Total receipts = $39,500
Expenses = $5,600
Percentage expenses =
[tex]\frac{5600}{39500}\times100=\frac{5600\times100}{39500}=14.177\approx14.2\text{ \%}[/tex][tex]\text{ \% expenses =}\frac{Actual\text{ expenses}}{\text{Total Receipts}}\times\frac{100}{1}[/tex]Solution: The percentage of expenses is 14.2%
Determine the equation of the line that passes through the point (-1, 2) and isperpendicular to the line y = -2.
1) In this question, let's find the equation, using the point-slope formula:
[tex](y-y_0)=m(x-x_0)[/tex]2) Notice that since we want a perpendicular line we can write a perpendicular line to y=2, as x=-1/2 for -1/2 is the opposite and reciprocal to 2 (the necessary condition to get a perpendicular line).
So, the slope of that perpendicular line is -1/2
3) Let's plug into that Point-Slope formula, the slope m= -1/2 and the point:
[tex]\begin{gathered} (y-2)=-\frac{1}{2}(x+1) \\ y-2=-\frac{1}{2}x-\frac{1}{2} \\ y=-\frac{1}{2}x-\frac{1}{2}+2 \\ y=-\frac{1}{2}x+\frac{3}{2} \end{gathered}[/tex]4) Thus, the answer is:
[tex]y=-\frac{1}{2}x+\frac{3}{2}[/tex]Kristi Yang borrowed $12,000. The term of the loan was 150 days, and the annual simple interest rate was 6.5%. Find the simple interest due on the loan. (Round your answer to the nearest cent.)
For an initial ammount borrowed I, an term of the loan t, and an annual interest rate r, the simple interest S due on the loan is given by:
[tex]S=I\cdot r\cdot\frac{t}{365}[/tex]For I = $12000, r = 0.065 and t = 150 days, we have:
[tex]\begin{gathered} S=12000\cdot0.065\cdot\frac{150}{365} \\ S=12000\cdot0.065\cdot0.41096 \\ S=\text{ \$320.55} \end{gathered}[/tex]Find the area and perimeter with these points.(-11,-8)(-11,0)(0,0)(0,-8)
we have the coordinates
(-11,-8)
(-11,0)
(0,0)
(0,-8)
step 1
plot the give points
using a graphing tool
see the attached figure below
The figure is a rectangle
where
L=0-(-11)=11 units (subtract x-coordinates)
W=0-(-8)=8 units (subtract y-coordinates)
step 2
Find out the area
A=L*W
A=11*8=88 unit2
step 3
Find out the perimeter
P=2(L+W)
P=2(11+8)
P=2(19)=38 units
Match each solid cone to it’s surface area. Answers are rounded to the nearest square unit
The surface area of a cone is given by the formula below:
[tex]S=\pi r^2+\pi rs[/tex]Where r is the base radius and s is the slant height.
So, calculating the surface area of first cone, we have:
[tex]\begin{gathered} s^2=21^2+6^2\\ \\ s^2=441+36\\ \\ s^2=477\\ \\ s=21.84\\ \\ S=\pi\cdot6^2+\pi\cdot6\cdot21.84\\ \\ S=525 \end{gathered}[/tex]The surface area of the second cone is:
[tex]\begin{gathered} s^2=8^2+12^2\\ \\ s^2=64+144\\ \\ s^2=208\\ \\ s=14.42\\ \\ S=\pi\cdot12^2+\pi\cdot12\cdot14.42\\ \\ S=996 \end{gathered}[/tex]The surface area of the third cone is:
[tex]\begin{gathered} s^2=15^2+8^2\\ \\ s^2=225+64\\ \\ s^2=289\\ \\ s=17\\ \\ S=\pi\cdot8^2+\pi\cdot8\cdot17\\ \\ S=628 \end{gathered}[/tex]And the surface area of the fourth cone is:
[tex]\begin{gathered} s^2=10^2+10^2\\ \\ s^2=100+100\\ \\ s^2=200\\ \\ s=14.14\\ \\ S=\pi\cdot10^2+\pi\cdot10\cdot14.14\\ \\ S=758 \end{gathered}[/tex]Find the missing factor. 8x2 - X - 9 = (x + 1)(
In this case, we'll have to carry out several steps to find the solution.
