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Andreas93 [3]
3 years ago
13

What is the surface area of 8 ft and 16 ft

Mathematics
1 answer:
Elina [12.6K]3 years ago
3 0
8ft times 16ft =128ft
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Please explain your answer and I'll mark the Brainliest! Which complex number is the graph showing?
zheka24 [161]

Answer:

A 2√2(cos 7π/4 + i sin 7π/4)

Step-by-step explanation:

A. 2√2(cos 7π/4 + i sin 7π/4)

2 sqrt(2) ( sqrt(2)/2 - sqrt(2)/2 i)

Distribute

  2-2i

This is in the fourth quadrant  

B. 2√2(cos 150° + i sin 150°)

2 sqrt(2) (-sqrt(3)/2 +1/2i)

-sqrt(6) +sqrt(2) i

This is in the third quadrant (NO)

C. 2(cos 7π/4 + i sin 7π/4)

2( ( sqrt(2)/2 - sqrt(2)/2 i))

sqrt(2) - sqrt(2) i

This is the fourth quadrant

D. 2(cos 90° + i sin 90°)

2(0+i)

2i

This is on the positive y axis  NO

Now we need to decide between the two in the fourth quadrant.

The point has an x coordinate of 2 and a y coordinate of -2

This aligns with point A

8 0
3 years ago
I need help on this what is RS
lana [24]

Answer:

11

Step-by-step explanation:

If RS=TV

5x-9=2x+3

5x-2x=3+9

3x=12

x=4

RS is 5x-9

5(4)-9

20-9

11

6 0
2 years ago
a cylinder and a cone have the same radius and height. The volume of the cylinder is 534ft3 . what volume of the cone?
DochEvi [55]

Answer: 178 ft^3

Step-by-step explanation:

A cylinder has the formula V=pi radius ^2 height

A cone has the formula V= 1/3 pi radius ^2 height

So 1/3 of the volume of the cylinder is the volume of a cone

1/3 of 534 = 178 ft^3

5 0
2 years ago
An urn contains 5 red balls and 2 green balls. Two balls are drawn one after the other. What is the probability that the second
AleksAgata [21]

Answer:

<em>The probability that the second ball is red is 71%</em>

Step-by-step explanation:

<u>Probabilities</u>

We know there are 5 red balls and 2 green balls. Let's analyze what can happen when two balls are drawn in sequence (no reposition).

The first ball can be red (R) or green (G). The probability that it's red is computed by

\displaystyle P(R)=\frac{5}{7}

The probability is's green is computed by

\displaystyle P(G)=\frac{2}{7}

If we have drawn a red ball, there are only 4 of them out of 6 in the urn, so the probability to draw a second red ball is

\displaystyle P(RR)=\frac{5}{7}\cdot \frac{4}{6}=\frac{10}{21}

If we have drawn a green ball, there are still 5 red balls out of 6 in the urn, so the probability to draw a red ball now is

\displaystyle P(GR)=\frac{2}{7}\cdot \frac{5}{6}=\frac{5}{21}

The total probability of the second ball being red is

\displaystyle P(XR)=\frac{10}{21}+\frac{5}{21}=\frac{5}{7}=0.71

The probability that the second ball is red is 71%

7 0
3 years ago
How do I solve this problem
Andrews [41]

All you do is add the numbers and that is how you get your answer!


3 0
3 years ago
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