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dlinn [17]
4 years ago
15

-6r+7r=5 ????????????

Mathematics
2 answers:
gtnhenbr [62]4 years ago
8 0

Answer:

r= 5

Step-by-step explanation:

telo118 [61]4 years ago
8 0
(-6+7)r=5

1r=5

Final: r=5
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How do I find the dimensions of one face
Sauron [17]

ok so dimensions is like area the dimensions of a cubes side is a one by one or the base and height of a cube is 1 inch each get it now or do you need me to explain more?

8 0
4 years ago
A projector displays an image on a wall. the area(in square feet) of the rectangular projection can be represented by
andrew-mc [135]

Answer:

The height of the projection will be (x - 5) feet.

Step-by-step explanation:

A projector displays an image on a wall. The area (in square feet) of the rectangular projection can be represented by (x² - 8x + 15).

Now to get the width and the height of the rectangular projection we have to factorize the expression (x² - 8x + 15).

Here, (x² - 8x + 15)

= x² - 3x - 5x + 15

= (x- 3)(x - 5)

Now, if the height of the projection is less than its width then, the height of the projection will be (x - 5) feet. (Answer)

3 0
3 years ago
Complete the square to rewrite the equation in the form (x−h)2=p.
soldier1979 [14.2K]

Answer:

second option

Step-by-step explanation:

Given

x² - 6x - 33 = 0 ( add 33 to both sides )

x² - 6x = 33

To complete the square

add ( half the coefficient of the x- term )² to both sides

x² + 2(- 3)x + 9 = 33 + 9

(x - 3)² = 42 → second option

4 0
4 years ago
What is 2x cubed plus 2x squared simplified too
Ostrovityanka [42]

Answer:That can’t be simplified further.

Step-by-step explanation:

5 0
3 years ago
Find a general solution of y" + 8y' + 16y=0.
Misha Larkins [42]

Answer:

The general solution: C_{1}e^{-4x} + xC_{2}e^{-4x}

Step-by-step explanation:

Differential equation: y'' + 8y' + 16y = 0

We have to find the general solution of the above differential equation.

The auxiliary equation for the above equation can be writtwn as:

m² + 8m +16 = 0

We solve the above equation for m.

(m+4)² = 0

m_{1} = -4, m_{2} = -4

Thus we have repeated roots for the auxiliary equation.

Thus, the general solution will be given by:

y = C_{1}e^{m_{1}x} + xC_{2}e^{m_{2}x}

y = C_{1}e^{-4x} + xC_{2}e^{-4x}

6 0
3 years ago
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