Step 01:
Data
8x² - x - 9 = (x + 1) ( ? )
Step 02:
We must find the missing root to solve the exercise.
x1 = - 1
x2 :
a = 8
b = -1
c = -9
[tex]x\text{ = }\frac{-(-1)\pm\sqrt[]{(-1)^2-4\cdot8\cdot(-9)}_{}}{2\cdot8}[/tex][tex]x\text{ = }\frac{1\pm17}{16}[/tex][tex]x\text{ = }\frac{1+17}{16}=\frac{18}{16}=\frac{9}{8}[/tex]x2 = 9 / 8
The answer is:
( x - 9/8)
8x² - x - 9 = (x + 1) ( x - 9/8 )
Given y=0.5x^2, describe the transformation (x,y) --> (x,4y) and sketch the graph of this image
We are given the equation y = 0.5x^2. To describe its transformation from (x, y) to (x, 4y), we can start by first graphing the given equation.
To graph, let's use sample points (x- and y-values):
x y
-2 2
-1 0.5
0 0
1 0.5
2 2
So we have the points (-2, 2), (-1, 0.5), (0, 0), (1, 0.5), and (2, 2) to help us graph the equation.
A transformation of (x, y) --> (x, ay) where a > 1 means a vertical stretch equal to |a|. In this case, because (x, y) is transformed to (x, 4y), the graph stretches vertically by a factor of 4.
To graph, let's use sample points (x- and y-values):
x y
-2 4(2) = 8
-1 4(0.5) = 2
0 4(0) = 0
1 4(0.5) = 2
2 4(2) = 8
The new graph would now look like this:
(7x10^1(4x10^-7
(5.55 x 10^4) - ( 3.41 x 10^4)
(9 x 10^7) divided (3 x 10^3)
Work needs to be shows !!!
Answer:
(5.55 * 10^4) - (3.41 * 10^4)
=21,400
(9 * 10^7) divided (3 * 10^3)
= 30,000
Step-by-step explanation:
(5.55 * 10^4) - (3.41 * 10^4)
= (5.55 * 10,000) - (3.41 * 10,000)
= 55,500 - 34,100
= 21,400
(9 * 10^7) divided (3 * 10^3)
= (9 * 10,000,000) ÷ (3 * 1,000)
= 90,000,000 ÷ 3,000
= 30,000
Sorry but i don't understand the "(7x10^1(4x10^7". Your question is invalid.
A paper is sold for Php60.00, which is 150% of the cost. How much is the store's cost?
The store's cost is php40
Let's call the store Cost = C
This means that this cost is elevated a 150% in order to get the price of php60
In an mathematical expression, this is:
C · 150% = php60
Then, let's convert the percentage to decimal. To do this, we just divide the percentage by 100:
150% ÷ 100 = 1.5
Now we can solve:
[tex]\begin{gathered} C\cdot1.5=60 \\ C=\frac{60}{1.5}=40 \end{gathered}[/tex]Then the store cost is C = php40
For the following two numbers, find two factors of the first number such that their product is the first number and their sum is the second number,40, 14
First we need to factorate the number 40:
[tex]40=2\cdot2\cdot2\cdot5[/tex]The possible numbers we can create using these factors are 2, 4, 5, 8, 10 and 20.
So If the product of the two factors (let's call them 'a' and 'b') is 40 and the sum is 14, we have:
[tex]\begin{gathered} a\cdot b=40 \\ a+b=14 \\ \\ \text{From the second equation:} \\ b=14-a \\ \\ \text{Using this value of b in the first equation:} \\ a(14-a)=40 \\ 14a-a^2=40 \\ a^2-14a+40=0 \end{gathered}[/tex]Using the quadratic formula to solve this equation, we have:
[tex]\begin{gathered} a_1=\frac{-b+\sqrt[]{b^2-4ac}}{2a}=\frac{14+\sqrt[]{196-160}}{2}=\frac{14+6}{2}=10 \\ a_2=\frac{-b-\sqrt[]{b^2-4ac}}{2a}=\frac{14-6}{2}=4 \\ \\ a=10\to b=14-10=4 \\ a=4\to b=14-4=10 \end{gathered}[/tex]So the factors which product is 40 and the sum is 14 are 4 and 10.
Doug travels 5 times as fast as Gloria. Traveling in opposite directions, they are 858 miles apart after 6.5 hours. Find their rates or travel.
