Help, please. This is related to mirror and shadow problems

Mathematics

1 answer:

nadya68 [22]2 years ago

5 0

Answer:

1. x/2=1.75/3.5 x=1

2. 4/2=y/3.5 y=3.5

3. P (Figure Left)=3+5+2+1+4+8+3+2=28 cm

A (Figure Left)=8x2-2x1-3x2=8 cm2

4. P (Figure Left) / P (Figure Right) = 1/1.75 = 28/ P (Figure Right)

P (Figure Right)= 49 cm

A (Figure Left) / A (Figure Right) = 1/1.752 = 8/ A (Figure Right)

A(Figure Right)= 24.5 cm2

Step-by-step explanation:

1. x/2=1.75/3.5 x=1

2. 4/2=y/3.5 y=3.5

3. P (Figure Left)=3+5+2+1+4+8+3+2=28 cm

A (Figure Left)=8x2-2x1-3x2=8 cm2

4. P (Figure Left) / P (Figure Right) = 1/1.75 = 28/ P (Figure Right)

P (Figure Right)= 49 cm

A (Figure Left) / A (Figure Right) = 1/1.752 = 8/ A (Figure Right)

A(Figure Right)= 24.5 cm2

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35. For the following exercises, given each set of information, find a linear equation satisfying the conditions, if possible.

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Answer:

The linear equation for the line which passes through the points given as (-2,8) and $(4,6), is written in the point-slope form as $y=-\frac{1}{3} x-\frac{26}{3}$ .

Step-by-step explanation:

A condition is given that a line passes through the points whose coordinates are (-2,8) and (4,6).

It is asked to find the linear equation which satisfies the given condition.

Step 1 of 3

Determine the slope of the line.

The points through which the line passes are given as (-2,8) and (4,6). Next, the formula for the slope is given as,

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Substitute $6 \& 8$ for $y_{2}$ and $y_{1}$ respectively, and 4&-2 for $x_{2}$ and $x_{1}$ respectively in the above formula. Then simplify to get the slope as follows, $m=\frac{6-8}{4-(-2)}$

$\begin{aligned}&m=\frac{-2}{6} \\&m=-\frac{1}{3}\end{aligned}$

Step 2 of 3

Write the linear equation in point-slope form.

A linear equation in point slope form is given as,

$y-y_{1}=m\left(x-x_{1}\right)$

Substitute $-\frac{1}{3}$ for m,-2 for $x_{1}$ , and 8 for $y_{1}$ in the above equation and simplify using the distributive property as follows, $y-8=-\frac{1}{3}(x-(-2))$\\ $y-8=-\frac{1}{3}(x+2)$\\ $y-8=-\frac{1}{3} x-\frac{2}{3}$$