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3 years ago

Factor. 18b−32 Enter your answer in the boxes. [ ]([ ])

2 answers:

3 0

Here, GCF of 18 & 32 is 2, so take it as common factor,
18b - 32 = 2(9b - 16)

In short, Your Answer would be: 2(9b - 16)

Hope this helps!

Oliga [24]3 years ago

3 0

18b-32
/2 /2

-2(-9b+16)

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Can you tell what is the answer to this math problem ?

Marrrta [24]

X^2-18 = 3x

x^2 - 3x - 18 = 0

(x-6)(x+3) = 0

x = {-3,6}

So -3 is the answer

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3 years ago

What are the zeros of f(x) = x2 + 3x – 10?

Nady [450]

Step One
Factor f(x)
f(x) = (x + 5)(x - 2)

Step Two
The zeros occur when either (x + 5) = 0 or (x - 2) = 0

x + 5 = 0
x = -5

x - 2 =0
x = 2

Step 3
Record the zeros
(-5,0) or (2,0) <<<<<< answer.

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3 years ago

State the following characteristics of the parab-

Damm [24]

Axis of symmetry- x=-1

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3 years ago

Find the area of the figure. (Sides meet at right angles.)

daser333 [38]

Answer:

48m squared

Step-by-step explanation:

- cut the problem in half horizontally

you get two squares one square is 4 by 4 and the other 8 by 4

4 times 4=16 8 times 4=32 32+16=48

6 0

3 years ago

Verify the identity 6cos^2(x) -3 = 3 - 6sin^2(x)

Lorico [155]

Answer:

It is proved that $6\cos ^{2}x -3 = 3 - 6\sin ^{2} x$ .

Step-by-step explanation:

We already have the identity of x as $\sin ^{2}x + \cos ^{2}x = 1$ .......... (1) .

So, from equation (1) we can write that

$\cos ^{2} x = 1 - \sin ^{2} x$

⇒ $6\cos ^{2} x = 6 - 6 \sin ^{2} x$

⇒ $6\cos ^{2} x -3 = 6 - 6 \sin ^{2}x -3$

⇒ $6\cos ^{2}x -3 = 3 - 6\sin ^{2} x$

Hence, it is proved that $6\cos ^{2}x -3 = 3 - 6\sin ^{2} x$ . (Answer)

5 0

3 years ago

Factor. ​ 18b−32 ​ Enter your answer in the boxes. [ ]([ ])

Factor. 18b−32 Enter your answer in the boxes. [ ]([ ])