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3 years ago

Show twelve in four different ways.use words, pictures and numbers

1 answer:

5 0

You can draw 12 lines
Draw a dime and two pennies
Draw a circle and cut it into 12 different pieces
Write out the number 12 Twelve
Write 12

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write an equation for the line that passes through the point (8,3) and parallel to -4x+8y=23. Use slope-intercept form.

kvasek [131]

(1/2)x+b=y because the equation that given the slope is 1/2 if you leave y alone and take the x’s number

put the point now , 4+b=3 so b is -1

the equation is (1/2)x-1=y

6 0

3 years ago

3 = –(–y + 6)<br> a. –9<br> b. –3<br> c. 3<br> d. 9

Vlad1618 [11]

3 = -(-y + 6)

First, simplify brackets. / Your problem should look like: 3 = y - 6
Second, add 6 to both sides. / Your problem should look like: 3 + 6 = y
Third, simplify 3 + 6 to 9. / Your problem should look like: 9 = y
Fourth, switch sides. / Your problem should look like: y = 9

The answer is D) 9.

7 0

3 years ago

Point F is located at (3, -2) on the grid. Which coordinate pair is located 5 units from Point F?

postnew [5]

(3,-8)
You move down from those coordinates
X stays the same while it just moves down only Y changes

3 0

2 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

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