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2 years ago

10

Find the volume of the cylinder.

1 answer:

Mnenie [13.5K]2 years ago

4 0

Answer:

120 i am sorry if i am wrong

Step-by-step explanation:

the volume of the cylinder is 120 cm
I have tried my hardest on studying this
question i am sorry if its wrong

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If you were to choose a card at random from a full deck, what is the probability of drawing the queen of hearts?

polet [3.4K]

Answer:

1/52

Step-by-step explanation:

A full deck is 52 cards, and there is only one queen of hearts, so the answer is 1/52

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3 years ago

How do 0.5 becomes a fraction i dont understand plz help

solniwko [45]

Answer:

it is 1/2 of 1

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Someone help? 25 points

kirill115 [55]

Answer:

A: scalene: non have the same angel

B: isosceles: two sides have the same angel

C: isosceles:two sides have the same angel

D: equilateral:all sides do

4 0

4 years ago

Read 2 more answers

in the figure below trianglePQM and triangleQRP are right triangles. the measure of lineQM is 6 and the measure of lineQP is 8.

Kipish [7]

Answer:

Option 4 is correct. The length of PR is 6.4 units.

Step-by-step explanation:

From the given figure it is noticed that the triangle PQR and triangle MQR.

Let the length of PR be x.

Pythagoras formula

$hypotenuse^2=base^2+perpendicular^2$

Use pythagoras formula for triangle PQM.

$PM^2=QM^2+PQ^2$

$PM^2=(6)^2+(8)^2$

$PM^2=36+64$

$PM^2=100$

$PM=10$

The value of PM is 10. The length of PR is x, so the length of MR is (10-x).

Use pythagoras formula for triangle PQR.

$PQ^2=QR^2+PR^2$

$(8)^2=QR^2+x^2$

$64-x^2=QR^2$ .....(1)

Use pythagoras formula for triangle MQR.

$MQ^2=QR^2+MR^2$

$(6)^2=QR^2+(10-x)^2$

$36=QR^2+x^2-20x+100$

$36-x^2+20x-100=QR^2$ .... (2)

From equation (1) and (2) we get

$36-x^2+20x-100=64-x^2$

$20x-64=64$

$20x=128$

$x=6.4$

Therefore length of PR is 6.4 units and option 4 is correct.

3 0

3 years ago

Find an equation of the tangent line to the graph of

Allisa [31]

Answer:

$y- y_o = m(x-x_o)$

For this case m represent the slope and is given by $m= g'(4) = 2$ and from the other condition given we know that $x_o = 4, y_o =-3$ . And replacing we got:

$y -(-3) = 2 (x-4)$

And if we reorganize the terms we have:

$y+3 = 2x -8$

And subtracting 3 in both sides we got:

$y = 2x -11$

Step-by-step explanation:

For this case we have the following function y=g(x) and we want to find the tangent line to this function using the conditions: $g(4) =-3, g'(4)=2$

So we need a function like this one:

$y- y_o = m(x-x_o)$

For this case m represent the slope and is given by $m= g'(4) = 2$ and from the other condition given we know that $x_o = 4, y_o =-3$ . And replacing we got:

$y -(-3)=2(x-4)$

And if we reorganize the terms we have:

$y+3 = 2x -8$

And subtracting 3 in both sides we got:

$y = 2x -11$

4 0

3 years ago