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2 years ago

Seven less than the product of 8 and a number equals 4

Mathematics

1 answer:

Vlad1618 [11]2 years ago

3 0

Answer:

11/8 or 1 3/8 or 1.375

Step-by-step explanation:

From the information given by the question, we can form an equation:

Let us call the number 'n'

8 × n - 7 = 4

We then move the 7 to the other side, changing it from minus to plus:

8 × n = 4 + 7

So,

8 × n = 11

We divide both sides by 8, to leave n on one side of the equation.

n = 11/8

Therefore, n = 11/8 or 1 3/8 or 1.375

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Select the correct answer. which expression is equivalent to the given expression? assume the denominator does not equal zero. a

Alexxandr [17]

If the denominator does not equal 0, the equivalent expression of $\frac{14x^4y^6}{7x^8y^2}$ is $\frac{2y^{4}}{x^{4}}$ $\frac{x^{10} y^{14}}{729}$

<h3>How to determine the equivalent expression?</h3>

The expression is given as:

$\frac{14x^4y^6}{7x^8y^2}$

Divide 14 by 7

$\frac{14x^4y^6}{7x^8y^2} = \frac{2x^4y^6}{x^8y^2}$

Apply the law of indices

$\frac{14x^4y^6}{7x^8y^2} = \frac{2y^{6-2}}{x^{8-4}}$

Evaluate the differences in the exponents

$\frac{14x^4y^6}{7x^8y^2} = \frac{2y^{4}}{x^{4}}$

Hence, the equivalent expression of $\frac{14x^4y^6}{7x^8y^2}$ is $\frac{2y^{4}}{x^{4}}$

Read more about equivalent expressions at:

brainly.com/question/2972832

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2 years ago

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3 years ago

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3 years ago

A 75-gallon tank is filled with brine (water nearly saturated with salt; used as a preservative) holding 11 pounds of salt in so

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Let $A(t)$ = amount of salt (in pounds) in the tank at time $t$ (in minutes). Then $A(0) = 11$ .

Salt flows in at a rate

$\left(0.6\dfrac{\rm lb}{\rm gal}\right) \left(3\dfrac{\rm gal}{\rm min}\right) = \dfrac95 \dfrac{\rm lb}{\rm min}$

and flows out at a rate

$\left(\dfrac{A(t)\,\rm lb}{75\,\rm gal + \left(3\frac{\rm gal}{\rm min} - 3.25\frac{\rm gal}{\rm min}\right)t}\right) \left(3.25\dfrac{\rm gal}{\rm min}\right) = \dfrac{13A(t)}{300-t} \dfrac{\rm lb}{\rm min}$

where 4 quarts = 1 gallon so 13 quarts = 3.25 gallon.

Then the net rate of salt flow is given by the differential equation

$\dfrac{dA}{dt} = \dfrac95 - \dfrac{13A}{300-t}$

which I'll solve with the integrating factor method.

$\dfrac{dA}{dt} + \dfrac{13}{300-t} A = \dfrac95$

$-\dfrac1{(300-t)^{13}} \dfrac{dA}{dt} - \dfrac{13}{(300-t)^{14}} A = -\dfrac9{5(300-t)^{13}}$

$\dfrac d{dt} \left(-\dfrac1{(300-t)^{13}} A\right) = -\dfrac9{5(300-t)^{13}}$

Integrate both sides. By the fundamental theorem of calculus,

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac1{(300-t)^{13}} A\bigg|_{t=0} - \frac95 \int_0^t \frac{du}{(300-u)^{13}}$

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac{11}{300^{13}} - \frac95 \times \dfrac1{12} \left(\frac1{(300-t)^{12}} - \frac1{300^{12}}\right)$

$\displaystyle -\dfrac1{(300-t)^{13}} A = \dfrac{34}{300^{13}} - \frac3{20}\frac1{(300-t)^{12}}$

$\displaystyle A = \frac3{20} (300-t) - \dfrac{34}{300^{13}}(300-t)^{13}$

$\displaystyle A = 45 \left(1 - \frac t{300}\right) - 34 \left(1 - \frac t{300}\right)^{13}$

After 1 hour = 60 minutes, the tank will contain

$A(60) = 45 \left(1 - \dfrac {60}{300}\right) - 34 \left(1 - \dfrac {60}{300}\right)^{13} = 45\left(\dfrac45\right) - 34 \left(\dfrac45\right)^{13} \approx 34.131$

pounds of salt.

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2 years ago