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3 years ago

PLS HELP !!!

Mathematics

2 answers:

SIZIF [17.4K]3 years ago

8 0

The answer you’re looking for is 3/10

cestrela7 [59]3 years ago

5 0

Answer:

3/5 of her time

Step-by-step explanation:

She spent 30 minutes out of 50 minutes on her math assignment. So, we can write this fraction:

30 minutes / 50 minutes

Divide both the numerator (top of the fraction) and denominator (bottom of the fraction) by 10 minutes to get

3 / 5

So, the answer is 3/5 of her time.

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they’d be the same when x=-56

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The length of AB¯¯¯¯¯ is 9 centimeters. A dilation with a scale factor of 2 is applied to AB¯¯¯¯¯ . What is the length o

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On which of these policies will an insurer pay if there’s a loss

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3 years ago

Solve 2x - 11 = k for x.

denis23 [38]

2x + -11 = k

Reorder the terms:
-11 + 2x = k

Solving
-11 + 2x = k

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '11' to each side of the equation.
-11 + 11 + 2x = 11 + k

Combine like terms: -11 + 11 = 0
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Divide each side by '2'.
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Simplifying

3 0

4 years ago

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:

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

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

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

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

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

5 0

4 years ago