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maw [93]
2 years ago
6

I need help please and thank you !

Mathematics
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

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Solve the inequality.<br> 5(w - 3) &lt; 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
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