Distance = rate x time
Gloria = X mph
Doug = 5X mph
Where X is the rate of speed
And we have that they are 858 miles apart after 6.5 hours
6.5 X + 5( 6.5 X) = 858 mph
6.5X + 32,4X= 858 mph
39X = 858mph
X = (858mph)/39 = 22
5X = 110
So Gloria is traveling at 22 mph and Doug is traveling at 110 mph. This meand that Gloria rate of speed is 22 and Doug is 110
11) -3(1 + 6r) = 14 - r
Distributing over parentheses,
[tex]\begin{gathered} -3\cdot1+(-3)\cdot6r=14-r \\ -3-18r=14-r \end{gathered}[/tex]Adding r at both sides,
[tex]\begin{gathered} -3-18r+r=14-r+r \\ -3-17r=14 \end{gathered}[/tex]Adding 3 at both sides,
[tex]\begin{gathered} -3-17r+3=14+3 \\ -17r=17 \end{gathered}[/tex]Dividing by -17 at both sides,
[tex]\begin{gathered} \frac{-17r}{-17}=\frac{17}{-17} \\ r=-1 \end{gathered}[/tex]
One year there was a total of 44 commercial and noncommercial orbital launches worldwide. In addition, the number number of commercial orbital launches. Determine the number of commercial and noncommercial orbital launches was two more thank twice the number of commercial orbital launches (HURRY I NEED ANSWER)
The number of commercial orbital is 14, and the number of noncommercial orbital is 30.
What is algebra?
When numbers and quantities are represented in formulas and equations by letters and other universal symbols.
Given that,
The total number of commercial and noncommercial orbital launches worldwide = 44
Also, the number of noncommercial orbital is two more than twice of commercial orbital
Let the number of commercial orbital =x
Then number of noncommercial orbital = 2x+2
Since, total number of commercial and noncommercial orbital = 44
x + 2x +2 = 44
3x = 42
x = 14
The number of commercial orbital = x = 14
The number of noncommercial orbital = 2x+2 = 2×14+2 = 30
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I need help solving this practice problem If you can , answer (a) and (b) separately so I can tell which is which
Step 1:
Write the expression
[tex](3x^5\text{ - }\frac{1}{9}y^3)^4[/tex]Step 2:
a)
[tex]\begin{gathered} (3x^5\text{ - }\frac{1}{9}y^3)^4 \\ =^4C_0(3x^5)^4(-\frac{1}{9}y^3)^0+^4C_1(3x^5)^3(-\frac{1}{9}y^3)^1+^4C_2(3x^5)^2(-\frac{1}{9}y^3)^2+ \\ +^4C_1(3x^5)^1(-\frac{1}{9}y^3)^3+^4C_0(3x^5_{})^0(-\frac{1}{9}y^3)^4 \end{gathered}[/tex]Step 3:
b) simplified terms of the expression
[tex]\begin{gathered} Note\colon \\ ^4C_0\text{ = 1} \\ ^4C_1\text{ = 4} \\ ^4C_2\text{ = 6} \\ ^4C_3\text{ = 4} \\ ^4C_4\text{ = 1} \end{gathered}[/tex]Next, substitute in the expression
[tex]\begin{gathered} =\text{ 1}\times81x^{20}\times1\text{ - 4}\times27x^{15}\text{ }\times\text{ }\frac{y^3}{9}\text{ + 6 }\times9x^{10}\times\frac{y^6}{81}\text{ - 4}\times3x^5\text{ }\times\text{ }\frac{y^9}{729} \\ +\text{ 1 }\times\text{ 1 }\times\frac{y^{12}}{6561}\text{ } \end{gathered}[/tex][tex]=81x^{20}-12x^{15}y^3\text{ + }\frac{2}{3}x^{10}y^6\text{ - }\frac{4}{243}x^5y^9\text{ + }\frac{1}{6561}y^{12}[/tex]Rounding the problem to the nearest tenth if necessary and find the missing length?
Step 1:
[tex]\text{Triangle PQR is similar to triangle GHP}[/tex]Step 2:
Write the corresponding sides of the similar triangle
[tex]\begin{gathered} \\ PQ\text{ }\cong\text{ PG} \\ RP\text{ }\cong\text{ PH} \\ \frac{PQ}{PG}\text{ = }\frac{RP}{PH} \\ \\ \frac{PQ}{91}=\frac{72}{56}\text{ } \end{gathered}[/tex]Next
Cross multiply
[tex]\begin{gathered} 56PQ\text{ = 72 }\times\text{ 91} \\ PQ\text{ = }\frac{6552}{56} \\ PQ\text{ = 117} \end{gathered}[/tex]Final answer
PQ ? = 117
14a) Kevin asked a random group of students what theirfavorite class was, and the results are below.If Kevin were to randomly select a boy to explain whichclass was his favorite, what is the probability that he willpick a boy who likes History?A. 19.6%BoysGirlsMath810B. 29.3%English1517C. 35.3%Science105History 189D. 52.9%
If Kevin were to randomly select a boy from this table, this means that we need to add up the total number of boys in this sample in order to find the denominator of the fraction we are dealing with.
