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Charra [1.4K]
3 years ago
14

During the week, Brittney earned 1/4 and 1/3 of her monthly bonus on two sales transactions. If those bonuses amounted to $840,

how much was her monthly bonus?
Mathematics
1 answer:
EleoNora [17]3 years ago
7 0

Answer:

$142.86

Step-by-step explanation:

Let the bonus be x

This implies that

\frac{1}{3}x +  \frac{1}{4}x = 840

Multiplying through by the L. C. M(that is 12)

12 \times \frac{1}{3}x + 12 \times  \frac{1}{4}x = 12 \times 840

\implies4x + 3x=10080

\implies7x=10080

Dividing through by 7

\implies \frac{7x}{7}  =  \frac{10080}{7}

\implies x=142.86

Hence her monthly bonus was $142.86

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we look at them that we have and look for matches

the only match we got is (-10,7) and (-10,-2)
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3 years ago
34. For the following exercises, given each set of information, find a linear equation satisfying the conditions, if possible.
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Answer:

The linear equation for the line which passes through the points given as (-1,4) and (5,2), is written in the point-slope form as $y=\frac{1}{3} x-\frac{13}{3}$.

Step-by-step explanation:

A condition is given that a line passes through the points whose coordinates are (-1,4) and (5,2).

It is asked to find the linear equation which satisfies the given condition.

Step 1 of 3

Determine the slope of the line.

The points through which the line passes are given as (-1,4) and (5,2). Next, the formula for the slope is given as,

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Substitute 2&4 for $y_{2}$ and $y_{1}$ respectively, and $5 \&-1$ for $x_{2}$ and $x_{1}$ respectively in the above formula. Then simplify to get the slope as follows,

m=\frac{2-4}{5-(-1)}$\\ $m=\frac{-2}{6}$\\ $m=-\frac{1}{3}$

Step 2 of 3

Write the linear equation in point-slope form.

A linear equation in point slope form is given as,

$y-y_{1}=m\left(x-x_{1}\right)$

Substitute $-\frac{1}{3}$ for m,-1 for $x_{1}$, and 4 for $y_{1}$ in the above equation and simplify using the distributive property as follows,

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Step 3 of 3

Simplify the equation further.

Add 4 on each side of the equation $y-4=\frac{1}{3} x-\frac{1}{3}$, and simplify as follows,

y-4+4=\frac{1}{3} x-\frac{1}{3}+4$\\ $y=\frac{1}{3} x-\frac{1+12}{3}$\\ $y=\frac{1}{3} x-\frac{13}{3}$

This is the required linear equation.

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