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3 years ago

Can y’all help me? :)

Mathematics

1 answer:

Airida [17]3 years ago

8 0

Answer:

Option A: $5\frac{1}{4} *4\frac{1}{5}$

Step-by-step explanation:

Area of a parallelogram is base times height so in this case it would be 5 1/4 times 4 1/5. 5 1/4 being the base and 4 1/5 being the height.

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Yukiko plays on the softball team at her school. The team won 9 of its first

Blababa [14]

Answer:

Step-by-step explanation:

9/12 is the ratio of wins. after simplifing it, you get 3/4. You can use that to find the wins after 64 games. Divide 64 by 4 and multiply that by 3. The answer is 48 game wins after 64 games played.

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3 years ago

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Help me please

vichka [17]

It has to be D because A and B are incorrect and Angle A is not equal to B.

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3 years ago

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Arrange the cones in order from lease volume to greatest volume

bearhunter [10]

Answer:

Volume of the cone in ascending order.

$V_{2}=270\pi\ units^{3}$

cone with DIAMETER of 18 & height of 10

cone with RADIUS of 10 & height of 9

cone with RADIUS of 11 & height of 9

cone with DIAMETER of 20 & height of 12

Step-by-step explanation:

Let $V_{2}. V_{3}. and\ V_{4}.$ be the volume of the cone.

Let d, r and h be the diameter, radius and height of the cone.

Given:

$d_{1} = 20\ and\ h_{1}=12$

$d_{2} = 18\ and\ h_{2}=10$

$r_{3} = 10\ and\ h_{3}=9$

$r_{4} = 11\ and\ h_{14}=9$

Arrange the cones in order from lease volume to greatest volume.

Solution:

The volume of the cone is given below.

$V=\pi r^{2} \frac{h}{3}$ ----------------(1)

where: r is radius of the base of cone.

and h is height of the cone.

The volume of the cone for $d_{1} = 20\ and\ h_{1}=12$

$r_{1} = \frac{d_{1}}{2}$

$r_{1} = \frac{20}{2}=10\ units$

$V_{1}=\pi (r_{1})^{2} \frac{h_{1}}{3}$

$V_{1}=\pi (10)^{2} \frac{12}{3}$

$V_{1}=\pi\times 100\times 4$

$V_{1}=400\pi\ units^{3}$

Similarly, for volume of the cone for $d_{2} = 18\ and\ h_{2}=10$

$r_{2} = \frac{d_{2}}{2}$

$r_{2} = \frac{18}{2}=9\ units$

$V_{2}=\pi (r_{2})^{2} \frac{h_{2}}{3}$

$V_{2}=\pi (9)^{2} \frac{10}{3}$

$V_{2}=\pi\times 81\times \frac{10}{3}$

$V_{2}=\pi\times 27\times 10$

$V_{2}=270\pi\ units^{3}$

Similarly, for volume of the cone for $r_{3} = 10\ and\ h_{3}=9$

$V_{3}=\pi (r_{3})^{2} \frac{h_{3}}{3}$

$V_{3}=\pi (10)^{2} \frac{9}{3}$

$V_{3}=\pi\times 100\times 3$

$V_{3}=\pi\times 300$

$V_{3}=300\pi\ units^{3}$

Similarly, for volume of the cone for $r_{4} = 11\ and\ h_{4}=9$

$V_{4}=\pi (r_{4})^{2} \frac{h_{4}}{3}$

$V_{4}=\pi (11)^{2} \frac{9}{3}$

$V_{4}=\pi\times 121\times 3$

$V_{4}=\pi\times 363$

$V_{4}=363\pi\ units^{3}$

So, the volume of the cone in ascending order.

$V_{2}=270\pi\ units^{3}$

7 0

3 years ago

I need help on 8 9 and 10

Lerok [7]

9. x + (y - 3 - 4x + 2( 3x - (-y - 1) + 2y) - 3x)

= x + (y - 3 - 4x + 2 (3x + y - 1) + 2y - 3x)
= x + (y - 3 - 4x + 6x + 2y - 2 + 2y - 3x)
= (x - 4x + 6x - 3x) + (y + 2y + 2y) - (3+2)
= 0 + 5y - 5
= 5y - 5 (or simplified as 5(y-1)

10. 6 + (-a -3(b + 6(a - b)+4a) - 2b)
= 6 + (-a - 3(b + 6a - 6b + 4a) -2b)
= 6 + (-a - 3b + 18a - 18b + 12a - 2b)
= 6 + (-a + 18a + 12a) + (-3b - 18b - 2b)
= 6 + 29a - 23b

Lmk if it's wrong, I'll try to edit it.

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2 years ago

What percent of 50 is 44

PtichkaEL [24]

Answer:

here u go

Step-by-step explanation:

88%

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3 years ago

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