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ivolga24 [154]
3 years ago
10

A=2b^3-1 what's the value of b if a is 13

Mathematics
2 answers:
skelet666 [1.2K]3 years ago
5 0
\sf~A=2b^3-1

Plug in A = 13

\sf~13=2b^3-1

Add 1 to both sides:

\sf~14=2b^3

Divide 2 to both sides:

\sf~7=b^3

Find the cube root of both sides:

\sf~b=\sqrt[3]{7}

\sf~b\approx~1.91
vlabodo [156]3 years ago
3 0
A = 2b³ - 1
13 = 2b³ - 1
<u>+ 1         + 1</u>
<u>14</u> = <u>2b³
</u> 2       2<u>
</u> 7 = b³
∛(7) = b
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Diego has two jobs. He works 3 hours at the library each day, making $8 per hour. In addition, he works at a coffee shop, making
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Answer:

  • D.  l = 3  ||  4 < c < y   ||  8l + 7.5c ≥ 70

Step-by-step explanation:

<h3>Work at library</h3>
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  • Payment = $8 per hour
  • Total payment = 8l
<h3>Work at coffee shop</h3>
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Target income = at least $70 daily  

<u>So the system as per above parameters:</u>

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3 years ago
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tia_tia [17]

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2 years ago
How many zeros should have 5 to have 72 dividers?
xenn [34]
Hello,

nice as problem.

| If\ a\ number\ x=a_1^{p_1}*a_2^{p_2}*a_3^{p_3}*....*a_n^{p_n}\\ the\ number\ of\ his\ dividers is:\\{p_1+1}*{p_2+1}*{p_3+1}*....*{p_n+1}
Here x has a power of 5 as divider and a power of 2 as divider (in order to make a power of 10.

5=5^1*2^0

50=5^1*(5^1*2^1)=5^2*2^1

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50 000 000=5^8*2^7

Numbers of dividers: (8+1)*(7+1)=9*8=72








3 0
3 years ago
4(x-a)+3b=5x+a solve for x
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4x - 4a + 3b = 5x + a

-4a + 3b= 5x + a - 4x

-4a + 3b - a = x

-5a + 3b = x

3 0
3 years ago
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= L 86.8614 = L 87

4 0
3 years ago
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