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3 years ago

Write a expression for this word phrase, the quotient of double a number and 4

Mathematics

1 answer:

Usimov [2.4K]3 years ago

6 0

The quotient

$\boxed{\ }:\boxed{\ }\\\\\boxed{\ }\div\boxed{\ }\\\\$

Double a number

n - a number

$2n$

Answer: $\boxed{2n:4}$

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Which expression is equivalent to (c-2d)(3c-2d)

Zanzabum

Answer: A

Step-by-step explanation:

(c-2d)(3c-2d)

= 3c^2 - 8dc + 4d^2

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2 years ago

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Suppose PV =NRT, where P=8, V=2X, N=2 and R=2X.you are given X=3. Solve for T

FromTheMoon [43]

Answer: T = 4

Step-by-step explanation:

1. Write all the variables down

P = 8 V = 2X N = 2 R = 2X X = 3

2. Since you know that X = 3 substitute it in to find V and R

V = 2X = 2(3) = 6

R = 2X = 2(3) = 6

3. Find PV

PV = P x V

= 8 x 6

= 48

4. Find NRT

NRT = N x R x T

= 2 x 6 x T

= 12 x T

= 12T

5. Find T

PV = NRT

48 = 12T

12T = 48

divide both sides by 12

T = 48 ÷ 12

T = 4

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2 years ago

PLEASE HELP! A city has a population of 390,000 people. Suppose that each year the population grows by 6.25%. What will the popu

Vaselesa [24]

Answer: The population will be 536,250 people

Step-by-step explanation:

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3 years ago

25) Wal-Mart is having a sale on packs of pencils. A pack of 24 pencils is on sale for $0.48. I plan to buy 20 packs of pencils.

Bezzdna [24]

Multiply the 20 packs by the 0.48 per pack
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2 years ago

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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.

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2 years ago