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3 years ago

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You have a 4 x 6 photo of you and your friend. a. You order a 5 x 7 print of the photo. Is the new photo similar to the original

? b. You enlarge the original photo to three times its size on your computer. Is the new photo similar to the original?

1 answer:

STALIN [3.7K]3 years ago

6 0

Answer:

NOPE, NOT EQUAL!!

Step-by-step explanation:

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X+(-3)=-7 Solve the equation and enter the value of x below

Phoenix [80]

Answer:

x=-4

Step-by-step explanation:

x+(-3)=-7

x-3=-7

x=-7+3=3-7

x=-4

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3 years ago

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On her way to visit her parents,jennifer drives 265 miles in 5 hours. What is her average rate of speed in miles per hour?

Nutka1998 [239]

Her total distance (so 265 miles) over the number of hours she drives will give miles per hour.
265miles/5hours=53 miles/hour

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3 years ago

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How do you write 4.40 × 10^3 in standard form?

Sindrei [870]

Answer:

4400 is the answer

Step-by-step explanation: You move the decimal point to the right 3 spaces. If it was 10^-3 you would move the decimal point to the left 3 spaces.

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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

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

Need help !!!!!!!!!!!!!!!!!!!

Tems11 [23]

Rectangle. it’s a 2d 4 sided shape with 2 sets of parallel sides

4 0

2 years ago