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3 years ago

Help me with this problem please!

Mathematics

2 answers:

Hatshy [7]3 years ago

8 0

Ok so you need to find the x and y- intercepts. to do this you simply plug 0 in for x and then for y, giving you the value of each variable.

15x + 20y = 1,800

15x + 20(0) = 1,800
15x = 1,800
x = 120

15(0) + 20y = 1,800
20y = 1,800
y = 90

your x-intercept is (120,0)
your y-intercept is (0,90)

BartSMP [9]3 years ago

6 0

First you need to get the equation into y - intercept form in order to graph it. The equation needs to be in the form y=mx+b. do you know how to convert that equation to y=mx+b?

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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3 years ago

In the standard (x,y) coordinate plane a right triangle has vertices at (-3,4)(3,4)(3,-4) what is the length in coordinates unit

MAXImum [283]

Given:

The vertices of a right triangle are:

$(-3,4),(3,4),(3,-4)$

To find:

The length of the hypotenuse of the given right triangle.

Solution:

Let the vertices of the right triangle are $A(-3,4),B(3,4),C(3,-4)$ .

The distance formula is:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Using distance formula, we get

$AB=\sqrt{(3-(-3))^2+(4-4)^2}$

$AB=\sqrt{(3+3)^2+(0)^2}$

$AB=\sqrt{(6)^2+0}$

$AB=\sqrt{36}$

$AB=6$

Similarly,

$BC=\sqrt{(3-3)^2+(-4-4)^2}$

$BC=\sqrt{(0)^2+(-8)^2}$

$BC=\sqrt{64}$

$BC=8$

And,

$AC=\sqrt{(3-(-3))^2+(-4-4)^2}$

$AC=\sqrt{(6)^2+(-8)^2}$

$AC=\sqrt{36+64}$

$AC=\sqrt{100}$

$AC=10$

Now, taking sum of squares of two smaller sides, we get

$AB^2+BC^2=6^2+8^2$

$AB^2+BC^2=36+64$

$AB^2+BC^2=100$

$AB^2+BC^2=AC^2$

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3 years ago

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