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makvit [3.9K]
3 years ago
6

How do I solve this?

Mathematics
2 answers:
Naddika [18.5K]3 years ago
8 0

Answer:

15

Step-by-step explanation:

5^2 +14^2=C^2

25+196=C^2

221=C^2

√221=√C^2

14.86=C

appr 15 =C

maria [59]3 years ago
6 0

14.87

Step-by-step explanation:

use the Pythagorean theorem

a^2+b^2=c^2

C being the longest side of the triangle

14^2+5^2=C^2

196+25=C^2

221=C^

find the square root of 221

C=14.866

rounded to nearest tenth

C=14.87

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Start with 180. 
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<span>Is 45 divisible by 2? No, so try a bigger divisor. </span>
<span>Is 45 divisible by 3? Yes, so write "3" as a prime factor, then work with the quotient, 15 </span>

<span>Is 15 divisible by 3? [Note: no need to revert to "2", because we've already divided out all the 2's] Yes, so write "3" (again) as a prime factor, then work with the quotient, 5. </span>

<span>Is 5 divisible by 3? No, so try a bigger divisor. </span>
Is 5 divisible by 4? No, so try a bigger divisor (actually, we know it can't be divisible by 4 becase it's not divisible by 2)
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<span>Once you end up with a quotient of "1" you're done. </span>

<span>In this case, you should have written down, "2 * 2 * 3 * 3 * 5"</span>
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(1)           (12, 18, 27, ...)

The common ratio is:

r=\dfrac{a_{n+1}}{a_n}\quad r =\dfrac{18}{12}=\boxed{\dfrac{3}{2}}\quad \rightarrow \quad r=\dfrac{27}{18}=\boxed{\dfrac{3}{2}}

The equation is:

a_n=a_o(r)^{n-1}\\\\Given:a_o=12,\  r=\dfrac{3}{2}\\\\\\Equation:\\a_n =12\bigg(\dfrac{3}{2}\bigg)^{n-1}\\\\\\\\9th\ term:\\a_9=12\bigg(\dfrac{3}{2}\bigg)^{9-1}\\\\\\a_9=12\bigg(\dfrac{3}{2}\bigg)^{8}\\\\\\.\quad =\large\boxed{\dfrac{19643}{64}}

(2)\qquad \bigg(\dfrac{1}{16},\dfrac{1}{8},\dfrac{1}{4},\dfrac{1}{2}\bigg)\\\\\\\text{The common ratio is}:\\\\r=\dfrac{a_{n+1}}{a_n}\quad  r=\dfrac{\frac{1}{8}}{\frac{1}{16}}=\boxed{2}\quad \rightarrow \quad r=\dfrac{\frac{1}{4}}{\frac{1}{8}}=\boxed{2}

The equation is:

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3 0
3 years ago
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