Let's add the number of boys up:
[tex]8+15+10+18=51[/tex]Now, the probability that a randomly selected boy likes History is the number of boys who said they like History divided by the total number of boys.
We already have the total number of boys, so now we need to find the number of boys who said they like History, which is 18 based off of the table.
The probability that Kevin will pick a boy who likes History is C) 35.3%.
[tex]\frac{18}{51}=.353[/tex]Yolanda has a rectangular poster that is 16 cm long and 10 cm wide what is the area of the poster in square meters do not round your answer is sure to include the correct unit in your answer
The area of a rectangle can be calculated as the height times the wide.
But be careful, the problem asks it in square meters! So let's use meters instead of centimeters.
Remember that : 1 m = 100 cm ----> 1 cm = 0.01 m
[tex]\begin{gathered} A=b\cdot h \\ \\ A=0.16\cdot0.10 \end{gathered}[/tex]Doing the multiplication
[tex]A=0.016\text{ m}^2[/tex]Therefore the area of the poster is 0.016 square meters
a new car is purchased for 24800. the value of the car depreciates at 12% per year what is the Y intercept of starting value
EXPLANATION
Let's see the facts:
Purchase Price = $24,800
Depretiation = 12%/year
We should apply the formula for exponential decay wich is expressed as:
A = P(1-r)^t
A= value of the car after t years
t= number of years
P = Initial Value
r= rate of decay in decimal form
We have that A=unknown t=? P=24,800 r=0.12
Replacing terms:
A=24,800(1-0.12)^t
Now, the y-intercept is the value obtained when t=0, so substituting this on the equation give us the following result:
A=24,800(1-0.12)^0 = 24,800(0.88)^0= 24,800*1=24,800
The starting value is $24,800
I need help quick with a math question !
Step-by-step explanation:
The area of a triangle is 2312 . Two of the side lengths are 93 and 96 and the included angle is obtuse. Find the measure of the included angle, to the nearest tenth of a degree.
to facilitate the exercise we will draw the triangle
We start using the area
[tex]A=\frac{b\times h}{2}[/tex]where A is the area, b the base and h the height
if we replace A=2312 and b=96 we can calculate the height(h)
[tex]\begin{gathered} 2312=\frac{96\times h}{2} \\ \\ h=\frac{2312\times2}{96} \\ \\ h=\frac{289}{6} \end{gathered}[/tex]now to calculate the measure of the angles we can solve the red triangle
first we find Y using trigonometric ratio of the sine
[tex]\sin (\alpha)=\frac{O}{H}[/tex]where alpha is the reference angle, O the opposite side from the angle and H the hypotenuse of the triangle
using Y like reference angle and replacing
[tex]\sin (y)=\frac{\frac{289}{6}}{93}[/tex]simplify
[tex]\sin (y)=\frac{289}{558}[/tex]and solve for y
[tex]\begin{gathered} y=\sin ^{-1}(\frac{289}{558}) \\ \\ y=31.2 \end{gathered}[/tex]value of angle y is 31.2°
Y and X are complementary because make a right line then if we add both numbers the solution is 180°
[tex]\begin{gathered} y+x=180 \\ 31.2+x=180 \end{gathered}[/tex]and solve for x
[tex]\begin{gathered} x=180-31.2 \\ x=148.8 \end{gathered}[/tex]measure of the included angle is 148.8°
DAN have coordinates D(-6, -1) the altitude drawn to side DN
Explanation:
The slope of the altitude drawn to side DN is the reciprocal and opposite to the slope of side DN, because the altitude is perpendicular to the side.
