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2 years ago

Which formula is correct in computing 50% of the volume of the shape below?

Mathematics

1 answer:

elixir [45]2 years ago

8 0

Answer:

Step-by-step explanation:

volume of a cylinder: V=πr^2h

so for 1/2 volume we can just multiply by 1/2

V = 1/2 (πr^2h)

plug in values r = 4.6 and h = 12.4

V= 1/2 π (4.6)^2 (12.4)

so the answer is D

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WILL GIVE BRAINLEST!!!

dusya [7]

Answer:

18 12

6 22

Step-by-step explanation:

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8 0

2 years ago

1 4/8 in a improper fraction?

stiv31 [10]

<h2>Answer: $\frac{3}{2}$ <u><em></em></u></h2>

Step-by-step explanation:

Step 1: Multiply the whole number part (1) by the denominator (8).

1 × 8 = 8

Step 2: Add the product from Step 1 (8) to the numerator (4).

8 + 4 = 12

Step 3: Write that result (12) above the denominator. So,

$\frac4{8}{1}$

Step 4: The fraction $\frac{12}{8}$

Can be reduced by dividing both numerator and denominator by the GCD(12,8) = 4. Thus,

$\frac1{4}{8} =\frac{3}{2}$

8 0

2 years ago

Write a fraction for the portion of the grid that is NOT shaded.

Aneli [31]

because there are a total of 15 grid and the question asked to pick the grid that doesn't shaded

5 0

2 years ago

Read 2 more answers

A juggler tosses a ball into the air . The balls height, h and time t seconds can be represented by the equation h(t)= -16t^2+40

malfutka [58]

PART A

The given equation is

$h(t) = - 16 {t}^{2} + 40t + 4$

In order to find the maximum height, we write the function in the vertex form.

We factor -16 out of the first two terms to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t) + 4$

We add and subtract

$- 16(- \frac{5}{4} )^{2}$

to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t) + - 16( - \frac{5}{4})^{2} - -16( - \frac{5}{4})^{2} + 4$

We again factor -16 out of the first two terms to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t + ( - \frac{5}{4})^{2} ) - -16( - \frac{5}{4})^{2} + 4$

This implies that,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t + ( - \frac{5}{4}) ^{2} ) + 16( \frac{25}{16}) + 4$

The quadratic trinomial above is a perfect square.

$h(t) = - 16 ( t- \frac{5}{4}) ^{2} +25+ 4$

This finally simplifies to,

$h(t) = - 16 ( t- \frac{5}{4}) ^{2} +29$

The vertex of this function is

$V( \frac{5}{4} ,29)$

The y-value of the vertex is the maximum value.

Therefore the maximum value is,

$29$

PART B

When the ball hits the ground,

$h(t) = 0$

This implies that,

$- 16 ( t- \frac{5}{4}) ^{2} +29 = 0$

We add -29 to both sides to get,

$- 16 ( t- \frac{5}{4}) ^{2} = - 29$

This implies that,

$( t- \frac{5}{4}) ^{2} = \frac{29}{16}$

$t- \frac{5}{4} = \pm \sqrt{ \frac{29}{16} }$

$t = \frac{5}{4} \pm \frac{ \sqrt{29} }{4}$

$t = \frac{ 5 + \sqrt{29} }{4} = 2.60$

or

$t = \frac{ 5 - \sqrt{29} }{4} = - 0.10$

Since time cannot be negative, we discard the negative value and pick,

$t = 2.60s$

8 0

2 years ago

How many square corners does a pentagon have

SIZIF [17.4K]

I assume that by square corners you mean 90 degree angles. if this is the case then there are 2 in a pentagon. Hope I helped :)

3 0

2 years ago