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3 years ago

I need help asap!!! pleaseee

Mathematics

1 answer:

givi [52]3 years ago

6 0

Answer: 625ft sq.

Explanation
Answer using A=bh (area=base x height)
Plug in information:
A=25x25
Solve:
A=625 ft sq.

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Ava, Grace, Jayden, and Zoe together stacked a total of 50 brownies on a tray. Ava stacked 0.14 of the brownies, Grace stacked 4

OLEGan [10]

1. Zoe stacked the most brownies. She stacked 27 brownies out of the 50 brownies.

2. 15

3. 43 + 3n ≥ 70, so n ≥ 9

4. D. 58x + 13 + 58y

5. The answer would be ≤4

6. 9.94 + 10.01p ≤ 70, so p ≤ 6

7. C 40 89

8. D. x+(x+5)=77

Hope I helped you! Please consider marking my answer the brainliest :)

4 0

3 years ago

Need help ASAP!!!!!!!!

egoroff_w [7]

Answer:

24/35

Step-by-step explanation:

8 0

3 years ago

Read the following and answer the questions that follow.

user100 [1]

Answer:

Step-by-step explanation:

(7/12) of 840 attended the concert:

(7/12)*(840) = 490 students attended the Spring Concert.

(840 - 490) = 350 did not attend

(350/840) is the fraction of students who did not attend. This can be reduced to (35/84)

(350/840) = 0.4167 or 41.67% did not attend.

3 0

2 years ago

2/7 divided by 7/6 brainlyest on the line

raketka [301]

12/49! hope this helps!

6 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