First we have to find the slope of side DN. The formula for the slope of a line with points (x1, y1) and (x2, y2) is:
[tex]m=\frac{y_1-y_2}{x_1-x_2}[/tex]In this problem the points are D(-6, -1) and N(-3, 10). The slope of side DN is:
[tex]m_{DN}=\frac{-1-10}{-6-(-3)}=\frac{-11}{-6+3}=\frac{-11}{-3}=\frac{11}{3}[/tex]Therefore the slope of the altitude is:
[tex]m_{\text{altitude}}=-\frac{1}{m_{DN}}=-\frac{1}{\frac{11}{3}}=-\frac{3}{11}[/tex]Answer:
The slope of the altitude is -3/11
Which expression is equivalent to sin(71(1) cos (72) - cos () sin (77.)?1?O cos (5)O sin (5)COS2012sin
Let:
[tex]\begin{gathered} A=\frac{\pi}{12} \\ B=\frac{7\pi}{12} \end{gathered}[/tex]Using the sine difference identity:
[tex]\begin{gathered} \sin (A)\cos (B)-\cos (A)\sin (B)=\sin (A-B) \\ so\colon \\ \sin (\frac{\pi}{12})\cos (\frac{7\pi}{12})-\cos (\frac{\pi}{12})\sin (\frac{7\pi}{12})=\sin (\frac{\pi}{12}-\frac{7\pi}{12}) \\ \sin (\frac{\pi}{12}-\frac{7\pi}{12})=\sin (-\frac{6\pi}{12}) \\ \sin (-\frac{\pi}{2}) \end{gathered}[/tex]Answer:
[tex]\sin (-\frac{\pi}{2})[/tex]3. A coin is tossed 140 times. The probability of getting tails is p = 0.500. Would a result of 55heads out of the 140 trials be considered usual or unusual? Why?Unusual, because the result is less than the maximum usual value.O Usual, because the result is between the minimum and maximum usual values.Unusual, because the result is less than the minimum usual value.Unusual, because the result is more than the maximum usual value
In order to calculate the minimum and maximum usual values, first let's calculate the mean and standard deviation of this distribution:
[tex]\begin{gathered} \mu=n\cdot p=140\cdot0.5=70\\ \\ \sigma=\sqrt{np(1-p)}=\sqrt{140\cdot0.5\cdot0.5}=5.92 \end{gathered}[/tex]Now, calculating the minimum and maximum usual values, we have:
[tex]\begin{gathered} minimum=\mu-2\sigma=70-11.84=58.16\\ \\ maximum=\mu+2\sigma=70+11.84=81.84 \end{gathered}[/tex]Since the given result is 55, it is an unusual reslt, because it is less tahan the minimum usual value.
Correct option: third one.
Cris pays a total of $11 for every 6 Gatoraid bottles. Circle the graph models a relationship with the same unit rate?
The line that describes this relationship goes from (0,0) to the point (6,11),we can draw it like this:
Given:• AJKL is an equilateral triangle.• N is the midpoint of JK.• JL 24.What is the length of NL?L24JKNO 12O 8V3O 12V2O 1213
Answer:
12√3
Explanation:
First, we know that JL = 24.
Then, the triangle JKL is equilateral. It means that all the sides are equal, so JK is also equal to 24.
Finally, N is the midpoint of segment JK, so it divides the segment JK into two equal parts. Therefore, JN = 12.
Now, we have a right triangle JLN, where JL = 24 and JN = 12.
Then, we can use the Pythagorean theorem to find the third side of the triangle, so NL is equal to:
[tex]\begin{gathered} NL=\sqrt[]{(JL)^2-(JN)^2} \\ NL=\sqrt[]{24^2-12^2} \end{gathered}[/tex]Because JL is the hypotenuse of the triangle and JN and NL are the legs.
So, solving for NL, we get:
[tex]\begin{gathered} NL=\sqrt[]{576-144} \\ NL=\sqrt[]{432} \\ NL=\sqrt[]{144(3)} \\ NL=\sqrt[]{144}\cdot\sqrt[]{3} \\ NL=12\sqrt[]{3} \end{gathered}[/tex]Therefore, the length of NL is 12√3
how do you solve for x in the following problem... 4 (x + 3) -2x + 8 = 28
Given the expression
[tex]4(x+3)-2x+8=28[/tex]To solve it for x, the first step is to calculate the term in parentheses, for this you have to apply the distributive property of multiplication.
[tex]\begin{gathered} (4\cdot x)+(4\cdot3)-2x+8=28 \\ 4x+12-2x+8=28 \end{gathered}[/tex]Next order the alike terms toghether and calculate:
[tex]\begin{gathered} 4x-2x+12+8=28 \\ 2x+20=28 \end{gathered}[/tex]Subtract 20 to both sides of the equation:
[tex]\begin{gathered} 2x+20-20=28-20 \\ 2x=8 \end{gathered}[/tex]And finally divide by 2 to reach the value of x:
[tex]\begin{gathered} \frac{2x}{2}=\frac{8}{2} \\ x=4 \end{gathered}[/tex]For this equation x=4
During a snowstorm, Grayson tracked the amount of snow on the ground. When the storm began, there were 4 inches of snow on the ground. For the first 3 hours of the storm, snow fell at a constant rate of 1 inch per hour. The storm then stopped for 5 hours and then started again at a constant rate of 3 inches per hour for the next 2 hours. As soon as the storm stopped again, the sun came out and melted the snow for the next 2 hours at a constant rate of 4 inches per hour. Make a graph showing the inches of snow on the ground over time using the data that Grayson collected.
