds 24 boxes of crayons each box holds 8 crayons how many crayons did the center receive
Answer:
The center recieved 432 crayons
Explanation:
Given the following information:
There are 2 and 1/4 cases
Each case holds 24 boxes of crayons
Each box holds 8 crayons.
The number of crayons the center receive is:
8 * 24 * (2 + 1/4)
= 8 * 24 * (8/4 + 1/4)
= 192 * (9/4)
= 1728/4
= 432
Help me with my schoolwork what is the slope of line /
The two points given on the line are
[tex]\begin{gathered} (x_1,y_1)\Rightarrow(-2,9) \\ (x_2,y_2)\Rightarrow(6,1) \end{gathered}[/tex]The slope of line that passes through (x1,y1) and (x2,y2) is gotten using the formula below
[tex]\begin{gathered} m=\frac{\text{change in y}}{\text{change in x}} \\ m=\frac{y_2-y_1}{x_2-x_1} \end{gathered}[/tex]By substituting the values, we will have
[tex]\begin{gathered} m=\frac{y_2-y_1}{x_2-x_1} \\ m=\frac{1-9}{6-(-2)} \\ m=-\frac{8}{6+2} \\ m=-\frac{8}{8} \\ m=-1 \end{gathered}[/tex]Therefore,
The slope of the line = -1
You are scuba diving at 120 feet below sea level. You begin to ascend at a rate of 4 feet per second.a. Where will you be 10 seconds after you begin your ascension? b. How long will it take to reach the surface?
The ascension can be modeled using the function:
[tex]d(t)=d_0-r\cdot t[/tex]Where d is the number of feet below the sea level at time t (in seconds), d₀ is the initial "depth", and r is the ascension rate.
From the problem, we identify:
[tex]\begin{gathered} r=4\text{ feet per second} \\ d_0=120\text{ feet} \end{gathered}[/tex]Then:
[tex]d(t)=120-4t[/tex]a)
After 10 seconds, we have t = 10:
[tex]\begin{gathered} d(10)=120-4\cdot10=120-40 \\ \\ \Rightarrow d(10)=80\text{ feet} \end{gathered}[/tex]After 10 seconds, we will be 80 feet below sea level.
b)
To find how long will it take to reach the surface, we need to solve the equation d(t) = 0.
[tex]\begin{gathered} d(t)=0 \\ 120-4t=0 \\ 4t=120 \\ \\ \therefore t=30\text{ seconds} \end{gathered}[/tex]We will reach the surface after 30 seconds.
Solve for x:
A
+79
X
Answer: -11
Step-by-step explanation: 66+46=112
180-112=68
79+?=68
79+-11=68
In scalene triangle ABC shown in the diagram below, m2C = 90°.B.Which equation is always true?sn A = sin Bcos sn A = cos BCanAB4 5 678 9 1011
inNote: To know which equation is true, then we will have to TEST for each of the choices we are to pick from.
From the tirangle in the image.
[tex]\begin{gathered} 1)\sin \text{ A =}\frac{\text{ Opp}}{\text{Hyp}}\text{ = }\frac{a}{c} \\ \cos \text{ B = }\frac{\text{ADJ}}{\text{HYP}}\text{ = }\frac{a}{c} \\ So\text{ from the above, we can s}ee\text{ that: SinA = Cos B :This mean the choice are equal} \\ \end{gathered}[/tex][tex]\begin{gathered} 2)\text{ To test for the second choice we have..} \\ \text{ Cos A = Cos B} \\ \text{for Cos A =}\frac{\text{Adj}}{\text{Hyp}}\text{ =}\frac{b}{c} \\ \\ \text{for Cos B = }\frac{Adj}{\text{Hyp}}\text{ = }\frac{a}{c} \\ \text{from here we can s}ee\text{ that Cos A }\ne\text{ Cos B : meaning Cos A is not equal to Cos B} \\ \end{gathered}[/tex]3) To test for the third choice: Sin A = Cos A
[tex]\begin{gathered} \sin \text{ A=}\frac{opp}{\text{Hyp}}\text{ = }\frac{a}{c} \\ \cos \text{ A = }\frac{Adj}{\text{Hyp}}\text{ = }\frac{b}{c} \\ we\text{ can s}ee\text{ that sinA }\ne\text{ cos }A,\text{ This mean they are not equal} \end{gathered}[/tex][tex]\begin{gathered} 4)\text{ To test if: tan A = sin B} \\ \text{ }tan\text{ A = }\frac{opp}{\text{Adj}}\text{ = }\frac{a}{b} \\ \\ \text{ sin B = }\frac{Opp}{\text{Hyp}}\text{ = }\frac{b}{c} \\ so\text{ from what we have, w can s}ee\text{ that tan A }\ne\text{ sinB: Meaning they are not equal.} \end{gathered}[/tex]Meaning the first choice is the answer that is sin A = CosB
I need help solving an optimization math problem please :)
Answer:
Explanation:
Let the side opposite the river = x
Let the adjacent side to the river = y
StatusExam9 ft.15 ft.The volume ofthe figure iscubic feet.15 ft.15 ft.
Step 1:
The figure is a composite figure with a square base pyramid and a cube.
Step 1:
The volume of the composite shape is the sum of the volume of a square base pyramid and a cube.
[tex]\text{Volume = L}^3\text{ + }\frac{1}{2}\text{ base area }\times\text{ height}[/tex]Step 3:
Given data
Cube
Length of its sides L = 15 ft
Square base pyramid
Height h = 9 ft
Length of the square base = 15 ft
Step 4:
Substitute in the formula.
