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3 years ago

Please help me, if correct get brainliest

Mathematics

1 answer:

Solnce55 [7]3 years ago

3 0

Answer:

If I am correct the answer should be graph it !!!! first quadrant

corners

(0,0)

(0,3)

(1.5 , 3.5) intersection

(5,0)

evaluate 6x-4y at those corners

-12

-5

makes sense, need big x and small y :)

Step-by-step explanation:

I apologize if I am incorrect I am only in 7th grade but I am pretty intelligent so please try it out!

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Complete the equation of the line whose slope is -2 and y-intercept is (0,-8) y=

Alexandra [31]

Answer:

y = -2x - 8

Step-by-step explanation:

We want to find the slope-intercept form of this equation. Slope-intercept form is denoted by: y = mx + b, where m is the slope and b is the y-intercept.

We already know that the slope is -2, so that means m = -2.

We also know the y-intercept (or the point where the line crosses the y-axis) is (0, -8), which means b = -8.

Thus, our equation is: y = -2x - 8.

~ an aesthetics lover

7 0

3 years ago

Read 2 more answers

On a coordinate plane, triangle A B C is shown. Point A is at (0, 0), point B is at (3, 4), and point C is at (3, 2). What is th

Elena L [17]

Answer:

The area of triangle for the given coordinates is 1.5 $\sqrt{4.6}$

Step-by-step explanation:

Given coordinates of triangles as

A = (0,0)

B = (3,4)

C = (3,2)

So, The measure of length AB = a = $\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

Or, a = $\sqrt{(3-0)^{2}+(4-0)^{2}}$

Or, a = $\sqrt{9+16}$

Or, a = $\sqrt{25}$

∴ a = 5 unit

Similarly

The measure of length BC = b = $\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

Or, b = $\sqrt{(3-3)^{2}+(2-4)^{2}}$

Or, a = $\sqrt{0+4}$

Or, b = $\sqrt{4}$

∴ b = 2 unit

And

So, The measure of length CA = c = $\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

Or, c = $\sqrt{(3-0)^{2}+(2-0)^{2}}$

Or, c = $\sqrt{9+4}$

Or, c = $\sqrt{13}$

∴ c = $\sqrt{13}$ unit

Now, area of Triangle written as , from Heron's formula

A = $\sqrt{s\times (s-a)\times (s-b)\times (s-c)}$

and s = $\frac{a+b+c}{2}$

I.e s = $\frac{5+2+\sqrt{13}}{2}$

Or. s = $\frac{7+\sqrt{13}}{2}$

So, A = $\sqrt{(\frac{(7+\sqrt{13})}{2})\times ((\frac{(7+\sqrt{13})}{2})-5)\times (\frac{7+\sqrt{13}}{2}-2)\times (\frac{7+\sqrt{13}}{2}-\sqrt{13})}$