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3 years ago

What’s 4 and 3/16 + 1 and 1/10 in simplest form?? PLEASE NEED THIS ASSAP pleaseeee

Mathematics

2 answers:

Neporo4naja [7]3 years ago

5 0

Answer:

423/80 or 5 and 23/80

Step-by-step explanation:

First convert to improper fractions. Then find common denominator, then you would add them.

mrs_skeptik [129]3 years ago

3 0

5 and 23/80 is the answer

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The following amounts were deposited in a savings account each month.

Ksju [112]

Answer: 20+44=64

32+56=88

88+64=$152

Step-by-step explanation:

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3 years ago

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Write this ratio in one other way 7/13

Ierofanga [76]

You can write this ratio in two other ways:

7 to 13

7 : 13

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3 years ago

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The perimeter of a rectangle is 300 feet. The width of the rectangle is 10 feet more than the length. What is the width of the r

gavmur [86]

300 = 2(x + x + 10)
300 = 2(2x +10)
300 = 4x + 20
4x = 280
x = 70
x + 10 = 80
width = 80ft
answer <span>D) 80 feet</span>

7 0

3 years ago

Solve for x 3x+2 4x-2

Blizzard [7]

Answer:

-2/3 or 1/2

Step-by-step explanation:

both eqn equal to zero

then we know that

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