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4 years ago

Which is a perfect square? 72 81 90 99

Mathematics

2 answers:

Tpy6a [65]4 years ago

8 0

81 is the perfect square because 9 * 9

GREYUIT [131]4 years ago

5 0

81 is the answer beacuse if you times 9*9 you get 81 aka 81 is your square hope this helped:)

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Help please I don’t get this

Bess [88]

Answer:

y = -2x - 1

Step-by-step explanation:

Formula is y = mx + b

m = slope

slope = change in y/change in x = (-1 - 3) / (0 - -2) = -4/2 = -2

So equation is now y = -2x + b

To find b, substitute any value from the table into the equation. Let's use (-2,3)

y = -2x + b

3 = -2(-2) + b

3 = 4 + b

-1 = b

So equation is y = -2x - 1

3 0

4 years ago

Read 2 more answers

On a cross country trip the Anderson family plan to average 500 miles in 10 hours of driving each day on average how many miles

PolarNik [594]

Answer:

50 miles per hour

Step-by-step explanation:

500 miles in 10 hours

= 500/10

= 50 miles per hour

7 0

3 years ago

Manny has a garden that is 12 feet long and 9 feet wide. What is the perimeter of

Julli [10]

Answer:

42 ft

Step-by-step explanation:

2(L + W ) = P

2 (12 + 9) = P

2 (21) = P

42 = P

P = 42

4 0

2 years ago

Please answer this question! 40 points and brainliest!

algol [13]

Answer:

2x^2+2x

Step-by-step explanation:

The large squares are x^2.

The rectangles are x

The small squares are 1

We have 2 large squares = 2x^2

and 2 rectangles = 2x

2x^2+2x

8 0

4 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