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2 years ago

1. Olivia jumps from an Olympic diving platform down to the pool below. Her height above

Mathematics

1 answer:

puteri [66]2 years ago

8 0

Answer:

You should provided some numbers to go off

Step-by-step explanation:

so I couldn't really give you a straight up answer

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Mark draws a card 50 times from a standard deck of 52 cards and gets a Heart 8 times. What is the experimental probability that

marysya [2.9K]

Good for Mark he’s a good man

5 0

3 years ago

Read 2 more answers

Endpoint: (9,-10), midpoint: (4,8)

Alika [10]

Answer:

( -1 , 26)

Step-by-step explanation:

So since you need to find the other endpoint, you would follow these steps:

1.) $4 = \frac{9 + x}{2\\}$

2.) 8 = 9 + x ( you just multiplied the 2 to the 4 to get 8)

3.) -1 = x (just solve it like a regular equation, so just subtract 9 on both sides to get rid of it and that leaves you with -1 = x)

You took the x values of both points and put them in the equation.

And its the same for y

1.) $8 = \frac{-10 + y}{2}$

2.) 16 = -10 + y

3.) 26 = y (you added the 10 on both sides because the 10 was negative and that took the 10 out and so it left you with 26 = y)

5 0

3 years ago

Which polyhedron has more faces than a hexahedron, but fewer faces than an dodecahedron?

timurjin [86]

The best and most correct answer among the choices provided by your question is the fourth choice or letter D.

Octahedron (8) is a <span>polyhedron has more faces than a hexahedron, but fewer faces than an dodecahedron.</span>

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

5 0

2 years ago

Read 2 more answers

Let a1, a2, a3, ... be a sequence of positive integers in arithmetic progression with common difference

Bezzdna [24]

Since $a_1,a_2,a_3,\cdots$ are in arithmetic progression,

$a_2 = a_1 + 2$

$a_3 = a_2 + 2 = a_1 + 2\cdot2$

$a_4 = a_3+2 = a_1+3\cdot2$

$\cdots \implies a_n = a_1 + 2(n-1)$

and since $b_1,b_2,b_3,\cdots$ are in geometric progression,

$b_2 = 2b_1$

$b_3=2b_2 = 2^2 b_1$

$b_4=2b_3=2^3b_1$

$\cdots\implies b_n=2^{n-1}b_1$

Recall that

$\displaystyle \sum_{k=1}^n 1 = \underbrace{1+1+1+\cdots+1}_{n\,\rm times} = n$

$\displaystyle \sum_{k=1}^n k = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}2$

It follows that

$a_1 + a_2 + \cdots + a_n = \displaystyle \sum_{k=1}^n (a_1 + 2(k-1)) \\\\ ~~~~~~~~ = a_1 \sum_{k=1}^n 1 + 2 \sum_{k=1}^n (k-1) \\\\ ~~~~~~~~ = a_1 n + n(n-1)$