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Degger [83]
3 years ago
9

PLEASE HELP ASAP!!!!!!

Mathematics
1 answer:
Masja [62]3 years ago
8 0
The answer is that they will meet at 7.20 pm ! hope this helped
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Identify the diameter of the circular base created by folding the figure into a right cone. HELP ASAP PLEASE!!
Akimi4 [234]

let's notice something, we have a circle with a radius of 12 and one 90° sector is cut off, so only three 90° sectors of the circle are left shaded, so namely the cone will be using 3/4 of that circle.

think of it as, this shaded area is some piece of paper, and you need to pull it upwards and have the cutoff edges meet, and when that happens, you'll end up with a cone-shaped paper cup, and pour in some punch.

now, once we have pulled up the center of the circle to make our paper cup, there will be a circular base, its diameter not going to be 24, it'll be less, but whatever that base is, we know that is going to have the same circumference as those in the shaded area.  Well, what is the circumference of that shaded area?

\bf \textit{circumference of a circle}\\\\ C=2\pi r~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=12 \end{cases}\implies C=2\pi 12\implies C=24\pi \implies \stackrel{\textit{three quarters of it}}{24\pi \cdot \cfrac{3}{4}} \\\\\\ 6\pi \cdot 3\implies 18\pi

well then, the circumference of that circle at the bottom will be 18π, so, what is the diameter of a circle with a circumferenc of 18π?

\bf \textit{circumference of a circle}\\\\ C=2\pi r~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ C=18\pi \end{cases}\implies 18\pi =2\pi r\implies \cfrac{18\pi }{2\pi }=r\implies 9=r \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{\textit{diameter is twice the radius}}{d=18}~\hfill

3 0
3 years ago
im stupid. honestly. just help me pls. this is probably really easy but i aint done any maths in over a month so ion remember ho
KATRIN_1 [288]

Answer:

19 coins

Step-by-step explanation:

We know that mean is the sum of all the elements divided by the number of elements there are: m = t/n, where m is the mean, t is the total sum, and n is the number of elements.

Here, we are given the mean number of coins for all 27 + 18 = 45 children is 16, so m = 16 and n = 45. Then:

m = t/n

16 = t/45

t = 720

This t means that the total number of coins that all the boys and girls have is 720 coins.

Now, we also know the mean number of coins for the boys only, which is 14. Since there are 27 boys, then:

m = t/n

14 = t/27

t = 378

This t means the total number of coins all the boys have.

Subtracting 378 from 720, we get 342, which is the total number of coins the girls have. We also know there are 18 girls, so:

m = t/n

m = 342 / 18 = 19

Thus, the mean is 19 coins.

7 0
3 years ago
Read 2 more answers
Function or not a function
Shkiper50 [21]

Answer:

  • Not a function.

Step-by-step explanation:

<u>Functions have graphs or slopes.</u>

<u>This is not a function.</u>

<u>This is not on a graph or on a slope.</u>

Your answer is not a function.

7 0
2 years ago
Nan buys a greeting card for 2 dollars and 29 cents. She pays with three $1 bills.
Alexandra [31]

Answer:

71 cents

Step-by-step explanation:

7 0
2 years ago
81 POINTS
Jobisdone [24]

Base Case: plug in n = 1 (the smallest positive integer)

If n = 1, then 3n-2 = 3*1-2 = 1. Square this and we see that (3n-2)^2 = 1^2 = 1

On the right hand side, plugging in n = 1 leads to...

n*(6n^2-3n-1)/2 = 1*(6*1^2-3*1-1)/2 = 1

Both sides are 1. So that confirms the base case.

-------------------------------

Inductive Step: Assume that

1^2 + 4^2 + 7^2 + ... + (3k-2)^2 = k*(6k^2-3k-1)/2

is a true statement for some positive integer k. If we can show the statement leads to the (k+1)th case being true as well, then we will have sufficiently proven the overall statement to be true by induction.

1^2 + 4^2 + 7^2 + ... + (3k-2)^2 = k*(6k^2-3k-1)/2

1^2 + 4^2 + 7^2 + ... + (3k-2)^2 + (3(k+1)-2)^2 = (k+1)*(6(k+1)^2-3(k+1)-1)/2

k*(6k^2-3k-1)/2 + (3(k+1)-2)^2 = (k+1)*(6(k^2+2k+1)-3(k+1)-1)/2

k*(6k^2-3k-1)/2 + (3k+3-2)^2 = (k+1)*(6k^2+12k+6-3k-3-1)/2

k*(6k^2-3k-1)/2 + (3k+1)^2 = (k+1)*(6k^2+9k+2)/2

k*(6k^2-3k-1)/2 + 9k^2+6k+1 = (k+1)*(6k^2+9k+2)/2

(6k^3-3k^2-k)/2 + 2(9k^2+6k+1)/2 = (k*(6k^2+9k+2)+1(6k^2+9k+2))/2

(6k^3-3k^2-k + 2(9k^2+6k+1))/2 = (6k^3+9k^2+2k+6k^2+9k+2)/2

(6k^3-3k^2-k + 18k^2+12k+2)/2 = (6k^3+9k^2+2k+6k^2+9k+2)/2

(6k^3+15k^2+11k+2)/2 = (6k^3+15k^2+11k+2)/2

Both sides simplify to the same expression, so that proves the (k+1)th case immediately follows from the kth case

That wraps up the inductive step. The full induction proof is done at this point.

7 0
3 years ago
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