Write an equation for the line

Mathematics

1 answer:

elena55 [62]4 years ago

4 0

Answer:

y = $\frac{-2}{3}$ x + $\frac{11}{3}$

Step-by-step explanation:

Step 1: Use the slope formula to determine the slope

We know that the y value decrease by 2 and the x increases by 3 so the formula of $\frac{deltaY}{deltaX}$ can also be written as $\frac{-2}{3}$

<em>(delta Y and delta X represent the change in the x and y)</em>

Step 2: Find the y-intercept

We substitute the slope and the point(-2, 5) into y=mx+b to solve for b

(5)=( $\frac{-2}{3}$ )(-2) + b

5 = $\frac{4}{3}$ + b

5 - $\frac{4}{3}$ = b

b = $\frac{11}{3}$

Step 3: Write the equation out

y = mx+ b

y = $\frac{-2}{3}$ x + $\frac{11}{3}$

Therefore the equation of a line that passes through the point (-2, 5) with a change in the x value of 3 and y value of -2 is y = $\frac{-2}{3}$ x + $\frac{11}{3}$

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Plz show work and solve quick please need help!!!!!

Simora [160]

Surface area is just the area of all these 4 triangles plus the rectangle.

First we can find the area of the rectangle.

$\sf A=lw$

Half of the length is 28 cm, so the full length must be 28 * 2 = 56 cm.

$\sf A=(56)(27)$

$\sf A=1512cm^2$

The base for the left and right triangles are 27. The heights would be the net length minus half the length of the rectangle:

$\sf 82.8-28=54.8\? cm$

Calculate the area:

$\sf A=\dfrac{1}{2}bh$

$\sf A=\dfrac{1}{2}(27)(54.8)$

$\sf A=739.8\?cm^2$

We have two of these triangles.

$\sf 739.8\times 2=1479.6\? cm^2$

Now do the other two pair of triangles. The bases for them are 28 + 28 = 56 cm. The heights would be the net width minus the width of the rectangle:

$\sf 81.2-27=54.2\?cm$

Now find the area:

$\sf A=\dfrac{1}{2}bh$

$\sf A=\dfrac{1}{2}(56)(54.2)$

$\sf A=1517.6\? cm^2$

We have two of these triangles.

$\sf 1517.6\times 2=3035.2\?cm^2$

Add all the areas together:

$\sf 1512+1479.6+3035.2=\boxed{\sf 6026.80\? cm^2}$