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Nastasia [14]
3 years ago
8

Which description is true about ​pre-image ABC​ and ​image A′B′C′​ ?

Mathematics
1 answer:
scoray [572]3 years ago
7 0

Answer:

Option (C)

Step-by-step explanation:

From the graph attached,

In ΔABC and ΔA'B'C',

∠A ≅ ∠A'

∠B ≅ ∠B'

Therefore, ΔABC ~ ΔA'B'C'

Corresponding sides of these triangles will be proportional.

\frac{\text{AB}}{\text{A'B'}}=\frac{\text{BC}}{\text{B'C'}}=\frac{\text{AC}}{\text{A'C'}}=\frac{2}{6}

Therefore, ratio of the sides, AC : A'C' = 1 : 3 shows that image triangle A'B'C' is a dilated form of pre-image ABC with a scale factor of 3.

Option (C) will be the correct option.

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What is the solution of the equation?
nadya68 [22]

Answer:

q = 4

Fully simplifying this equation gives you answer (A) q = 4

Hope this helps!

8 0
3 years ago
-(6x+6)+2=-6x-3 please solve
makkiz [27]

Answer:

There is no solution.

Step-by-step explanation:

To solve, use the order of operations. Follow them until you solve for x or just numbers remain.

-(6x + 6) + 2 = -6x - 3

-6x - 6 + 2 = -6x - 3

-6x - 4 = -6x - 3

-4 = -3

The answer is no solution since this is a false statement.

3 0
3 years ago
Lines AB and CD are straight lines. Find x and y. Give reasons to justify your solutions. I need both answers.
Kazeer [188]

Answer:

1) x = 14°, y = 5°

2) x = 18.5°, y = 37°

Step-by-step explanation:

1) ∠AOC and ∠BOD are vertical angles, then ∠BOD = 25°

∠MOD = ∠MOB + ∠BOD = 90°

3x + 23° + 25° = 90°

3x = 90° - 23° - 25°

x = 42°/3

x = 14°

∠LOB = ∠LOM + ∠MOB = 90°

5y + 3x + 23° = 90°

5y = 90° - 23° - 3(14°)

y = 25°/5

y = 5°

2) ∠AOC and ∠BOD are vertical angles, then ∠BOD = 16°

∠EOB = ∠EOD + ∠DOB =  90°

2y + 16° = 90°

y = (90° - 16°)/2

y = 37°

∠DOF = ∠BOF + ∠DOB =  90°

4x + 16° = 90°

x = (90° - 16°)/4

x = 18.5°

4 0
3 years ago
What is the answer 4b-7-15b+3
Molodets [167]
Simplify

4b−7−15b+3

=4b+−7+−15b+3

Combine Like Terms:

=4b+−7+−15b+3

=(4b+−15b)+(−7+3)

=−11b−4

7 0
3 years ago
Read 2 more answers
Let x be the average number of employees in a group health insurance plan, and let y be the average administrative cost as a per
Step2247 [10]

A scatter diagram has points that show the relationship between two sets of data.

We have the following data,

\left\begin{array}{ccccccc}\mathrm{x}&3&7&15&32&74\\\mathrm{y}&40&35&30&25&17\end{array}\right

where <em>x</em> is the average number of employees in a group health insurance plan and <em>y</em> is the average administrative cost as a percentage of claims.

To make a scatter diagram you must, draw a graph with the independent variable on the horizontal axis (<em>in this case x</em>) and the dependent variable on the vertical axis (<em>in this case y</em>). For each pair of data, put a dot or a symbol where the x-axis value intersects the y-axis value.

Linear regression is a way to describe a relationship between two variables through an equation of a straight line, called line of best fit, that most closely models this relationship.

To find the line of best fit for the points, follow these steps:

Step 1: Find X\cdot Y and X\cdot X as it was done in the below table.

Step 2: Find the sum of every column:

\sum{X} = 131 ~,~ \sum{Y} = 147 ~,~ \sum{X \cdot Y} = 2873 ~,~ \sum{X^2} = 6783

Step 3: Use the following equations to find intercept a and slope b:

\begin{aligned}        a &= \frac{\sum{Y} \cdot \sum{X^2} - \sum{X} \cdot \sum{XY} }{n \cdot \sum{X^2} - \left(\sum{X}\right)^2} =             \frac{ 147 \cdot 6783 - 131 \cdot 2873}{ 5 \cdot 6783 - 131^2} \approx 37.05 \\ \\b &= \frac{ n \cdot \sum{XY} - \sum{X} \cdot \sum{Y}}{n \cdot \sum{X^2} - \left(\sum{X}\right)^2}        = \frac{ 5 \cdot 2873 - 131 \cdot 147 }{ 5 \cdot 6783 - \left( 131 \right)^2} \approx -0.292\end{aligned}

Step 4: Assemble the equation of a line

\begin{aligned} y~&=~a ~+~ b \cdot x \\y~&=~37.05 ~-~ 0.292 \cdot x\end{aligned}

6 0
3 years ago
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