All categories

3 years ago

Four more than seven times a number is 12.” Write an equation that represents the statement. Then solve.

Mathematics

1 answer:

Leona [35]3 years ago

3 0

7x + 4 = 12

7x = 12-4

7x = 8

x = 7/8

You might be interested in

Consider the enlargement of the rectangle. Use the proportion to find the missing dimension bod the original rectangle. (Sorry t

fgiga [73]

Here's the right way:

$\dfrac{x}{\frac 3 5} = \dfrac{20}{12}$

$x = \dfrac{20}{12} \cdot \dfrac{3}{5} = 1$

Here's the way they want you to do it:

$\dfrac{x}{\frac 3 5} = \dfrac{20}{12}$

$12 x = 20 \times \dfrac{3}{5} = 12 \textrm{ FIRST BOX}$

$x = 12/12 = 1 \textrm{ SECOND BOX}$

6 0

3 years ago

Read 2 more answers

Find the output, y, when the input x, is 30.<br> y = 14 – 0.5x<br> y =

Mumz [18]

Answer:

y=-1

Step-by-step explanation:

14-(.5*30) same as dividing 30 by 2

14-15=-1 combine

6 0

3 years ago

An ordered pair is

yarga [219]

Answer:

Step-by-step explanation:

The m to the y can’t equal to m so it’s b

4 0

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

Debora [2.8K]

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.

7 0

2 years ago

Avery and Caden have saved $27,000 towards a down payment on a house. They want to keep some of the money in a bank account that

Thepotemich [5.8K]

Answer:

$6750 in the bank account and $20,250 in the stock fund

Step-by-step explanation:

If B is the money they put in the bank and S is the amount they put in the stock fund, then:

B + S = 27000

1.024 B + 1.072 S = 1.06 × 27000

Solving the system of equations:

1.024 (27000 − S) + 1.072 S = 28620

27648 − 1.024 S + 1.072 S = 28620

0.048 S = 972

S = 20250

B = 27000 − S

B = 6750

They should put $6750 in the bank account and $20,250 in the stock fund.

6 0

3 years ago