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erma4kov [3.2K]
4 years ago
10

If, in 7 years, Susan will be 2 times as old as she was 3 years ago, what is Susan’s present age? PLEAZE EXPLAIN HOW AND DO IT W

ITH USING X
Mathematics
2 answers:
Solnce55 [7]4 years ago
8 0

Answer:

Susan is 13 years old.

Step-by-step explanation:

Susan=x

After 7 years she will be 7+x  then she will be 2x as old as she was 3 year ago.

age 3 yrs ago=x-3

2x-6=7+x

if you simplify....

now she is 13

hope it helps!!!!

belka [17]4 years ago
5 0
Let Susan's present age be x

Now, after 7 years ( her age will become 7 +x then) she'll be 2 as old she was 3 years ago.

Her age 3 years ago must be x-3

Now according to question,

twice Her age 3 years ago = her age after 7 years

now,

2(x-3) = 7+x
2x-6 = 7+x
2x-x =7+6
x = 13
x = 13



Therefore, her age presenr age is 13 years.
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\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\ldots \ldots \ldots \ldots \ldots+\frac{1}{\sqrt{8}+\sqrt{9}}1+21+2+31+3+41+……………+8+91

Rationalizing the denominator, we get

\Rightarrow\left(\frac{1}{1+\sqrt{2}} \times \frac{1-\sqrt{2}}{1-\sqrt{2}}\right)+\left(\frac{1}{\sqrt{2}+\sqrt{3}} \times \frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}}\right)+\left(\frac{1}{\sqrt{3}+\sqrt{4}} \times \frac{\sqrt{3}-\sqrt{4}}{\sqrt{3}-\sqrt{4}}\right)+\cdots \ldots+\left(\frac{1}{\sqrt{8}+\sqrt{9}} \times \frac{\sqrt{8}-\sqrt{9}}{\sqrt{8}-\sqrt{9}}\right)⇒(1+21×1−21−2)+(2+31×2−32−3)+(3+41×3−43−4)+⋯…+(8+91×8−98−9)

We know that,

\left(a^{2}-b^{2}\right)=(a+b)(a-b)(a2−b2)=(a+b)(a−b)

Now, on substituting the formula, we get,

=\frac{1-\sqrt{2}}{1-2}+\frac{\sqrt{2}-\sqrt{3}}{2-3}+\frac{\sqrt{3}-\sqrt{4}}{3-4}+\cdots \ldots \cdot \frac{(\sqrt{8}-\sqrt{9})}{8-9}=1−21−2+2−32−3+3−43−4+⋯…⋅8−9(8−9)

\Rightarrow \frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\cdots+\frac{1}{\sqrt{8}+\sqrt{9}}=(\sqrt{2}-1)+(\sqrt{3}-\sqrt{2})+(\sqrt{4}-\sqrt{3})+\cdots+(\sqrt{9}-\sqrt{8})⇒1+21+

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