The product of 8, and a number increased by 7. is 104. What is the number?

Mathematics

1 answer:

Lynna [10]3 years ago

4 0

Answer:
12.125
Step-by-step
8x+7=104
-7 -7
8x=97
/8 /8
x=12.125

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Show all work below to answer the following for a certain city that had a population of 150,000 in the year 2010. In 2020, the p

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The constant rate of continuous growth, k, for this population is equal to 2.11935%. And the population will reach 250,000 people in 24.36 years.

For solving this question, you should apply the Population Growth Equation.

<h3>Population Growth Equation</h3>

The formula for the Population Growth Equation is:

$P_f=P_o*(1+\frac{R}{100} )^t$

Pf= future population

Po=initial population

r=growth rate

t= time (years)

STEP 1 - Find the constant rate of continuous growth, k, for this population.

For this exercise, you have:

Pf= future population= 185,000 in 2020.

Po=initial population =150,000 in 2010.

r=growth rate= ?

t= time (years)=2020-2010=10

Then,

$P_f=P_o*(1+\frac{R}{100} )^t\\ \\ 185000=150000\cdot \left(1+\frac{R}{100}\:\right)^{10}\\ \\ \left(1+\frac{R}{100}\right)^{10}=\frac{185000}{150000} \\ \\ \left(1+\frac{R}{100}\right)^{10}=\frac{37}{30}\\ \\ R=100\sqrt[10]{\frac{37}{30}}-100=2.11935\%$

STEP 2 - Find the <em>t</em> for population 250,000 people.

$P_f=P_o*(1+\frac{R}{100} )^t\\ \\ 250000=150000\cdot \left(1+\frac{2.11935}{100}\:\right)^{10}\\ \\ \left(1+\frac{2.11935}{100}\right)^{10}=\frac{250000}{150000} \\ \\ \left(1+\frac{2.11935}{100}\right)^t=\frac{5}{3}\\ \\ t\ln \left(1+\frac{2.11935}{100}\right)=\ln \left(\frac{5}{3}\right)\\ \\ t=\frac{\ln \left(\frac{5}{3}\right)}{\ln \left(\frac{102.11935}{100}\right)}\\ \\ t=24.36$