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solniwko [45]
3 years ago
6

What are the points for -3x+6y+5=-7

Mathematics
1 answer:
Novay_Z [31]3 years ago
7 0
-3x + 6y + 5 = -7
<u>                -5    -5</u>
      -3x + 6y = -12
-3x + 3x + 6y = -12 + 3x
<u>6y</u> = <u>-12 + 3x</u>
 6           6
  y = -2 + 1/2x
-3x + 6(-2 + 1/2x) = -12
-3x - 12 + 3x = -12
-3x + 3x - 12 = -12
0x - 12 = -12
<u>     +12   +12</u>
       <u>0x</u> = <u>0</u>
        0     0
         x = 0
-3(0) + 6y = -12
0 + 6y = -12
<u>+0           +0</u>
      <u>6y</u> = <u>-12</u>
       6       6
        y = -2
  (x, y) = (0, -2)
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Tems11 [23]

Answer:

4.32ft^2

Step-by-step explanation:

A of a rhombus = l x w

=2.4x1.8

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∴ the area of the solar panel is 4.32ft^2

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Which of the following is true about the right triangle?
katrin2010 [14]

Option 1. cos (a) = \frac{3}{5}, option 3. sin (a) = \frac{4}{5}, and option 5. tan (a) = \frac{4}{3} are true about the right triangle.

Step-by-step explanation:

Step 1:

According to the Pythagorean theorem, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Here the hypotenuse measures 15 units. Assume the other side measures x.

15^{2} = x^{2} +9^{2} , x^{2} = 15^{2} -9^{2} .

x^{2} = 144, x= 12.

Step 2:

sin \theta = \frac{oppositeside}{hypotenuse} ,cos \theta = \frac{adjacentside}{hypotenuse} , tan \theta = \frac{oppositeside}{adjacentside}.

For angle A, the opposite side is 12 units long, the adjacent side is 9 units long and the hypotenuse is 15 units long.

sin (a) = \frac{12}{15} ,cos(a) = \frac{9}{15} , tan(a) = \frac{12}{9}.

sin (a) = \frac{4}{5} ,cos (a) = \frac{3}{5} , tan (a) = \frac{4}{3}.

So option 1. cos (a) = \frac{3}{5}, option 3. sin (a) = \frac{4}{5}, and option 5. tan (a) = \frac{4}{3} are true about the right triangle.

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Step-by-step explanation:


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Ede4ka [16]

Answer:

Part a) The quadratic function is 4x^{2} +20x-39=0

Part b) The value of x is 1.5\ in

Part c) The photo and frame together are 7\ in wide

Step-by-step explanation:

Part a) Write a quadratic function to find the distance from the edge of the photo to the edge of the frame

Let

x----> the distance from the edge of the photo to the edge of the frame

we know that

(6+2x)(4+2x)=63\\24+12x+8x+4x^{2}=63\\ 4x^{2} +20x+24-63=0\\4x^{2} +20x-39=0

Part b) What is the value of x?

Solve the quadratic equation 4x^{2} +20x-39=0

The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}

in this problem

we have

4x^{2} +20x-39=0

so

a=4\\b=20\\c=-39

substitute in the formula

x=\frac{-20(+/-)\sqrt{20^{2}-4(4)(-39)}} {2(4)}

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x=\frac{-20(+/-)32} {8}

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x=\frac{-20(-)32} {8}=-6.5\ in

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