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3 years ago

What is the exact volume of the cylinder?

Mathematics

2 answers:

Cloud [144]3 years ago

3 0

The Answer is A. 5070π in3

den301095 [7]3 years ago

3 0

The answer is A. 5070 $\pi$ ³.

The volume of a cylinder can be calculated by using the following formula:

V= $\pi$ r²h

V= $\pi$ ·13² in·30 in

V= $\pi$ ·169 in·30 in

V=5070 $\pi$ in³

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one rectangle measures 2 units by 7 units. a second rectangle measures 11 units by 37 units. are these two figures scaled versio

Amanda [17]

Answer: The both triangle are not scale version to each other.

Explanation:

It is given that the one rectangle measures 2 units by 7 units. a second rectangle measures 11 units by 37 units.

The two shapes are scaled version of each other if their corresponding sides are in same proportion. In other words the if two shapes scaled version of each then the shapes are similar.

The sides or first rectangle are 2 and 7. The measures of second rectangle are 11 and 37.

The proportion of small side is,

$\frac{2}{11}$

The proportion of large side is,

$\frac{7}{37}$

So,

$\frac{2}{11}\neq \frac{7}{37}$

Since the proportion are not equal, therefore both triangle are not scale version to each other.

6 0

3 years ago

List the LCM of the numbers. (MCC.6.NS.4)<br> 3, 11, 12

Alekssandra [29.7K]

Answer:

ans=132

Step-by-step explanation:

7 0

3 years ago

A tank contains 30 lb of salt dissolved in 300 gallons of water. a brine solution is pumped into the tank at a rate of 3 gal/min

sesenic [268]

$A'(t)=(\text{flow rate in})(\text{inflow concentration})-(\text{flow rate out})(\text{outflow concentration})$
$\implies A'(t)=\dfrac{3\text{ gal}}{1\text{ min}}\cdot\left(2+\sin\dfrac t4\right)\dfrac{\text{lb}}{\text{gal}}-\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{300+(3-3)t\text{ gal}}$
$A'(t)+\dfrac1{100}A(t)=6+3\sin\dfrac t4$

We're given that $A(0)=30$ . Multiply both sides by the integrating factor $e^{t/100}$ , then

$e^{t/100}A'(t)+\dfrac1{100}e^{t/100}A(t)=6e^{t/100}+3e^{t/100}\sin\dfrac t4$
$\left(e^{t/100}A(t)\right)'=6e^{t/100}+3e^{t/100}\sin\dfrac t4$
$e^{t/100}A(t)=600e^{t/100}-\dfrac{150}{313}e^{t/100}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+C$
$A(t)=600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+Ce^{-t/100}$

Given that $A(0)=30$ , we have

$30=600-\dfrac{150}{313}\cdot25+C\implies C=-\dfrac{174660}{313}\approx-558.02$

so the amount of salt in the tank at time $t$ is

$A(t)\approx600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)-558.02e^{-t/100}$