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2 years ago

Brett is filling up his swimming pool. It is a cylindrical shape. How many

Mathematics

1 answer:

Otrada [13]2 years ago

8 0

Answer:

Step-by-step explanation:

height h=1.2 m

radius r= D/2=4.5=2.25 m

Volume V=πr²h=π(2.25)²×1.2 ≈3.18m³

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Bernice bought 3/1/2 pound of apples and 5/2/3 pounds of oranges and 11/12 pounds of bananas what was the total weight of fruit

torisob [31]

10 1/12

3 1/2 + 5 2/3 + 11/12 = 10 1/12

4 0

2 years ago

Read 2 more answers

What is the triangles' perimeter

DENIUS [597]

To find the perimeter of any shape, you find the sum of their side lengths.

In this case, this triangle has side lengths of 1.197, 0.764, and 1.74.
Add these, and the sum is your answer.

1.197 + 0.764 + 1.74 = 3.701

Therefore 3.701 is the correct answer :)

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2 years ago

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Which choice is the equation of the line that passes through the point (5, 19) and has a slope of

Liono4ka [1.6K]

Answer:

The equation is;

y = 12x-41

Step-by-step explanation:

Mathematically;

slope intercept form is;

y = mx + b

where m is slope and b is intercept

Since we already have the slope , the equation is;

y = 12x + b

To get b which is the y intercept, we simply substitute the value of the points given

Thus;

19 = 5(12) + b

19 = 60+ b

b = 19-60

b = -41

So the equation is;

y = 12x - 41

3 0

2 years ago

Find derivative problem<br> Find B’(6)

dalvyx [7]

Answer:

$B^\prime(6) \approx -28.17$

Step-by-step explanation:

We have:

$\displaystyle B(t)=24.6\sin(\frac{\pi t}{10})(8-t)$

And we want to find B’(6).

So, we will need to find B(t) first. To do so, we will take the derivative of both sides with respect to x. Hence:

$\displaystyle B^\prime(t)=\frac{d}{dt}[24.6\sin(\frac{\pi t}{10})(8-t)]$

We can move the constant outside:

$\displaystyle B^\prime(t)=24.6\frac{d}{dt}[\sin(\frac{\pi t}{10})(8-t)]$

Now, we will utilize the product rule. The product rule is:

$(uv)^\prime=u^\prime v+u v^\prime$

We will let:

$\displaystyle u=\sin(\frac{\pi t}{10})\text{ and } \\ \\ v=8-t$

Then:

$\displaystyle u^\prime=\frac{\pi}{10}\cos(\frac{\pi t}{10})\text{ and } \\ \\ v^\prime= -1$

(The derivative of u was determined using the chain rule.)

Then it follows that:

$\displaystyle \begin{aligned} B^\prime(t)&=24.6\frac{d}{dt}[\sin(\frac{\pi t}{10})(8-t)] \\ \\ &=24.6[(\frac{\pi}{10}\cos(\frac{\pi t}{10}))(8-t) - \sin(\frac{\pi t}{10})] \end{aligned}$