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katovenus [111]
3 years ago
9

What is the area of this figure

Mathematics
2 answers:
vodomira [7]3 years ago
5 0

area = 4 \times 4  +  \frac{4 \times 6}{2}   \\ = 16 + 12 = 28 \:  {yd}^{2}

Alla [95]3 years ago
5 0

~Hello There!~

area of triangle = \frac{b *h}{2}

area of triangle = \frac{6*8}{2}

area of triangle = 24

area of square = b*h

area of square = 4 * 4

area of square = 16

total area = 24 + 16

total area = 40yd^{2}

Hope This Helps You!

Good Luck :)

Have A Great Day ^_^

- Hannah ❤

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I think you are referring to this:
The volume of the sink shaped like a half sphere is [2000/3] π in^2.

A. The volume of a cylinder =area of the base * height
For this part the cup has a diameter, d, of 4 in and height, h, of 8 in.
[2000 / 3]π / [32π] = 20.83
Answer: 21 cups.

B. For this part the diameter of the cup is 8 in and the height is 8 in
And the number of cups needed is [2000π/3] / (128π) = 5.2
Answer: 5 cups

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Is this math or algebra?
8 0
2 years ago
A construction elevator can carry at most 2,320 pounds. If the elevator operator weighs 195 pounds and each palette of roofing m
asambeis [7]

Answer:

5

Step-by-step explanation:

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3 0
3 years ago
A train travels at a speed of 45 kmph. In 48 minutes, how much distance will it travel? ( with explanation )
aalyn [17]

Answer:

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5 0
3 years ago
Read 2 more answers
Please help! I've been working on this for a few days and I just don't understand, it's due in a few hours. Thank you.
never [62]

Answer:

Part A: α = arc tan (y/x) = tan⁻¹ (y/x)

Part B: quadrant II α = arc tan (- y/x) = arc (- tan (y/x)) = - tan⁻¹ (y/x) 180°>α>90°

            quadrant III α = arc tan (-y)/(-x) = arc tan (y/x) = tan⁻¹ (y/x) 270°>α>180°

            quadrant IV α = arc tan (- y/x) = arc (- tan (y/x)) = - tan⁻¹ (y/x) 360°>α>270°

Part C: quadrant II 180°>α>90°  α = tan⁻¹ (-6/1) = - tan⁻¹ 6 = 180° - 80.53° = 99.47°

Step-by-step explanation:

PART A:

In this case we will use trigonometric function tanα to calculate angle α:

tanα = y/x => α = arc tan (y/x) = tan⁻¹ (y/x)

This formula is use in general way and in first quadrant  90°>α>0°

Part B:

But in the other quadrants you must know to use unit circle to reduce angle from II, III and IV quadrant to the first quadrant.

If angle is in the quadrant II  180°>α>90° then

tanα = - tan (180°-α)

If angle is in the quadrant III  270°>α>180° then

tanα = tan (α-180°)

If angle is in the quadrant IV  360°>α>270° then

tanα =- tan (360°-α)

Part C:

vector w (x,y) = (-1,6) this vector lies in quadrant II      180°>α>90°

180°- α = tan⁻¹ (-6/1) = - tan⁻¹ 6 = 80.53°  => 180° - α = 80.53°

α = 180° - 80.53 = 99.47°

It's not easy to understand this, but it's not easy for me to explain.

God with you!!!

5 0
3 years ago
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