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3 years ago

9

I 1:00 p.m. the water level in the pool is 13 inches. at 1:30 PM the water level is 18 inches. at 2:30 PM to water level is 28 i

nches. what is the constant rate of change?

2 answers:

Sonja [21]3 years ago

8 0

5 inches per 30 minutes
10 inches per 1 hour

LiRa [457]3 years ago

3 0

5 inches for 30 minutes and 10 inches every hour

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2 years ago

Read 2 more answers

Divide 800ml into ratio 3:7

Black_prince [1.1K]

Simply take the numerator (3) and denominator (7) and add them together.

3+7 = 10, which means that the ratio will provide parts that are in multiples of 10 depending on the total

so take 10x = 800

x = 80.

Then multiply this by the numerator and denominator

3*80/7*80 = 240/560.

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3 years ago

Find u, v , u , v , and d(u, v) for the given inner product defined on Rn. u = (0, −4), v = (5, 3), u, v = 3u1v1 + u2v2

tigry1 [53]

Answer:

$=-12$

$d(u,v)=2\sqrt{31}$

Step-by-step explanation:

We are given that inner product defined on $R_n$

$=3u_1v_1+u_2v_2$

u=(0,-4),v=(5,3)

We have to find the value of <u,v> and d(u,v)

We have $u_1=0,u_2=-4,v_1=5,v_2=3$

Substitute the value then we get

$=3(0\cdot5)+(-4)(3)=-12$

Now, $d(u,v)=\left \|v-u\right \|$

Using this formula we get

$d(u,v)=\left \| (5,7) \right \|=\sqrt{3(5)^2+(7)^2}=\sqrt{75+49}=\sqrt{124}$

$d(u,v)=2\sqrt{31}$

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3 years ago

Gravel is being dumped from a conveyor belt at a rate of 10 cubic feet per minute. It forms a pile in the shape of a right circu

Sunny_sXe [5.5K]

Answer:

0.0221 feet per minute.

Step-by-step explanation:

$\text{Volume of a cone}=\dfrac{1}{3}\pi r^2 h$

If the Base Diameter = Height of the Cone

The radius of the Cone = h/2

Therefore,

$\text{Volume of the cone}=\dfrac{\pi h}{3} (\dfrac{h}{2}) ^2 \\V=\dfrac{\pi h^3}{12}$

Rate of Change of the Volume, $\dfrac{dV}{dt}=\dfrac{3\pi h^2}{12}\dfrac{dh}{dt}$

Since gravel is being dumped from a conveyor belt at a rate of 10 cubic feet per minute. Therefore, the Volume of the cone is increasing at a rate of 10 cubic feet per minute.

$\dfrac{dV}{dt}=10$ ft^3/min$

We want to determine how fast is the height of the pile is increasing when the pile is 24 feet high.

We have:

$\dfrac{3\pi h^2}{12}\dfrac{dh}{dt}=10\\\\$When h=24$ feet$\\\dfrac{3\pi *24^2}{12}\dfrac{dh}{dt}=10\\144\pi \dfrac{dh}{dt}=10\\ \dfrac{dh}{dt}= \dfrac{10}{144\pi}\\ \dfrac{dh}{dt}=0.0221$ feet per minute$

When the pile is 24 feet high, the height of the pile is increasing at a rate of 0.0221 feet per minute.

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3 years ago

Alloy, c, weighing 500 ounces and having 35% aluminum, is obtained from melting together x ounces of alloy, a, containing 25% al

boyakko [2]

From the given scenario in this item, we can say that the sum of x and y is 500. By the aluminum balance, the equation that we can formulate from this,
0.25x + 0.40y = (500)(0.35)
The system of linear equations that can best express the given is,
x + y = 500
0.25x + 0.40y = 175

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3 years ago