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Dima020 [189]
3 years ago
6

Find the value of y for the given value of x. y=x2+9;x=−12

Mathematics
1 answer:
DiKsa [7]3 years ago
7 0

Answer:

0

Step-by-step explanation:

0 is what i got

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Hey please help me with this one.<br><br>- Find the value fd a
valkas [14]

Answer: a = 15

Step-by-step explanation:

6a + 10 = 3a + 55

Subtract 10 from both sides

6a = 3a + 45

Subtract 3a from both sides

3a = 45

Divide each side by 3

a = 15

5 0
3 years ago
Use the method of completing the square to transform the quadratic equation into the equation form (x + p)^2 = q. 6+12x+3x^2=0
Mrac [35]
Hello : 
<span>3x²+12x+6=0
3(x²+4x +2) =0
x²+4x+2 = 0
(x²+4x+4)-4+2 = 0
(x+2)² = 2
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3 0
3 years ago
write the absolute value equations in the form |x-b|=c (where b is a number and c can be either number or an expression) that ha
Katena32 [7]

Answer:

b= 2x c=14x

Step-by-step explanation:

3 0
2 years ago
¡Help please!
pshichka [43]
= m^4 + 28m^2 - 36m^4 + 8m^5 
= 8m^5 - 35m^4 + 28m^2

Answer
polynomial in standard form
8m^5 - 35m^4 + 28m^2
5 0
3 years ago
The figure below shows part of a stained-glass window depicting the rising sun. Which function can be used to find the area of t
Kryger [21]

Answer:

A(w) = w^2 + 5w - \frac{1}{8}\pi w^2

Step-by-step explanation:

A = the area of the region outside the semicircle but inside the rectangle

w = the width of the rectangle or diameter of the semicircle

Since "A" is determined by "w", therefore, "A" is a function of "w" = A(w).

A(w) = (area of rectangle) - (area of semicircle)

A(w) = (l*w) - (\frac{1}{2} \pi r^2)

Where,

lenght of rectangle (l) = w + 5

width of rectangle (w) = w

r = ½*w = \frac{w}{2}

Plug in the values:

A(w) = ((w + 5)*w) - (\frac{1}{2} \pi (\frac{w}{2})^2)

A(w) = ((w + 5)*w) - (\frac{1}{2} \pi (\frac{w}{2})^2)

Simplify

A(w) = (w^2 + 5w) - (\frac{1}{2} \pi (\frac{w^2}{4})

A(w) = w^2 + 5w - \frac{1}{2}*\pi*\frac{w^2}{4}* \pi

A(w) = w^2 + 5w - \frac{1*\pi*w^2}{2*4}

A(w) = w^2 + 5w - \frac{1*\pi w^2}{8}

A(w) = w^2 + 5w - \frac{1}{8}\pi w^2

3 0
3 years ago
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