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2 years ago

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Mathematics

1 answer:

andreyandreev [35.5K]2 years ago

3 0

$2\cdot 10^{3x}=26\\10^{3x}=13\\\log 10^{3x}= \log 13\\3x\log 10= \log 13\\3x= \log 13\\\\$

$x=\frac{\log 13}{3}$

lmk if you have questions

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Read 2 more answers

What is the solution set of x^2+y^2=26 and x-y=6?

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3 years ago

The Henley's took out a loan for $195,000 to purchase a home. At a 4.3% interest rate

ella [17]

Interest paid after 30 years is $494,546.99.

Solution:

Principal (P) = $195,000

Interest rate (r) = 4.3%

Time (t) = 30 years

n = number of times interest calculated per year

n = 1

Compound interest formula:

$$A=P\left(1+\frac{r}{n}\right)^{n t}$

where A is the final amount

$$A=195000\left(1+\frac{4.3\%}{1}\right)^{1\times 30}$

$$A=195000\left(1+\frac{4.3}{100}\right)^{30}$

$$A=195000\left(\frac{100+4.3}{100}\right)^{30}$

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A = 689546.99

Interest = Amount - Principal

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Interest paid after 30 years is $494,546.99.

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3 years ago

H(x)=1/8x^3-x^2<br><br> Over which interval does h have a positive average rate of change?

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Answer:

$(-\infty,0)\cup(\frac{16}{3},\infty)\\That \ is x\frac{16}{3}$

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$h(x)=\frac{1}{8}x^{3} -x^{2} \\\\Differentiate \ h(x) \ with \ respect \ to \ x\\h'(x)=\frac{3}{8}x^{2} -2x$

For positive rate of change $h'(x)>0$

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6 0

3 years ago