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tiny-mole [99]
2 years ago
9

A square has side length 4 cm. What is its area in cm squared?

Mathematics
2 answers:
Ymorist [56]2 years ago
7 0

Answer:

16 cm squared (small 2 at the top of the number)

stellarik [79]2 years ago
6 0

Answer:

16

Step-by-step explanation:

4x4=16

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-9/3 + 4

-3 + 4 = 1

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Dr. Spike is a big fan of Bojangles and is particularly interested in the popularity of its celebrated wings dinner. Starting fr
Effectus [21]

Answer:

c

Step-by-step explanation:

If data is skewed then the upper and lower half have different amount of spread. Here there is no proof to say that data is not spread around mean or there are outliers. Therefore we can't say that the distribution of wings dinners sales in not normal.

6 0
3 years ago
PLEASE HELP ME ASAP I REALLY NEED HELP!!!
padilas [110]

1) \overline{UP}

2) \overline{UP} \cong \overline{UP}

3) \overline{UR}

4) \overline{PR}

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7 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

pounds of salt.

7 0
2 years ago
A 100-foot rope from the top of a tree house to the ground forms a 45∘ angle of elevation from the ground. How high is the top o
DochEvi [55]

Answer:

The height of tree house is 70.71 feet

Step-by-step explanation:

We are given that A 100-foot rope from the top of a tree house to the ground forms a 45∘ angle of elevation from the ground

Refer the attached figure

Length of rope AC = Hypotenuse =100 feet

The top of a tree house to the ground forms a 45∘ angle of elevation from the ground =\angle ACB = 45^{\circ}

We are supposed to find the height of tree house i.e.AB = Perpendicular

So, Using trigonometric ratio

Sin \theta = \frac{perpendicular}{Hypotenuse}\\Sin 45= \frac{AB}{AC}\\\frac{1}{\sqrt{2}}=\frac{AB}{100}\\100 \times \frac{1}{\sqrt{2}}=AB\\70.71=AB

Hence The height of tree house is 70.71 feet

8 0
3 years ago
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