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likoan [24]
3 years ago
12

Rob is saving to buy a new MP3 player. For every $16 he earns babysitting, he saves $6. On Saturday, Rob earned $32 babysitting.

How much money did he save?
Mathematics
2 answers:
dsp733 years ago
8 0

Answer:

For every $16 Rob would get $6.

So for $32 it would be double since 32 is 2x more than 16 ¯\_(ツ)_/¯

The answer would be 6 x 2 or $12.

$12 is the answer.

Step-by-step explanation:

If you can please mark brainliest :3

marta [7]3 years ago
6 0

Answer:

He saved $12.

Step-by-step explanation:

32/16 = 2

2 x 6 = 12

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Show work and explain with formulas:
Marina86 [1]

17 Answer:  205

<u>Step-by-step explanation:</u>

\{1\dfrac{2}{3}+1\dfrac{5}{6}+2+...+8\dfrac{1}{3}\}\implies a_1=1\dfrac{2}{3},\ d=\dfrac{1}{6}\\\\\\a_n=a_1+d(n-1)\qquad solve\ for\ n\\\\8\dfrac{1}{3}=1\dfrac{2}{3}+\dfrac{1}{6}(n-1)\\\\\\\dfrac{25}{3}=\dfrac{5}{3}+\dfrac{1}{6}n-\dfrac{1}{6}\\\\\\\dfrac{50}{6}=\dfrac{10}{6}+\dfrac{1}{6}n-\dfrac{1}{6}\\\\\\\dfrac{41}{6}=\dfrac{1}{6}n\\\\\\\dfrac{41}{6}\cdot 6=n\\\\41=n

\text{Now use the sum formula:}\\\\S_n=\dfrac{a_1+a_n}{2}\cdot n\\\\\\S_{41}=\dfrac{1\dfrac{2}{3}+8\dfrac{1}{3}}{2}\cdot 41\\\\\\.\quad =\dfrac{10}{2}\cdot 41\\\\\\.\quad =5\cdot 41\\\\.\quad =\large\boxed{205}

18 Answer:  1968

<u>Step-by-step explanation:</u>

a_1=-6,\ d=2,\ n=48, \quad \text{solve for }a_{48}\\\\a_{n}=a_1+d(n-1)\\\\a_{48}=-6+2(48-1)\\\\.\quad =-6+2(47)\\\\.\quad =-6+94\\\\.\quad =88

\text{Now use the sum formula:}\\\\S_n=\dfrac{a_1+a_n}{2}\cdot n\\\\\\S_{48}=\dfrac{-6+88}{2}\cdot 48\\\\\\.\quad =\dfrac{82}{2}\cdot 48\\\\\\.\quad =41\cdot 48\\\\.\quad =\large\boxed{1968}

19 Answer:  -116

<u>Step-by-step explanation:</u>

\{3, -2, -7, ...\}\\a_1=3,\ d=-5,\ n=8, \quad \text{solve for }a_{8}\\\\a_{n}=a_1+d(n-1)\\\\a_{8}=3-5(8-1)\\\\.\quad =3-5(7)\\\\.\quad =3-35\\\\.\quad =-32\\\\\text{Now use the sum formula:}\\\\S_8=\dfrac{a_1+a_8}{2}\cdot 8\\\\\\S_{8}=\dfrac{3-32}{2}\cdot 8\\\\\\.\quad =\dfrac{-29}{2}\cdot 8\\\\\\.\quad =-29\cdot 4\\\\.\quad =\large\boxed{-116}

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3 years ago
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Solve for the x using elimination <br> 3x-2y=13x−2y=1 <br> 2x+2y=42x+2y=4
Nikitich [7]

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Divide the first equation by 5

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Subtract the first equation from the second

x        = 1

x + 2y = 3

Subtract the first equation from the second again

x        = 1

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Divide the second equation by 2

x        = 1

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<h3>So, the solution is  x = 1  and  y = 1  {or: (1, 1)} </h3>
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\int \dfrac{du}{u}=\log|u|+C    [\because \int \dfrac{dx}{x}=\log |x|+C]

b.

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c.

\int \dfrac{du}{u\sqrt{u^2-a^2}}=\dfrac{1}{a}\csc^{-1}(\dfrac{x}{a})+C        [\because \dfrac{adx}{x\sqrt{x^2-a^2}}=\csc^{-1}(\dfrac{x}{a})+C]

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