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nata0808 [166]
3 years ago
7

Must the product of three polynomials again be a polynomial?

Mathematics
1 answer:
melomori [17]3 years ago
5 0

Answer:

Yes

Step-by-step explanation:

The product of three polynomials will be a polynomial regardless of the signs of the leading coefficients of the polynomials.

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Name the four basic steps to take when solving a word problem.
Grace [21]

answer

Step-by-step explanation:

3 0
2 years ago
2. Write the expression 23/5 as a radical.<br><br> Need help!!!!!!!!!!!!
amm1812

Answer:

OPTION D

2^³/5

= 5√2^3

Option D is legit.

4 0
3 years ago
What is the relation between the trigonometric ratios tan(R) and tan(J)? Please answer ASAP!
Ilya [14]

Step-by-step explanation:

∆PQR similar to ∆ JKL

JK/KL= PQ/QR

6/12=3/QR

QR=3×2= 6

therefore, tan(R) = PQ/QR = 3/6=1/2

and, tan(J) = KL/JK =12/6=2

\frac{ \tan(R) }{ \tan(J) }  =  \frac{ \frac{1}{2} }{2}

\frac{ \tan(R) }{ \tan(J) }   =  \frac{1}{4}

tan(J)=4.tan(R)

option D

6 0
3 years ago
The value of X is:<br> A 11.3<br> B 8<br> C 16
diamong [38]

Answer:

x ≈ 11.3

Step-by-step explanation:

Using the sine ratio in the right triangle

sin45° = \frac{opposite}{hypotenuse} = \frac{8}{x} ( multiply both sides by x )

x × sin45° = 8 ( divide both sides by sin45° )

x = \frac{8}{sin45} ≈ 11.3

6 0
3 years ago
I just need a little help on this ??! Please.
lawyer [7]

1. Since it's m - 7 you would have 7 to both sides so you would in fact have m < 13. If you double check your answer, you see that if m is say 12 (because 12 is obviously less than 13), 12 - 7 < 6

2. Again, use the same process on this problem as the first one. Add 8 to each sides because it's you're subtracting 8 from n. So you end up with n > 13. Check your answer. Say n is 14. 14 - 8 > 5

3. This one is different because you are adding 5 to p. So in order to get p by itself, you need to subtract 5 from both sides. p < 5. Say p is 4, 4 + 5 < 10.

When working with problems like these, you need to isolate the variable on one side and get it by itself.

5 0
3 years ago
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