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ddd [48]
3 years ago
5

Help please!!!

Mathematics
2 answers:
astra-53 [7]3 years ago
7 0

Answer: 46 is your answer hope this helped

Step-by-step explanation:

Artist 52 [7]3 years ago
3 0

Answer:

4 \sqrt{11}

Step-by-step explanation:

<em><u>According</u></em><em><u> </u></em><em><u>to</u></em><em><u> </u></em><em><u>Pythagoras</u></em><em><u> </u></em><em><u>Theorem</u></em><em><u>, </u></em>

{(hypotenuse)}^{2}  -  {height}^{2}  =  {(base)}^{2}

=  >  {24}^{2}  -  {20}^{2}  =  {x}^{2}

=  > \:  {x}^{2}  = 576 - 400

=  >  {x}^{2}  = 176

=  > x =  \sqrt{176}

=  > x =  \sqrt{2 \times 2 \times 2 \times 2 \times 11}

=  > x = 2 \times 2 \sqrt{11}

=  > x = 4 \sqrt{11} (ans)

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The infinite geometric series S=1+( 2/3 )+( 2/3 )^ 2 +( 2/3 )^ 3 .... equal to:
weqwewe [10]

SOLUTION

The question simply means that we should find the sum to infinity of the geometric series.

The formula of sum to infinity of a geometric serie is given by

S_{\infty}=\frac{a}{1-r}

Where

\begin{gathered} S_{\infty}\text{ is the sum to infinity} \\  \\ a\text{ is the first term = 1} \\  \\ r\text{ is the common ratio = }\frac{2}{3} \end{gathered}

So, this becomes

\begin{gathered} S_{\infty}=\frac{a}{1-r} \\  \\ S_{\infty}=\frac{1}{1-\frac{2}{3}} \\  \\ S_{\infty}=\frac{1}{\frac{3-2}{3}} \\  \\ S_{\infty}=\frac{1}{\frac{1}{3}} \\  \\ S_{\infty}=3 \end{gathered}

Therefore, option b is the correct answer

5 0
1 year ago
2<img src="https://tex.z-dn.net/?f=%5Csqrt%7Bx%2B5%7D%2B4%5C%5C" id="TexFormula1" title="\sqrt{x+5}+4\\" alt="\sqrt{x+5}+4\\" al
Andreyy89

2 \sqrt{x + 9}

8 0
3 years ago
Read 2 more answers
I need help with number 3
olya-2409 [2.1K]

Answer: He would earn $14

Step-by-step explanation: Since Sam had 11 games but only 9 of them were working just take 11 - 9 which = 2 then take 7 x 2 to get $14

Hope this helps

3 0
3 years ago
Read 2 more answers
5/7 + 3/8?
Rudiy27

I don't know if anyone know say him answer

7 0
2 years ago
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