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Fudgin [204]
3 years ago
12

Marshall divides his money into four categories. He saves 1/3 of his money. He gives 1/6 of his money to a charity. He uses 1/14

of his money to buy snacks at school, and he spends 3/7 of his money on fun. What fraction of his money does he save and spend on fun?
Mathematics
2 answers:
Len [333]3 years ago
7 0
1/3 + 3/7
save + fun + extra

(7/21 + 9/21) = 16/21 on save and fun
frutty [35]3 years ago
4 0

Marshall divides his money into four categories.

He saves 1/3 of his money.

He gives 1/6 of his money to a charity.

He uses 1/14 of his money to buy snacks at school,

and he spends 3/7 of his money on fun.

Money he saves = \frac{1}{3}

Money he spends for fun = \frac{3}{7}

Fraction of his money he save and spend on fun = \frac{1}{3} + \frac{3}{7}

LCD of (3 and 7) is 21

\frac{1*7}{3*7} + \frac{3*3}{7*3}

\frac{7}{21} + \frac{9}{21}

Denominators are same so we add the numerators

\frac{16}{21}

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ollegr [7]

Answer:

<u>6</u>

10

=<u>3</u>

5

Step-by-step explanation:

3/5 you just need to divide

5 0
3 years ago
I need help ASAP please
pychu [463]
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The formula for the area of a triangle is 1/2bh=A. Which of the following is not a possible first step to rearrange this formula
yarga [219]

We want to put b alone on one side of the equation.

If we multiply by 1/2, we get 1/4bh = A/2. Not only did we not remove a variable from the left side, we made it even more complicated.

So the answer is A.

7 0
3 years ago
The necklace charm shown has two parts, each shaped like a trapezoid with identical dimensions. What is the total area, in squar
pashok25 [27]
The area of the trapezoid can be calculated through the equation, 
                               A = (b₁ + b₂)h / 2
where b₁ and b₂ are the bases and h is the height. Substituting the known values from the given, 
                              A = (25mm + 32mm)(15 mm) / 2 
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Since there are two trapezoids in the necklace, the area calculated is to be multiplied by two to get the total area. 
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3 0
3 years ago
Read 2 more answers
For the following telescoping series, find a formula for the nth term of the sequence of partial sums
gtnhenbr [62]

I'm guessing the sum is supposed to be

\displaystyle\sum_{k=1}^\infty\frac{10}{(5k-1)(5k+4)}

Split the summand into partial fractions:

\dfrac1{(5k-1)(5k+4)}=\dfrac a{5k-1}+\dfrac b{5k+4}

1=a(5k+4)+b(5k-1)

If k=-\frac45, then

1=b(-4-1)\implies b=-\frac15

If k=\frac15, then

1=a(1+4)\implies a=\frac15

This means

\dfrac{10}{(5k-1)(5k+4)}=\dfrac2{5k-1}-\dfrac2{5k+4}

Consider the nth partial sum of the series:

S_n=2\left(\dfrac14-\dfrac19\right)+2\left(\dfrac19-\dfrac1{14}\right)+2\left(\dfrac1{14}-\dfrac1{19}\right)+\cdots+2\left(\dfrac1{5n-1}-\dfrac1{5n+4}\right)

The sum telescopes so that

S_n=\dfrac2{14}-\dfrac2{5n+4}

and as n\to\infty, the second term vanishes and leaves us with

\displaystyle\sum_{k=1}^\infty\frac{10}{(5k-1)(5k+4)}=\lim_{n\to\infty}S_n=\frac17

7 0
3 years ago
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