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3 years ago

The diagram shows corresponding lengths in two similar figures. Find the ratio of the perimeters and the ratio of the areas.

Mathematics

1 answer:

GREYUIT [131]3 years ago

7 0

15:20

divide each by 5

3:4

to find the ratio of the area's we square the ratio's of the sides

3^3 : 4^2

9:16

Choice B

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50 POINTS PLEASE HELP!!

Vinvika [58]

Answer: A. 35

Step-by-step explanation: The largest prime number less than 50. The largest number less than 50 is 49, but it is not prime, because 49=7·7. The previous number is 48, but it is also composite, because 48=2·2·2·2·3. The previous number is 47, it is prime.

The smallest composite number greater than 10. The smallest number greater than 10 is 11, but it is prime. The next number is 12, it is composite, because 12=2·2·3.

The difference between the largest prime number less than 50 and the smallest composite number greater than 10 is 47-12=35.

6 0

3 years ago

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Find the 6th term of the expansion of (2p - 3q)11. a. -7,185,024p4q7 c. -7,185p4q7 b. -7,185,024p6q5 d. -7,185p6q5

bezimeni [28]

$\bf (2p-3q)^{11}\implies 
\begin{array}{llll}
term&coefficient&value\\
-----&-----&-----\\
1&&(2p)^{11}(-3q)^0\\
2&+11&(2p)^{10}(-3q)^1\\
3&+55&(2p)^9(-3q)^2\\
4&+165&(2p)^8(-3q)^3\\
5&+330&(2p)^7(-3q)^4\\
6&+462&(2p)^6(-3q)^5
\end{array}$

the coefficient for the first term is 1, the next is 11 and so on... now, notice, the elements of the binomial, the 1st element starts off with 11, and every term it goes down by 1, the 2nd element starts off at 0, and goes up by 1 in each term.

now, to get the next coefficient, you simply, "get the product of the current coefficient and the exponent of the 1st element, and divide that by the exponent of the 2nd element in the next term".

for example, how did we get 165 for the 4th term.... well (55*9)/3

how did we get 462 for the 6th term? well (330*7)/5.

and then you can just expand it from there.

8 0

3 years ago

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Laura bought 23 pounds of sugar for $12. How many dollars did she pay per pound of sugar?

Vikki [24]

Answer:

≈ $0.52 per pound

Step-by-step explanation:

12/23 ≈ $0.52

6 0

2 years ago

Let P(n) be the statement that n! < nn where n is an integer greater than 1.

Ira Lisetskai [31]

Answer: See the step by step explanation.

Step-by-step explanation:

a) First, Let P(n) be the statement that n! < n^n

where n ≥ 2 is an integer (This is because we want the statement of P(2).

In this case the statement would be (n = 2): P(2) = 2! < 2^2

b) Now to prove this, let's complet the basis step:

We know that 2! = 2 * 1 = 2

and 2^2 = 2 * 2 = 4

Therefore: 2 < 4

c) For this part, we'll say that the inductive hypothesis would be assuming that k! < k^k for some k ≥ 1

d) In this part, the only thing we need to know or prove is to show that P(k+1) is also true, given the inductive hypothesis in part c.

e) To prove that P(k+1) is true, let's solve the inductive hypothesis of k! < k^k:

(k + 1)! = (k + 1)k!

(k + 1)k! < (k + 1)^k < (k + 1)(k + 1)^k

Since k < k+1 we have:

= (k + 1)^k+1

f) Finally, as the base and inductive steps are completed, the inequality is true for any integer for any n ≥ 1. If we had shown P(4)

as our basis step, then the inequality would only be proven for n ≥ 4.

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3 years ago

Thank u to da person who is willing to help me will get this Brainliest! :3

kumpel [21]

1) use distributive property
3x - 6 + 5x + 4
8x - 2

2) 4-2x - 14 - 3
-14 - 2x

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3 years ago

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