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3 years ago

Need help please stressing out over this problem!?!?!

Mathematics

1 answer:

ludmilkaskok [199]3 years ago

6 0

$v = \frac{4}{3} \times \pi \times {r}^{3}$

Finding the radius,

$d = 2r, \: r = \frac{d}{2} = \frac{8}{2} = 4inches$

Plugging in our values,

$v = \frac{4}{3} \times 3.14 \times {4}^{3} = \frac{4}{3} \times 3.14 \times 64 = 267.95 {inches}^{3} \\$

Rounding to the nearest whol number,

$v = 268 {inches}^{3}$

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When running a 100 meter race Bill reaches his maximum speed when he is 45 meters from the starting line and 5 seconds have elap

Mkey [24]

Bill's speed is constant after 5 seconds and 45 meters into the race.

Correct responses:

a. Bill's maximum speed is 8 m/s

b. 24 meters

c. 57 meters

<h3>Methods used to calculate speed and distance traveled</h3>

Given parameters are;

The distance of the race = 100 meter

Distance at which Bill reaches maximum speed = 45 meters

Speed Bill maintains after 5 seconds = The maximum speed

Time at which Bill is 85 meters from the starting line = 10 seconds after start

a. Required;

Bill's maximum speed in meters.

Solution:

Distance Bill runs at maximum speed, d = 85 m - 45 m = 40 m

Time at which Bill runs the 40 m at maximum speed, t = 10 s - 5 s = 5 s

$Speed = \mathbf{\dfrac{Distance}{Time}}$

Therefore;

$Bill's \ maximum \ speed = \dfrac{40 \ m}{5 \ s} = \underline{8 \ m/s}$

b. i. Required:

The distance Bill will run for 3 seconds at the maximum speed;

Solution:

Distance, s = Speed, v × Time, t

The distance traveled at maximum speed in 3 seconds is therefore;

Distance = 8 m/s × 3 s = 24 m

The distance Bill travels in 3 seconds at the maximum speed is 24 meters.

ii) The distance Bill travels at 8 seconds after start, is given as follows;

Distance traveled in the first 5 seconds = 45 meters

Distance traveled in the next 3 seconds = 24 meters

Therefore;

Bill's distance from the starting line 8 seconds after start is 45 meters + 25 meters = 69 meters

c. Bill's distance 6.5 seconds after the start of the race is given as follows;

Distance traveled in the first 5 seconds = 45 meters

Distance traveled in the next 1.5 seconds = 1.5 s × 8 m/s = 12 meters

Bill's distance from the starting line 6.5 seconds after start of the race = 45 meters + 12 meters = 57 meters

Learn more about distance, speed, time, relationship here:

https://brainly.in/question/49075584

6 0

2 years ago

HELP PLEASE:

sattari [20]

Answer:

Its 40 for the first and 20for the other one

Step-by-step explanation:

I just took the test.

3 0

3 years ago

Read 2 more answers

Please i really need it rn pls

Step2247 [10]

The 3rd option is correct

7 0

3 years ago

Read 2 more answers

(a) Let R = {(a,b): a² + 3b <= 12, a, b € z+} be a relation defined on z+)

grin007 [14]

Answer:

R is an equivalence relation, since R is reflexive, symmetric, and transitive.

Step-by-step explanation:

The relation R is an equivalence if it is reflexive, symmetric and transitive.

The order to options required to show that R is an equivalence relation are;

((a, b), (a, b)) ∈ R since a·b = b·a

Therefore, R is reflexive

If ((a, b), (c, d)) ∈ R then a·d = b·c, which gives c·b = d·a, then ((c, d), (a, b)) ∈ R

Therefore, R is symmetric

If ((c, d), (e, f)) ∈ R, and ((a, b), (c, d)) ∈ R therefore, c·f = d·e, and a·d = b·c

Multiplying gives, a·f·c·d = b·e·c·d, which gives, a·f = b·e, then ((a, b), (e, f)) ∈R

Therefore R is transitive

From the above proofs, the relation R is reflexive, symmetric, and transitive, therefore, R is an equivalent relation.

Reasons:

Prove that the relation R is reflexive

Reflexive property is a property is the property that a number has a value that it posses (it is equal to itself)

The given relation is ((a, b), (c, d)) ∈ R if and only if a·d = b·c

By multiplication property of equality; a·b = b·a

Therefore;

((a, b), (a, b)) ∈ R

The relation, R, is reflexive.

Prove that the relation, R, is symmetric

Given that if ((a, b), (c, d)) ∈ R then we have, a·d = b·c

Therefore, c·b = d·a implies ((c, d), (a, b)) ∈ R

((a, b), (c, d)) and ((c, d), (a, b)) are symmetric.

Therefore, the relation, R, is symmetric.

Prove that R is transitive

Symbolically, transitive property is as follows; If x = y, and y = z, then x = z

From the given relation, ((a, b), (c, d)) ∈ R, then a·d = b·c

Therefore, ((c, d), (e, f)) ∈ R, then c·f = d·e

By multiplication, a·d × c·f = b·c × d·e

a·d·c·f = b·c·d·e

Therefore;

a·f·c·d = b·e·c·d

a·f = b·e

Which gives;

((a, b), (e, f)) ∈ R, therefore, the relation, R, is transitive.

Therefore;

R is an equivalence relation, since R is reflexive, symmetric, and transitive.

Based on a similar question posted online, it is required to rank the given options in the order to show that R is an equivalence relation.

Learn more about equivalent relations here:

brainly.com/question/1503196

4 0

3 years ago

Suppose that 45% of people have dogs. If two people are randomly chosen, what is the probability that they both have a dog

mylen [45]

Answer:

$P(Dogs) = 0.2025$

Step-by-step explanation:

Given

$Proportion, p = 45\%$

Required

Probability of two people having dog

First, we have to convert the given parameter to decimal

$p = \frac{45}{100}$

$p = 0.45$

Let P(Dogs) represent the required probability;

This is calculated as thus;

P(Dogs) = Probability of first person having a dog * Probability of second person having a dog

$P(Dogs) = p * p$

$P(Dogs) = 0.45 * 0.45$

$P(Dogs) = 0.45^2$

$P(Dogs) = 0.2025$

Hence, the probability of 2 people having a dog is  $P(Dogs) = 0.2025$ 

4 0

4 years ago