How to solve problem in the document

Mathematics

1 answer:

jeyben [28]1 year ago

5 0

I’ve attached a photo, with the instructions. Hope it helps. If you have any questions don’t hesitate to ask

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What is the area of the above trapezoid? Please help<br> I will cash app if you help out

Leno4ka [110]

Answer:

b should be correct

Step-by-step explanation:

no need to cash app me its been a while since ive done math so i dont even know if this is right

7 0

3 years ago

5 years ago, my grandfather was five times as old as I was. Three years from now, my grandfather will be three times as old as I

balu736 [363]

<h3>Answer:</h3>

13 years old

<h3>Explanation:</h3>

Let g and m represent grandfather's age now and my age now. The relation 5 years ago was ...

... g -5 = 5(m -5)

The relation in 3 years will be ...

... g +3 = 3(m +3)

Subtracting the first equation from the second, we get ...

... 8 = -2m +34

... 2m = 26 . . . . . add 2m-8

... m = 13

My age now is 13.

4 0

3 years ago

Read 2 more answers

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:

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

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