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4 years ago

74.80 decreased by 45%

Mathematics

1 answer:

ASHA 777 [7]4 years ago

3 0

Use a multiplication problem of 74.80*0.45, and you get an answer of <span>33.66.

</span>

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A taxi cab in myrtle beach charges $2 per mile and $1 for every person. If a taxi cab ride for two people costs $12, how far did

Oxana [17]

Answer:

5 Miles

Step-by-step explanation:

5 Miles = 1 Miles = $2 X 5 = $10

1 Person = $1

Therefore 2 People = $1 X 2

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3 years ago

Re-write the quadratic function below in Standard Form<br> y =4(x + 5)2 – 3

Alenkasestr [34]

Answer:

I love algebra anyways

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Step-by-step explanation:

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3 years ago

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Is 1.875 and rational number

jonny [76]

Answer: 1.875 is a rational number because it can be represented as a ratio or a fraction.

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3 years ago

Graph a line that contains the point -6,1 and has a slope of 5

Stells [14]

Answer:

y=5x+31

Step-by-step explanation:

y=mx+b

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31=b

y=5x+31

(5*-6)+31

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3 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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3 years ago