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2 years ago

Consider the equation Ax+By=−36. If the x-intercept is (−3,0) and the y-intercept is (0,9), what are the values of A and B?

Mathematics

1 answer:

Grace [21]2 years ago

5 0

Answer:

A= 12

B = -4

Step-by-step explanation:

( -3,0) & (0,9) should satisfy the equation Ax + By = -36

when ( -3,0)

-3A+ 0 = -36

A = 12

when (0,9)

0 + 9B = -36

B = -4

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Point A is at (-1, -9) and point M is at (0.5, -2.5).

vladimir2022 [97]

Answer:

Point B is located at point (2, 4)

Step-by-step explanation:

Since point M is at the center, you find the distance from point A and add that to the other side.

| -1 - 0.5 | = | -1.5 | = 1.5

| -9 - -2.5 | = | -6.5 | = 6.5

Now we can add that to point M to find point B

0.5 + 1.5 = 2

-2.5 + 6.5 = 4

Point B should be at (2, 4)

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2 years ago

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Step-by-step explanation:

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2 years ago

Area of a triangle with points at (-9,5), (6,10), and (2,-10)

Ann [662]

First we are going to draw the triangle using the given coordinates.
Next, we are going to use the distance formula to find the sides of our triangle.
Distance formula: $d= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Distance from point A to point B:
$d_{AB}= \sqrt{[6-(-9)]^2+(10-5)^2}$
$d_{AB}= \sqrt{(6+9)^2+(10-5)^2}$
$d_{AB}= \sqrt{(15)^2+(5)^2}$
$d_{AB}= \sqrt{225+25}$
$d_{AB}= \sqrt{250}$
$d_{AB}=15.81$

Distance from point A to point C:
$d_{AC}= \sqrt{[2-(-9)]^2+(-10-5)^2}$
$d_{AC}= \sqrt{(2+9)^2+(-10-5)^2}$
$d_{AC}= \sqrt{11^2+(-15)^2}$
$d_{AC}= \sqrt{121+225}$
$d_{AC}= \sqrt{346}$
$d_{AC}= 18.60$

Distance from point B from point C
$d_{BC}= \sqrt{(2-6)^2+(-10-10)^2}$
$d_{BC}= \sqrt{(-4)^2+(-20)^2}$
$d_{BC}= \sqrt{16+400}$
$d_{BC}= \sqrt{416}$
$d_{BC}=20.40$

Now, we are going to find the semi-perimeter of our triangle using the semi-perimeter formula:
$s= \frac{AB+AC+BC}{2}$
$s= \frac{15.81+18.60+20.40}{2}$
$s= \frac{54.81}{2}$
$s=27.41$

Finally, to find the area of our triangle, we are going to use Heron's formula:
$A= \sqrt{s(s-AB)(s-AC)(s-BC)}$
$A=\sqrt{27.41(27.41-15.81)(27.41-18.60)(27.41-20.40)}$
$A= \sqrt{27.41(11.6)(8.81)(7.01)}$
$A=140.13$

We can conclude that the perimeter of our triangle is 140.13 square units.

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2 years ago

Only need help with #'s 10, 12, 14, 16

gregori [183]

Im just about to work one out for you and you just do the same on the rest

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3 years ago