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3 years ago

What value of x makes this proportion true 4/7=20/35

Mathematics

2 answers:

JulijaS [17]3 years ago

8 0

The answer would be x=5

Romashka [77]3 years ago

7 0

If you're asking me to find the value of x then its 5
4 x 5 = 20
7 x 5 = 35

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Add 2 to 5, then multiply 3 by the result

Nostrana [21]

Answer:

Step-by-step explanation:

Step 1: add 2 to 5

=> 2 + 5

=> 7

Step 2: Multiply the result by 3:

=> 7 * 3

=> 21

Therefore the final answer = 21

Hope this helps!

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2 years ago

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Sandy mixed together 9 gallons of one brand of juice and 8 gallons of a second brand of juice, which contains 48 percent real fr

Gekata [30.6K]

Answer:

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Step-by-step explanation:

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2 years ago

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Lostsunrise [7]

Answer: 16

Step-by-step explanation:

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2 years ago

The school play earned $11,271 in ticket sales. If the cost of each ticket was $13, how many tickets were sold?

mojhsa [17]

Answer:

867 tickets were sold because if you divide how much they earned by how much it cost you get 867

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3 years ago

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Write an expression for the 12th partial sum of the series 3/2+7/3+19/6+... using summation notation

lapo4ka [179]

Answer:

$S_{12}=\sum_{i=1}^{12} [\frac{3}{2}+(i-1)\times \frac{5}{6}]$

$S_{12}=73$

Step-by-step explanation:

$First\ term\ of\ the\ series(a_1)=\frac{3}{2}\\\\Second\ term\ of\ the\ series(a_2)=\frac{7}{3}\\\\Third\ term\ of\ the\ series(a_3)=\frac{19}{6}\\\\a_2-a_1=\frac{7}{3}-\frac{3}{2}=\frac{5}{6}\\\\a_3-a_2=\frac{19}{6}-\frac{7}{3}=\frac{5}{6}\\\\Hence\ it\ is\ an\ Arithmetic\ Series\ with\ first\ term=\frac{3}{2}\ and\ constant\ difference=\frac{5}{6}$

$a_1=\frac{3}{2}+0\times \frac{5}{6}\\\\a_2=\frac{3}{2}+1\times \frac{5}{6}\\\\a_3=\frac{3}{2}+2\times \frac{5}{6}\\\\.\\.\\.\\a_n=\frac{3}{2}+(n-1)\times \frac{5}{6}\\\\S_n=a_1+a_2+a_3+......+a_n\\\\S_n=(\frac{3}{2}+0\times \frac{5}{6})+(\frac{3}{2}+1\times \frac{5}{6})+(\frac{3}{2}+2\times \frac{5}{6})+....+(\frac{3}{2}+[n-1]\times \frac{5}{6})\\\\S_n=\sum_{i=1}^n [\frac{3}{2}+(i-1)\times \frac{5}{6}]\\\\S_n=(\frac{3}{2}+\frac{3}{2}+\frac{3}{2}+...n\ times)+\frac{5}{6}(1+2+3+4+...+(n-1))\\\\$

$S_n=\frac{3}{2}\times n+\frac{5}{6}\times \frac{n(n-1)}{2}\\\\$

$S_{12}=\sum_{i=1}^{12} [\frac{3}{2}+(i-1)\times \frac{5}{6}]$

$S_{12}=\frac{3}{2}\times 12+\frac{5}{6}\times \frac{(12)(12-1)}{2}\\\\S_{12}=18+55\\\\S_{12}=73$

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3 years ago