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SSSSS [86.1K]
3 years ago
5

Factor the polynomial by it’s greatest common monomial factor. 9b^8 + 24b^3

Mathematics
2 answers:
docker41 [41]3 years ago
8 0

Answer:

9b^8 + 24b^3 = (3b^3) [ 3b^5  +  (8)]

greatest common monomial is  3*b^{3}

Step-by-step explanation:

we see  9b^8 + 24b^3

there is a common factor here.

3b^3  so

9b^8 + 24b^3 = (3b^3) * 3b^5  +  (3b^3) *(8)

9b^8 + 24b^3 = (3b^3) [ 3b^5  +  (8)]

so...

monitta3 years ago
8 0

Answer:

3b^3(3b^5+8)

Step-by-step explanation:

that is the khan academy answer

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Subtract these polynomials. (3x^22+2x+4)-(x^2+2x+1)
mixas84 [53]

Answer:

3x^22−x^2+3

Step-by-step explanation:

Let's simplify step-by-step.

3x^22+2x+4−(x2+2x+1)

Distribute the Negative Sign:

=3x^22+2x+4+−1(x2+2x+1)

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Combine Like Terms:

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Pauline invested $10,000 at two different
Dahasolnce [82]

Answer:

So for this, I will be doing a system of equations. Let x = money invested at 4% and y = money invested at 6%. We know that x and y will total up to 10k since that's how much he initially invested, so that'll be one of our equations.

Our other equation will represent the total balance he has after investing. We know that x's interest is 4% (since the money is increasing, this would translate to 1.04 in decimal form) and y's interest is 6% (aka 1.06 in decimal form since its increasing) and that the total balance is 10470 (initial balance + interest income).

Step-by-step explanation:

Using all this info, these are our two equations:

<em>X+Y=10000</em>

<em>1.04x+1.06=10470</em>

<em>Next, substitute x in the second equation for (10000 - y) and solve for y as such:</em>

<em>1.04(10000-y) +1.06y =10470</em>

<em>10400 - 1.04y + 1.60y = 10470</em>

<em>10400 + 0.02y = 10470</em>

<em>0.02y =70</em>

<em>y = 3500</em>

<em>Now that we have the value of y, substitute it into either equation to solve for x:</em>

<em>x + 3500 = 10000</em>

<em>x= 6500</em>

<em>1.04x + 1.06(3500)= 10470</em>

<em>1.04x = 6500</em>

<em>Answer</em>

<em>In short:</em>

<em>$6,500 was invested in the 4% rate</em>

<em>$3,500 was invested in the 6% rate</em>

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