[tex]\begin{gathered} \text{Volume = 15}^3\text{ + }\frac{1}{3}\text{ }\times15^2\text{ }\times\text{ 9} \\ \text{= 3375 + 675} \\ =4050ft^3 \end{gathered}[/tex]A chocolate factory has a goal to produce10121012pounds of chocolate frogs per day. If the machines operate for712712hours per day making215215pounds of chocolate frogs per hour, will the chocolate factory make it’s goal?The chocolate factory meet their goal with the total being10121012pounds of chocolate frogs produced.
First, rewrite all the mixed fractions as impropper fractions:
[tex]\begin{gathered} 10\frac{1}{2}=10\times\frac{2}{2}+\frac{1}{2}=\frac{20}{2}+\frac{1}{2}=\frac{21}{2} \\ \\ 7\frac{1}{2}=7\times\frac{2}{2}+\frac{1}{2}=\frac{14}{2}+\frac{1}{2}=\frac{15}{2} \\ \\ 2\frac{1}{5}=2\times\frac{5}{5}+\frac{1}{5}=\frac{10}{5}+\frac{1}{5}=\frac{11}{5} \end{gathered}[/tex]Next, multiply the rate of chocolate production over time by the the operating time of the machines to find the total amount of pounds of chocolate frogs produced in one day:
[tex]7\frac{1}{2}\times2\frac{1}{5}=\frac{15}{2}\times\frac{11}{5}=\frac{15\times11}{2\times5}=\frac{3\times11}{2}=\frac{33}{2}=16\frac{1}{2}[/tex]Then, the chocolate factory can produce 16 1/2 pounds of chocolate frogs per day.
Since 16 1/2 is greater than 10 1/2, then the chocolate factory will meet their goal with the total being over 10 1/2 pounds of chocolate frogs produced.
Determine which of the lines are parallel and which of the lines are perpendicular. Select all of the statements that are true.
Line a passes through (-1, -17) and (3, 11).
Line b passes through (0,4) and (7,-5).
Line c passes through (7, 1) and (0, 2).
Line d passes through (-1,-6) and (1, 8).
Answers:
Line A is parallel to line D.
Line A is perpendicular to line C.
Line C is perpendicular to line D.
=====================================================
Explanation:
Let's use the slope formula to calculate the slope of the line through (-1,-17) and (3,11)
[tex](x_1,y_1) = (-1,-17) \text{ and } (x_2,y_2) = (3,11)\\\\m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}\\\\m = \frac{11 - (-17)}{3 - (-1)}\\\\m = \frac{11 + 17}{3 + 1}\\\\m = \frac{28}{4}\\\\m = 7\\\\[/tex]
The slope of line A is 7
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Now let's find the slope of line B.
[tex](x_1,y_1) = (0,4) \text{ and } (x_2,y_2) = (7,-5)\\\\m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}\\\\m = \frac{-5 - 4}{7 - 0}\\\\m = -\frac{9}{7}\\\\[/tex]
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Now onto line C.
[tex](x_1,y_1) = (7,1) \text{ and } (x_2,y_2) = (0,2)\\\\m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}\\\\m = \frac{2 - 1}{0 - 7}\\\\m = \frac{1}{-7}\\\\m = -\frac{1}{7}\\\\[/tex]
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Lastly we have line D.
[tex](x_1,y_1) = (-1,-6) \text{ and } (x_2,y_2) = (1,8)\\\\m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}\\\\m = \frac{8 - (-6)}{1 - (-1)}\\\\m = \frac{8 + 6}{1 + 1}\\\\m = \frac{14}{2}\\\\m = 7\\\\[/tex]
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Here's a summary of the slopes we found
[tex]\begin{array}{|c|c|} \cline{1-2}\text{Line} & \text{Slope}\\\cline{1-2}\text{A} & 7\\\cline{1-2}\text{B} & -9/7\\\cline{1-2}\text{C} & -1/7\\\cline{1-2}\text{D} & 7\\\cline{1-2}\end{array}[/tex]
Recall that parallel lines have equal slopes, but different y intercepts. This fact makes Line A parallel to line D.
Lines A and C are perpendicular to one another, because the slopes 7 and -1/7 multiply to -1. In other words, -1/7 is the negative reciprocal of 7, and vice versa. These two lines form a 90 degree angle.
Lines C and D are perpendicular for the same reasoning as the previous paragraph.
Line B unfortunately is neither parallel nor perpendicular to any of the other lines mentioned.
You can use a graphing tool like Desmos or GeoGebra to verify these answers.
With aging body fat increases in muscle mass declines the graph to the right shows the percent body fat in a group of adult women and men as they age from 25 to 75 years age is represented along the X-axis and percent body fat is represented along the Y-axis use interval notation to give the domain and range for the graph of the function for women
Step 1
The domain and range of a function is the set of all possible inputs and outputs of a function respectively. The domain is found along the x-axis, the range on the other hand is found along the y-axis.
Find the domain of the graph of the function of women using interval notation.
[tex]\text{Domain:\lbrack}25,75\rbrack[/tex]Step 2
Find the range of the graph of the function of women using interval notation.
[tex]\text{Range:}\lbrack32,40\rbrack[/tex]Therefore, the domain and range in interval notation for the women respectively are;
[tex]\begin{gathered} \text{Domain:\lbrack}25,75\rbrack \\ \text{Range:}\lbrack32,40\rbrack \end{gathered}[/tex]