Ask question

Ask question

All categories

2 years ago

6

I need help please and thank you !

1 answer:

Vera_Pavlovna [14]2 years ago

8 0

Answer:

π(12x^3y^5z)^2

Step-by-step explanation:

I doubt your teacher would want you to evaluate the whole thing but if they do leave a comment

You might be interested in

Solve the inequality.<br> 5(w - 3) < 5w + 20

saveliy_v [14]

Answer:

-15 < 20

Step-by-step explanation:

5(w - 3) < 5w + 20

Distribute;

5w - 15 < 5w + 20

Subtract 5w from both sides;

-15 < 20

7 0

2 years ago

Read 2 more answers

I just want to see if my answer is correct, I haven't done volume in a long time!

motikmotik

You did it right because 12 x 2 = 24 x 8 = 192

7 0

3 years ago

Read 2 more answers

What is the true solution to 2In e sqrt in 5x = 2 In 15?

swat32

Answer:

B.

Step-by-step explanation:

I have attached the work to your problem.

Please see the attachment below.

I hope this helps!

3 0

3 years ago

Identify, whether the given polynomial is a monomial, binomial, or a trinomial.

Finger [1]

That's a monomial; it has one variable, one coefficient, and one degree.

8 0

3 years ago

The expression (secx + tanx)2 is the same as _____.

trapecia [35]

<u>Answer:</u>

The expression $\bold{(\sec x+\tan x)^{2} \text { is same as } \frac{1+\sin x}{1-\sin x}}$

<u>Solution:</u>

From question, given that $\bold{(\sec x+\tan x)^{2}}$

By using the trigonometric identity $(a + b)^{2} = a^{2} + 2ab + b^{2}$ the above equation becomes,

$(\sec x+\tan x)^{2} = \sec ^{2} x+2 \sec x \tan x+\tan ^{2} x$

We know that $\sec x=\frac{1}{\cos x} ; \tan x=\frac{\sin x}{\cos x}$

$(\sec x+\tan x)^{2}=\frac{1}{\cos ^{2} x}+2 \frac{1}{\cos x} \frac{\sin x}{\cos x}+\frac{\sin ^{2} x}{\cos ^{2} x}$

$=\frac{1}{\cos ^{2} x}+\frac{2 \sin x}{\cos ^{2} x}+\frac{\sin ^{2} x}{\cos ^{2} x}$

On simplication we get

$=\frac{1+2 \sin x+\sin ^{2} x}{\cos ^{2} x}$

By using the trigonometric identity $\cos ^{2} x=1-\sin ^{2} x$ ,the above equation becomes

$=\frac{1+2 \sin x+\sin ^{2} x}{1-\sin ^{2} x}$

By using the trigonometric identity $(a+b)^{2}=a^{2}+2ab+b^{2}$

we get $1+2 \sin x+\sin ^{2} x=(1+\sin x)^{2}$

$=\frac{(1+\sin x)^{2}}{1-\sin ^{2} x}$

$=\frac{(1+\sin x)(1+\sin x)}{1-\sin ^{2} x}$

By using the trigonometric identity $a^{2}-b^{2}=(a+b)(a-b)$ we get $1-\sin ^{2} x=(1+\sin x)(1-\sin x)$

$=\frac{(1+\sin x)(1+\sin x)}{(1+\sin x)(1-\sin x)}$

$= \frac{1+\sin x}{1-\sin x}$

Hence the expression $\bold{(\sec x+\tan x)^{2} \text { is same as } \frac{1+\sin x}{1-\sin x}}$

8 0

3 years ago