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3 years ago

What is 6.3-2(1.5c+4.1)

2 answers:

7 0

$Expand\;-2\left(1.5c+4.1\right)$
$\mathrm{Distribute\:parentheses\:using}: \:a\left(b+c\right)=ab+ac$
$\;a=-2,\:b=1.5c,\:c=4.1$

$Simplify\;-2\cdot \:1.5c-2\cdot \:4.1 \ \textgreater \ \mathrm{Multiply\:the\:numbers:}\:2\cdot \:1.5=3$
$-3c-2\cdot \:4.1$

$\mathrm{Multiply\:the\:numbers:}\:2\cdot \:4.1=8.2 \ \textgreater \ -3c-8.2 \ \textgreater \ 6.3-3c-8.2$

$\mathrm{Subtract\:the\:numbers:}\:6.3-8.2=-1.9 \ \textgreater \ -3c-1.9$

Hope this helps!

Anna35 [415]3 years ago

7 0

Answer: -3c - 1.9

6.3 - 2(1.5c + 4.1) Distributive Property => (-2 x 1.5c) + (-2 x 4.1)
6.3 - 3c - 8.2 Combine like terms
-3c - 1.9 Answer!

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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.

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3 years ago

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trapecia [35]

Answer:

<h3>The answer is option C.</h3>

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4 0

3 years ago

Read 2 more answers

The median weight for a 5 foot tall male to enlist in the US Army is 114.5 lbs. This weight can vary by 17.5 lbs. Write and solv

myrzilka [38]

An absolute value inequality that represents the weight of a 5-foot male who would not meet the minimum or maximum weight requirement allowed to enlist in the Army is 97 lbs < x < 132 lbs.

<h3>What are inequalities?</h3>

Inequalities help us to compare two unequal expressions. Also, it helps us to compare the non-equal expressions so that an equation can be formed.

It is mostly denoted by the symbol <, >, ≤, and ≥.

The median weight for a 5 foot tall male to enlist in the US Army is 114.5 lbs. This weight can vary by 17.5 lbs. Therefore, the inequality can be written as,

(114.5 - 17.5) lbs < x < (114.5 + 17.5) lbs

97 lbs < x < 132 lbs

Hence, an absolute value inequality that represents the weight of a 5-foot male who would not meet the minimum or maximum weight requirement allowed to enlist in the Army is 97 lbs < x < 132 lbs.

Learn more about Inequality:

brainly.com/question/19491153

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