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3 years ago

13

What is the radius of a hemisphere with a volume of 83838 in, to the nearest tenth of an inch?

1 answer:

ipn [44]3 years ago

3 0

Answer:34.2

Step-by-step explanation: I just did it in delta math

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What is 6/5 of 7003=<br> PLEASE HELP AGAIIIN!!!!!!!!!

ikadub [295]

Answer:

8,403.6

Step-by-step explanation:

6/5 of 7003

= 6/5 × 7003

= 42,018/5

= 8,403.6

Hence 6/5 of 7003 is 8,403.6

4 0

2 years ago

Find the coordinates of the missing endpoint if P is the midpoint of Line Segment NQ.

dimaraw [331]

Midpoint formula is $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$ .

<h3>19.</h3>

So starting with this one, we will be solving for the coordinates of the unknown endpoint separately. Starting with the x-coordinate, since we know that the midpoint x-coordinate is 5 and the x-coordinate of N is 2, our equation is set up as such: $\frac{x+2}{2}=5$ From here we can solve for the x-coordinate of Q.

Firstly, multiply both sides by 2: $x+2=10$

Next, subtract both sides by 2 and your x-coordinate is $x=8$

With finding the y-coordinate, it's a similar process as with the x-coordinate except that we are using the y-coordinates of the midpoint and endpoint N.

$\frac{y+0}{2}=2\\ y=4$

<u>Putting it together, the missing endpoint is (8,4).</u>

<em>(The process is pretty much the same with the other problems, so I'll go through them real quickly.)</em>

<h3>20.</h3>

$\frac{x+5}{2}=6\\ x+5=12\\ x=7$

$\frac{y+4}{2}=3\\ y+4=6\\ y=2$

<u>The missing endpoint is (7,2).</u>

<h3>21.</h3>

$\frac{x+3}{2}=-1\\ x+3=-2\\ x=-5$

$\frac{y+9}{2}=5\\ y+9=10\\y=1$

<u>The missing endpoint is (-5,1).</u>

8 0

2 years ago

If there are 26 people in a room, how many of these people would you EXPECT to have a name that starts with a vowel?

avanturin [10]

Answer:

follow xxflirtykingxx

8 0

2 years ago

If you have a cubic polynomial of the form y = ax^3 + bx^2 + cx + d and lets say it passes through the points (2,28), (-1, -5),

nikklg [1K]

Step-by-step explanation:

<u>Step 1: Solve using the first point</u>

<em>(2, 28)</em>

$28 = a(2)^3 + b(2)^2 + c(2) + d$

$28 = 8a + 4b + 2c + d$

<u>Step 2: Solve using the second point</u>

<em>(-1, -5)</em>

$-5 = a(-1)^3 + b(-1)^2 + c(-1) + d$

$-5 = -a + b - c + d$

<u>Step 3: Solve using the third point</u>

<em>(4, 220)</em>

$220 = a(4)^3 + b(4)^2 + c(4) + d$

$220 = 64a + 16b + 4c + d$

<u>Step 4: Solve using the fourth point</u>

<em>(-2, -20)</em>

$-20 = a(-2)^3 + b(-2)^2 + c(-2) + d$

$-20 = -8a + 4b - 2c + d$

<u>Step 5: Combine the first and fourth equations</u>

<u /> $28 - 20 = 8a - 8a + 4b + 4b + 2c - 2c + d + d$

$8 = 8b + 2d$

$8 - 8b = 8b - 8b + 2d$

$(8 -8b)/2 = 2d/2$

$4 - 4b = d$

<u>Step 6: Solve for c in the second equation</u>

$-5 + 5 = -a + b - c + d + 5$

$0 + c = -a + b - c + c + d + 5$

$c = -a + b + d + 5$

<u>Step 7: Substitute d with the stuff we got in step 5</u>

$c = -a + b + (4 - 4b) + 5$

$c = -a + b + 4 - 4b + 5$

$c = -a - 3b + 9$

<u>Step 8: Substitute d and c into the first equation</u>

<u /> $28 = 8a + 4b + 2(-a - 3b + 9) + (4 - 4b)$

$28 = 8a + 4b - 2a - 6b + 18 + 4 - 4b$

$28 - 22 = 6a - 6b + 22 - 22$

$6 / 6 = (6a - 6b) / 6$

$1 + b = a - b + b$

$1 + b = a$

<u>Step 9: Substitute a, b, and c into the third equation</u>

$220 = 64(1 + b) + 16b + 4(-(1 + b) - 3b + 9) + (4 - 4b)$

$220 = 64 + 64b + 16b + 4(-1 - b - 3b + 9) + 4 - 4b$

$220 - 100 = 60b + 100 - 100$

$120 / 60 = 60b / 60$

$2 = b$

<u>Step 10: Find a using b = 2</u>

$a = b + 1$

$a = (2) + 1$

$a = 3$

<u>Step 11: Find c using a = 3 and b = 2</u>

$c = -a - 3b + 9$

$c = -(3) - 3(2) + 9$

$c = -3 - 6 + 9$

$c = 0$

<u>Step 12: Find d using b = 2</u>

$d = 4 - 4b$

$d = 4 - 4(2)$

$d = 4 - 8$

$d = -4$

Answer: $a = 3, b = 2, c = 0,d = -4$

6 0

2 years ago

Read 2 more answers

The sum of 14 and a number is equal to 17

alexdok [17]

N+14=17. n=3. So the missing number is 3.

3 0

3 years ago

Read 2 more answers