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3 years ago

8

Simplify (3/4ab^7)^3

2 answers:

dybincka [34]3 years ago

7 0

See picture for a reply to your question.

stira [4]3 years ago

3 0

Answer:

33a3b21

———————

26

Step-by-step explanation:

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What are all the real zeros of y=(x-12)^3-7?

atroni [7]

Answer:

$\sqrt[3]{7} +12$

Step-by-step explanation:

Use symbolab

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3 years ago

I'm supposed to find the error in this problem

igor_vitrenko [27]

Answer:

The height is 12 cm instead of 13 cm

Step-by-step explanation:

use 12 Cm to take out answer

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3 years ago

To all you misguided people out there:

Alla [95]

Answer:

See below.

Step-by-step explanation:

I think you're supposed to solve the problem explaining all steps, and then copy the answer to the top.

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3 years ago

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Within the set of whole numbers , one set of numbers satisfies the recursive formula an=(an-1)^2-an-1, when a4=870 what are the

andrew11 [14]

Answer: 3 and 6 will be the first and second term of the given series respectively.

Explanation: since it is given- $a_n=(a_{n-1} )^2-a_{n-1}$ and $a_4=870$

so, by replacing n by 4, we will get- $a_4=a_3^2-a_3$ ⇒870= $a_3^2-a_3$ ⇒ $a_3^2-a_3-870=0$

after solving this equation we will get, $a_3= -29$ and 30.

If $a_3= -29$ , but it is not a whole number.

Thus, $a_3=30$ will be the value of third term of the given series.

Again, by applying the given condition $a_n=(a_{n-1} )^2-a_{n-1}$ , $a_2= 6$ or -5 (not possible), so, $a_2=6$ will be the second term.

Similarly, $a_1=3$ or -2 but we will go to the whole number.

Thus, $a_1=3$ will be the first term.

8 0

4 years ago

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What is the complete factorization of the polynomial below?

aksik [14]

Option A: $(x+1)(x+3 i)(x-3 i)$ is the complete factorization of the polynomial $x^{3} +x^{2} +9x+9$

Explanation:

The polynomial is $x^{3} +x^{2} +9x+9$

Now, we shall find the complete factorization of the polynomial.

Let us group the common terms, we have,

$\left(x^{3}+x^{2}\right)+(9 x+9)$

Taking out the common terms,

$x^{2} (x+1)+9(x+1)$

Factor out $x+1$ from both the terms, we have,

$\left(x^{2}+9\right)(x+1)=0$

The term $\left(x^{2}+9)\right.$ can be factored as

$\begin{aligned}x^{2}+9 &=0 \\x^{2} &=-9 \\x &=\pm 3 i\end{aligned}$ and $\begin{aligned}x+1 &=0 \\x &=-1\end{aligned}$

Thus, the roots are $x=\pm 3 i$ and $x=-1$

These roots can be written as $(x+1)(x+3 i)(x-3 i)$

Thus, the complete factorization of the polynomial is $(x+1)(x+3 i)(x-3 i)$

8 0

3 years ago

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