Simplify 1/3 12x - 6 A. -2x-1 B.-2x-8 C.-x+4 D.3x+10 E. 4x-2

which expression is equivalent to 100 n2 − 1? (10n)2 − (1)2 (10n2)2 − (1)2 (50n)2 − (1)2 (50n2)2 − (1)2

givi [52]

Which expression is equivalent to 100 n2 − 1? (10n)2 − (1)2 (10n2)2 − (1)2 (50n)2 − (1)2 (50n2)2 − (1)2

Regroup:
100n^2-1

=10^2n^2-1
=(10n)^2-1^2

So the correct answer is :
(10n)^2-1^2

To attempt to factor a polynomial of four or more terms with no common factor, first rewrite it in groups. Each group may possibly be separately factored, and the resulting expression may possibly lend itself to further factorization if a greatest common factor or special form is created.

5 0

3 years ago

Read 2 more answers

The functions f (theta) and g (theta) are sine functions, where f (0) equals g (0) equals 0.The amplitude of f (theta) is twice

777dan777 [17]

If period of $f(\theta)$ is one-half the period of $g(\theta)$ and
 $g(\theta)$ has a period of 2π, then $T_{g} =2T_{f}=2 \pi$ and $T_{f}= \pi$ .

To find the period of sine function $f(\theta)=asin(b\theta+c)$ we use the rule $T_{f}= \frac{2\pi}{b}$ .

f is sine function where f (0)=0, then c=0; with period $\pi$ , then $f(\theta)=asin 2\theta$ , because $T_{f}= \frac{2 \pi }{2} = \pi$ .

To find a we consider the condition $f( \frac{ \pi }{4} )=4$ , from where $asin2* \frac{\pi}{4} =a*sin \frac{ \pi }{2} =a=4$ .

If the amplitude of $f(\theta)$ is twice the amplitude of $g(\theta)$ , then $g(\theta)$ has a product factor twice smaller than $f(\theta)$ and while period of $g(\theta)$ is 2π and g(0)=0, we can write $g(\theta)=2sin\theta$ .

8 0

3 years ago

PLSSS HELP WILL GIVE BRAINLIEST

klasskru [66]

I know a would be 40ft for Rachael and 50ft for Clare but I’m not completely sure about b

6 0

3 years ago

Factor: d2 + 16dm + 64m2

lutik1710 [3]

Answer:

$(d + 8m)^2$