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2 years ago

7

What is the volume of a cylinder with a base area of 23 cm2 and a height of 15 cm

2 answers:

mars1129 [50]2 years ago

8 0

Answer:

V =345 cm^3

Step-by-step explanation:

The volume of a cylinder is given by

V = Bh where B is the base and h is the height

V= 23 * 15

V =345 cm^3

lutik1710 [3]2 years ago

5 0

Answer:

345cm³

Step-by-step explanation:

AB h = 23 x15

= 345cm³

Your welcome!

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The table shows y as a function of x. Suppose a point is added to this table. Which choice gives a point that preserves the func

Hunter-Best [27]

Answer:

<em>A) (-5,7)</em>

Step-by-step explanation:

<u>Functions and Relations</u>

A set of values A can have a relation with another set B as long as at least one element of A has at least one image in B. Functions are special relations where each element of A (the domain of the function) has one and only one image on B (the range of the function).

By looking at the options, we can see that x=9, x=-8, and x=-1 already have defined values in Y, so if we define another value for any of them the relation will stop being a function. The only possible choice to preserve the function is the option

$\boxed{A)\ (-5,7)}$

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3 years ago

Marge and Kimo equally Shared 1/4 of a pie that was left over. What fraction of the original pie did each friend get? Use the pi

luda_lava [24]

Answer:

1/8

Step-by-step explanation:

1/4 divided by 2

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2 years ago

Guys help me what is 137x8 multiply with expanded form???????????? Please help!!!!!!!!!!!!!!!!!! 33 points for the person who he

Helen [10]

Answer:

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Step-by-step explanation:

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3 years ago

Match solutions and differential equations. Note: Each equation may have more than one solution. Select all that apply. (a) 7y''

Nonamiya [84]

Answer:

a- $e^x,e^{-x}$

b- $x^{-2}$

c- $x^3$

Step-by-step explanation:

a.7 y''-7 y =0

Auxillary equation

$D^2-1=0$

$(D-1)(D+1)=0$

D=1,-1

Then , the solution of given differential equation

$y=e^x,y=e^{-x}$

2. $7x^2y''+14xy'-14 y=0$

Y= $y=x^3$

$y=3x^2$

$y''=6x$

Substitute in the given differential equation

$42x^3+42x^3-14x^3\neq 0$

Hence, $x^3$ is not a solution of given differential equation

$e^x,e^{-x}$ are also not a solution of given differential equation.

y= $x^{-2}$

$y'=-2x^{-3}$

$y''=6x^{-4}$

Substitute the values in the differential equation

$7x^2(6x^{-4})+14x(-2x^{-3})-14 x^{-2}$

= $42x^{-2}-28x^{-2}-14x^{-2}=0$

Hence, $x^{-2}$ is a solution of given differential equation.

c. $7x^2y''-42y=0$

$y=x^3$

$y'=3x^2$

$y''=6x$

Substitute the values in the differential equation

$42x^3-42x^3=0$

Hence, $x^3$ is a solution of given differential equation.

a- $e^x,e^{-x}$

b- $x^{-2}$

c- $x^3$

6 0

3 years ago

Read 2 more answers

Determine whether the given lengths can be sides of a right triangle. Which of the following are true statements. A.The lengths

Stels [109]

Answer:

Given sides 12, 16 and 20 can be the sides of right triangle.

Step-by-step explanation:

Sides of right triangle always follow the Pythagoras theorem.

i.e $(base)^2 + (Height)^2 = (Hypotenuse)^2$

For the given Lengths 7, 40 and 41

We need to check if

$7^2 + 40^2 =41^2 \\or\\ 7^2 + 40^2 \neq41^2$

$Since \\7^2 + 40^2 = 1649\\and \\41^2 = 1681$

That means, $\\ 7^2 + 40^2 \neq 41^2$

hence 7,40 and 41 can not be the sides of right triangle.

Next,

Given sides 12,16 and 20.

Again follow the similar process used in the above problem.

$12^2 + 16^2 =400\\And \\20^2 = 400\\Since 12^2 + 16^2 = 20^2$

Therefore given sides 12,16 and 20 can be the sides of right triangle.

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2 years ago