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3 years ago

Solve using a2 + b2 = c2 show work please :)

Mathematics

1 answer:

mel-nik [20]3 years ago

4 0

Answer:

Step-by-step explanation:

h^2=x^2+y^2 which just means the the hypotenuse of a right triangle squared is equal to the sum of its squared sides.

$h^2=10^2+15^2\\ \\ h^2=100+225\\ \\ h^2=325\\ \\ h=\sqrt{325}\\ \\h=5\sqrt{13}\\ \\ h\approx 18.03$

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At a track meet , 10 different runners compete in the 1600m run.

Zanzabum

Answer:

There are 3,628,800 different ways for the runners to finish.

Step-by-step explanation:

Arrangments of x elements:

The number of possible arrangments of x elements is given by the following formula:

$A_{x} = x!$

In this question:

10 different runners, which means that the number of different ways that there are for the runners to finish is an arrangment of 10 elements. So

$A_{10} = 10! = 3628800$

There are 3,628,800 different ways for the runners to finish.

4 0

3 years ago

alex wants to buy the same number of stamps and envelopes dancer sold in packs of 14 and envelopes are sold and packs of 10 what

lianna [129]

5 packs of stamps and 7 packs of envelopes. There will be a total of 70 envelopes and stamps.

6 0

3 years ago

Prove the following by induction. In each case, n is apositive integer.<br> 2^n ≤ 2^n+1 - 2^n-1 -1.

frutty [35]

<h2>Answer with explanation:</h2>

We are asked to prove by the method of mathematical induction that:

$2^n\leq 2^{n+1}-2^{n-1}-1$

where n is a positive integer.

Let us take n=1

then we have:

$2^1\leq 2^{1+1}-2^{1-1}-1\\\\i.e.\\\\2\leq 2^2-2^{0}-1\\\\i.e.\\2\leq 4-1-1\\\\i.e.\\\\2\leq 4-2\\\\i.e.\\\\2\leq 2$

Hence, the result is true for n=1.

Let us assume that the result is true for n=k

i.e.

$2^k\leq 2^{k+1}-2^{k-1}-1$

Now, we have to prove the result for n=k+1

i.e.

<u>To prove:</u> $2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Let us take n=k+1

Hence, we have:

$2^{k+1}=2^k\cdot 2\\\\i.e.\\\\2^{k+1}\leq 2\cdot (2^{k+1}-2^{k-1}-1)$

( Since, the result was true for n=k )

Hence, we have:

$2^{k+1}\leq 2^{k+1}\cdot 2-2^{k-1}\cdot 2-2\cdot 1\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{k-1+1}-2\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-2$

Also, we know that:

$-2$

(

Since, for n=k+1 being a positive integer we have:

$2^{(k+1)+1}-2^{(k+1)-1}>0$ )

Hence, we have finally,

$2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Hence, the result holds true for n=k+1

Hence, we may infer that the result is true for all n belonging to positive integer.

i.e.

$2^n\leq 2^{n+1}-2^{n-1}-1$ where n is a positive integer.

6 0

3 years ago

Please help me pleaseee

Sonja [21]

Answer:

m=112

Step-by-step explanation:

16-2=4

m/8=14

14x8=112

6 0

3 years ago

5(4x−3)−(3x+1)=3(5x+9)+2x

Kruka [31]

20x - 15 - 3x - 1 = 15x + 27 + 2x

17x - 16 = 17x + 27

17x - 17x = 27 + 16

0 = 43

Since zero is not equal to 43, this equation has no solution.

3 0

3 years ago

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