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evablogger [386]
3 years ago
14

Please help! What is the log of 10,000,000,000.?

Mathematics
2 answers:
NARA [144]3 years ago
8 0

Answer:

10

Step-by-step explanation:

the Log of 10 000 000 000 is 10

Illusion [34]3 years ago
4 0

Answer:

the log of 10,000,000,000 is 10

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Can some one plz help I don't know how to do this
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This is a hard one so what you do is make a number line and put -5x, -9, and -24 hope this helps
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Solve each equation for the ind<br>2. 2(x-7)=10​
SVETLANKA909090 [29]

Answer:

x= 12

Step-by-step explanation:

2(x-7) = 10

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3 years ago
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Please Help !!Mr. Mudd gives each of his children $2000 to invest as part of a friendly family competition. The competition will
VikaD [51]

Answer:

Albert = $2159.07; Marie = $2244.99; Hans = $2188.35; Max = $2147.40

Marie is $10 000 richer

Step-by-step explanation:

Albert

(a) $1000 at 1.2 % compounded monthly

A = P\left(1 + \dfrac{r}{n}\right)^{nt}

A = 1000(1 + 0.001)¹²⁰ = $1127.43

(b) $500 losing 2%

0.98 × 500 = $490

(c) $500 compounded continuously at 0.8%

\begin{array}{rcl}A & = & Pe^{rt}\\& = & 500e^{0.008 \times 10}\\& = &\mathbf{\$541.64}\\\end{array}\\

(d) Balance

Total = 1127.43 + 490.00+ 541.64 = $2159.07

Marie

(a) 1500 at 1.4 % compounded quarterly

A = 1500(1 + 0.0035)⁴⁰ = $1724.99

(b) $500 gaining 4 %

1.04 × 500 = $520.00

(c) Balance

Total = 1724.99 + 520.00 = $2244.99

Hans

$2000 compounded continuously at 0.9 %

\begin{array}{rcl}A& = &2000e^{0.009 \times 10}\\& = &\mathbf{\$2188.35}\\\end{array}\\

Max

(a) $1000 decreasing exponentially at 0.5 % annually

A = 1000(1 - 0.005)¹⁰= $951.11

(b) $1000 at 1.8 % compounded biannually

A = 1000(1 + 0.009)²⁰ = $1196.29

(c) Balance

Total = 951.11 + 1196.29 = $2147.40

Marie is $ 10 000 richer at the end of the competition.

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