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4 years ago

Substitution method Y=0.2+2200 Y=-x+5500

1 answer:

4 0

Since both of the equations both equal why, substitute the equations to make them equal to each other. Like this:Y=0.2+2200 and Y=-x+5500. To this: 0.2+2200=-x+5500. Then solve.

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I need helpppp!! plzz

kumpel [21]

Answer:

Step-by-step explanation:

The slope is 3/2

the y-intercept is -4.5

so the equation is y = 3/2x + (-4.5)

(it’s a plus because the slope is positiv, meaning the line goes upwards to the right.)

8 0

3 years ago

NEED HELP 6m²+6m-9m²

Maru [420]

6m² + 6m - 9m²

Add like terms.

(6m² - 9m²) + 6m

-3m² + 6m

~Hope I helped!~

3 0

3 years ago

Read 2 more answers

Real world situation for 834+14s>978-10s

liraira [26]

834+14s=848>978-10s=968

a real life situation for that could be like how many apples 848apples/968total

6 0

3 years ago

Graph the image of this figure after a dilation with a scale factor of 13 centered at the point (4, −2) .

Sonja [21]

Answer:

From the given triangle figure;

Labelled the triangle as A , B and C.

The coordinates of this triangle ABC are;

A = (1, 10) ,

B = (-2, 4)

C = (7, 4).

Given : Scale factor(k) = $\frac{1}{3}$ and centered at point (4, -2).

The rule of dilation with $k= \frac{1}{3}$ and center at point (4,-2) is:

$(x, y) \rightarrow (\frac{1}{3}(x-4)+4 , \frac{1}{3}(y+2)-2)$

$(x, y) \rightarrow (\frac{1}{3}x-\frac{4}{3}+4 , \frac{1}{3}y+\frac{2}{3}-2)$

$(x, y) \rightarrow (\frac{1}{3}x+\frac{8}{3} , \frac{1}{3}y-\frac{4}{3})$

then, the dilation of the given figure are;

$A(1, 10) \rightarrow (\frac{1}{3}\cdot 1+\frac{8}{3} , \frac{1}{3} \cdot 10-\frac{4}{3})$ = $(\frac{1}{3}+\frac{8}{3} , \frac{10}{3}-\frac{4}{3})$ = $(\frac{9}{3} , \frac{6}{3})$ = A'(3 , 2)

$B(-2, 4) \rightarrow (\frac{1}{3}\cdot -2+\frac{8}{3} , \frac{1}{3} \cdot 4-\frac{4}{3})$ = $(-\frac{2}{3}+\frac{8}{3} , \frac{4}{3}-\frac{4}{3})$ = $(\frac{6}{3} , \frac{0}{3})$ =B'(2 , 0)

$C(7, 4) \rightarrow (\frac{1}{3}\cdot 7+\frac{8}{3} , \frac{1}{3} \cdot 4-\frac{4}{3})$ = $(\frac{7}{3}+\frac{8}{3} , \frac{4}{3}-\frac{4}{3})$ = $(\frac{15}{3} , \frac{0}{3})$ = C'(5 , 0)