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lakkis [162]
3 years ago
7

SOMEONE PLEASE HELP MEEEEE

Mathematics
1 answer:
natka813 [3]3 years ago
8 0

Answer:

Area = 23.12 cm²

perimeter = 24.28 cm

Step-by-step explanation:

top shaded region is circular arc with 90° sector angle

so perimeter of arc is 90 / 360 * 2πr

here radius of arc is half the length of square side

r = 6.8/2 = 3.4

top area shaded perimeter = 90/360 * 2 * 3.14 * 3.4

= 1/4 * 2 * 3.14 * 3.4

= 5.338

bottom region perimeter is also same

total perimeter of shaded region is = side + side + two arcs

= 6.8 + 6.8 + 2 * 5.338

= 13.6 + 10.676

= 24.276

= 24.28

Area of top shaded region is 90/360 * πr²

A = 1/4 * 3.14 * 3.4²

= 9.0746

bottom shaded region area is half of area of square - area of bottom arc

= 6.8 * 3.4  - 9.0746

= 23.12 - 9.0746

= 14.0454

so total area of shaded region is 9.0746 + 14.0454

= 23.12

or you can just calculate area as half of area of total square = 6.8² / 2

= 23.12

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Answer:

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Step-by-step explanation:

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3 years ago
Please please help!
Dimas [21]

Answer:

C

Step-by-step explanation:

Calculate the distance (d) using the distance formula

d = √ (x₂ - x₁ )² + (y₂ - y₁ )²

with (x₁, y₁ ) = (- 3, 9) and (x₂, y₂ ) = (3, 1)

d = \sqrt{(3+3)^2+(1-9)^2}

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3 years ago
4. a) A ping pong ball has a 75% rebound ratio. When you drop it from a height of k feet, it bounces and bounces endlessly. If t
Klio2033 [76]

First part of question:

Find the general term that represents the situation in terms of k.

The general term for geometric series is:

a_{n}=a_{1}r^{n-1}

a_{1} = the first term of the series

r = the geometric ratio

a_{1} would represent the height at which the ball is first dropped. Therefore:

a_{1} = k

We also know that the ball has a rebound ratio of 75%, meaning that the ball only bounces 75% of its original height every time it bounces. This appears to be our geometric ratio. Therefore:

r=\frac{3}{4}

Our general term would be:

a_{n}=a_{1}r^{n-1}

a_{n}=k(\frac{3}{4}) ^{n-1}

Second part of question:

If the ball dropped from a height of 235ft, determine the highest height achieved by the ball after six bounces.

k represents the initial height:

k = 235\ ft

n represents the number of times the ball bounces:

n = 6

Plugging this back into our general term of the geometric series:

a_{n}=k(\frac{3}{4}) ^{n-1}

a_{n}=235(\frac{3}{4}) ^{6-1}

a_{n}=235(\frac{3}{4}) ^{5}

a_{n}=55.8\ ft

a_{n} represents the highest height of the ball after 6 bounces.

Third part of question:

If the ball dropped from a height of 235ft, find the total distance traveled by the ball when it strikes the ground for the 12th time. ​

This would be easier to solve if we have a general term for the <em>sum </em>of a geometric series, which is:

S_{n}=\frac{a_{1}(1-r^{n})}{1-r}

We already know these variables:

a_{1}= k = 235\ ft

r=\frac{3}{4}

n = 12

Therefore:

S_{n}=\frac{(235)(1-\frac{3}{4} ^{12})}{1-\frac{3}{4} }

S_{n}=\frac{(235)(1-\frac{3}{4} ^{12})}{\frac{1}{4} }

S_{n}=(4)(235)(1-\frac{3}{4} ^{12})

S_{n}=910.22\ ft

8 0
3 years ago
What is the least common multiple of 32 24 and 18
noname [10]

Answer:

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3 0
3 years ago
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A right pyramid with a regular hexagon base has a height of 3 units If a side of the hexagon is 6 units long, then the apothem i
Setler79 [48]

Answer:

Part a) The slant height is 3\sqrt{2}\ units

Part b) The lateral area is equal to 54\sqrt{2}\ units^{2}

Step-by-step explanation:

we know that

The lateral area of a right pyramid with a regular hexagon base is equal to the area of its six triangular faces

so

LA=6[\frac{1}{2}(b)(l)]

where

b is the length side of the hexagon

l is the slant height of the pyramid

Part a) Find the slant height l

Applying the Pythagoras Theorem

l^{2}=h^{2} +a^{2}

where

h is the height of the pyramid

a is the apothem

we have

h=3\ units

a=3\ units

substitute

l^{2}=3^{2} +3^{2}

l^{2}=18

l=3\sqrt{2}\ units

Part b) Find the lateral area

LA=6[\frac{1}{2}(b)(l)]

we have

b=6\ units

l=3\sqrt{2}\ units

substitute the values

LA=6[\frac{1}{2}(6)(3\sqrt{2})]=54\sqrt{2}\ units^{2}

3 0
3 years ago
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