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2 years ago

The value of y when x=5

Mathematics

1 answer:

Molodets [167]2 years ago

8 0

D. 35 is the answer. Since we know that 14=2k based on the knowledge given in the first part of the equation, we can solve for K. We learn that K=7. Then, we can substitute K into the second equation: y=(7)(5) which is 35.

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Cylinder E has the following dimensions: h = 30 cm and r = 4 cm. Cylinder F has the same radius, but has a height of 5 cm . What

Anarel [89]

Answer:

1759.52cm^3

Step-by-step explanation:

Given data

Cylinder E

h = 30 cm and

r = 4 cm

The expression for the volume is

V= πr^2h

V= 3.142*4^2*30

V= 3.142*16*30

V=1508.16 cm^3

Cylinder F

h=5 cm

r = 4 cm

The expression for the volume is

V= πr^2h

V= 3.142*4^2*5

V= 3.142*16*5

V=251.36 cm^3

Hence the total volume is

=251.36+1508.16

= 1759.52cm^3

6 0

2 years ago

60 POINTS BRAINLIEST

Mariana [72]

60 POINTS BRAINLIEST

ANSWER QUESTIONS A B & C

PHOTO ATTACHED BELOW !!!!

A) Using the given spreadsheet as a basis, discuss the usefulness of a standard

deviation in a given dataset. You need to cite any literature sources you use.

B) Produce an appropriate fully-labeled chart to compare the gold, silver and

bronze medals totals of the 10 countries.

C) Use a suitable and fully-labeled chart to reflect the contribution of each

country to the overall medals total.

PLEASE HELP DUE TODAY!!60 POINTS BRAINLIEST

ANSWER QUESTIONS A B & C

PHOTO ATTACHED BELOW !!!!

A) Using the given spreadsheet as a basis, discuss the usefulness of a standard

deviation in a given dataset. You need to cite any literature sources you use.

B) Produce an appropriate fully-labeled chart to compare the gold, silver and

bronze medals totals of the 10 countries.

C) Use a suitable and fully-labeled chart to reflect the contribution of each

country to the overall medals total.

PLEASE HELP DUE TODAY!!60 POINTS BRAINLIEST

ANSWER QUESTIONS A B & C

PHOTO ATTACHED BELOW !!!!

A) Using the given spreadsheet as a basis, discuss the usefulness of a standard

deviation in a given dataset. You need to cite any literature sources you use.

B) Produce an appropriate fully-labeled chart to compare the gold, silver and

bronze medals totals of the 10 countries.

C) Use a suitable and fully-labeled chart to reflect the contribution of each

country to the overall medals total.

PLEASE HELP DUE TODAY!!

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5 0

3 years ago

Read 2 more answers

Solve : 3 under root a/b

Aleksandr [31]

Do you know the numbers a/b equals too

3 0

2 years ago

Tell whether each graph is a function and justify your answer. Which graph is not a good representation of a real-world situati

mina [271]

Using the function concept, it is found that:

Graph B is a function because each value of x corresponds to exactly one y-value.

In a function, one value of the input can be related to only one value of the output.

In a graph, it means that for each value of x(horizontal axis), there can be only one respective value of y(vertical axis).

In this problem, at Graph A, when x = 5, for example a vertical line crosses the function 3 times, hence there are 3 respective values of y for x = 5, and the same is valid for other values of x, hence it is not a function.

At Graph B, for each value of x, there is only one value of y, hence it is a function.

Hence:

Graph B is a function because each value of x corresponds to exactly one y-value.

To learn more about the function concept, you can take a look at brainly.com/question/12463448

5 0

2 years ago

If a factory continuously dumps pollutants into a river at the rate of the quotient of the square root of t and 45 tons per day,

julsineya [31]

<h2>Hello!</h2>

The answer is:

The first option, the amount dumped after 5 days is 0.166 tons.

To solve the problem, we need to integrate the given expression and evaluate using the given time.

So, integrating we have:

$\int\limits^5_0 {\frac{\sqrt{t} }{45} } \, dt=\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \, dt\\\\\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \ dt=\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt\\\\\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt=(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)\\\\(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)=(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)$

$(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)=(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)\\\\(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)=(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)\\\\(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)=(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})\\\\(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})=\frac{2}{135}*11.18-0=0.1656=0.166$

Hence, we have that the amount dumped after 5 days is 0.166 tons.

Have a nice day!

5 0

3 years ago