We can plot all that happened in the next graph:
This is the graph showing the inches of the snow on the ground over time using the data that Grayson collected.
During a coffee house's grand opening.350 out of the first 500 customers who visited ordered only one single item while the rest ordered multiple items. Among the 150 customs who left a tip 60 of them ordered multiple items?
We are given a two-way frequency table with some missing joint frequencies.
The frequencies in the total row and column are called "marginal frequencies"
The frequencies in the other rows and columns are called "joint frequencies"
Let us first find the joint frequency "Single Item and Tip"
[tex]\begin{gathered} x+60=150 \\ x=150-60 \\ x=90 \end{gathered}[/tex]So, the joint frequency "Single Item and Tip" is 90 (option B)
Now, let us find the joint frequency "Single Item and No Tip"
[tex]\begin{gathered} 90+x=360 \\ x=360-90 \\ x=270 \end{gathered}[/tex]So, the joint frequency "Single Item and No Tip" is 270 (option D)
Now first we need to find the marginal frequency as below
[tex]\begin{gathered} 360+x=500 \\ x=500-360 \\ x=140 \end{gathered}[/tex]Finally, now we can find the joint frequency "Multiple Items and No Tip"
[tex]\begin{gathered} 60+x=140 \\ x=140-60 \\ x=80 \end{gathered}[/tex]So, the joint frequency "Multiple Items and No Tip" is 80 (option A)
Therefore, the missing joint frequencies are
Option A
Option B
Option D
If the figure below were reflected across the waxis, what would be the new coordinates of point A
The coordinates of point A are (-2,3). A reflection across the y-axis is given by:
[tex](x,y)\rightarrow(-x,y)[/tex]Applying this rule to point A we have:
[tex](-2,3)\rightarrow(2,3)[/tex]Therefore, the image of point A is (2,3) and the correct option is B.
Use the graphing tool to determine the true statementsregarding the represented function. Check all that apply.f(x) > 0 over the interval (1,).Of(x) < 0 over the interval [1,0).Of(x) 0 over the interval (-∞, 1].Of(x) > 0 over the interval (-∞, 1).Of(x) > 0 over the interval (-∞o).Intro2010-202
The true statements are,
f(x) > 0 over the interval (1, ∞)
f(x) ≤ 0 over the interval (-∞, 1]
Interval of a function:
If the value of the function f (x) rises as the value of x rises, the function interval is said to be positive. Instead, if the value of the function f (x) drops as the value of x increases, the function interval is said to be negative.
If the endpoints are absent from an interval, it is referred to as being open. It's indicated by ( ). Examples are (1, 2), which denotes larger than 1 and less than 2. Any interval that contains all the limit points is said to be closed. The symbol for it is []. For instance, [2, 5] denotes a value greater or equal to 2 and lower or equal to 5. If one of an open interval's endpoints is present, it is referred to as a half-open interval.
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Given ABC shown below. Map ABC using the transformations given below. In each case, start with ABC , graph the image and state the Coodinates of the image's vertices.a) a reflection in the line x = 2 to produce A' B' C'b) a reflection in the line y= 1 to produce A" B"C"
a) A'(8,7), B' (10, -6) and C' (2,-3)
b) A" (-4,-5) B" (-6,8) C" (2,5)
1) Examining the graph, we can locate the following points of ABC
To reflect across line x=2 let's count to the left the same distance from x=2
Pre-image Reflection in the line x=2
A (-4, 7) (x+8, y) A'(8,7)
B (-6,-6) (x+16, y) B' (10, -6)
C (2,-3) (x,y) C' (2,-3) Remains the same since C is on x=2
b) A reflection about the line y=1 similarly we'll count the distances and then write new points over the line y=1.
So the Image of this is going to be
Pre-image Reflection in the line x=2
A (-4, 7) (x, y-12) A" (-4,-5)
B (-6,-6) (x, y) B" (-6, 8)
C (2,-3) (x,y) C' (2,5)
A" (-4,-5)
B" (-6,8)
C" (2,5)