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cluponka [151]
2 years ago
12

Reese is selling lemonade at the parade. He gets to keep 30% of the money he collects. A large lemonade is $6.00 and a small lem

onade is $2.00.
The expression represents 30% of the money he collects.

0.30(6l + 2s)

Use the Distributive Property to expand the expression.

The simplified expression is
Mathematics
1 answer:
TiliK225 [7]2 years ago
8 0

Answer:

Simplified expression is 1.8i+0.6s

Step-by-step explanation:

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zaharov [31]

Answer:

The answer to the problem is -15.52

3 0
3 years ago
Read 2 more answers
Find an equation for the perpendicular bisector of the line segment whose endpoints
TEA [102]

Answer:

y= -2x -8

Step-by-step explanation:

I will be writing the equation of the perpendicular bisector in the slope-intercept form which is y=mx +c, where m is the gradient and c is the y-intercept.

A perpendicular bisector is a line that cuts through the other line perpendicularly (at 90°) and into 2 equal parts (and thus passes through the midpoint of the line).

Let's find the gradient of the given line.

\boxed{gradient =  \frac{y1 -y 2}{x1 - x2} }

Gradient of given line

=  \frac{1 - ( - 5)}{3 - ( - 9)}

=  \frac{1 + 5}{3 + 9}

=  \frac{6}{12}

=   \frac{1}{2}

The product of the gradients of 2 perpendicular lines is -1.

(½)(gradient of perpendicular bisector)= -1

Gradient of perpendicular bisector

= -1 ÷(½)

= -1(2)

= -2

Substitute m= -2 into the equation:

y= -2x +c

To find the value of c, we need to substitute a pair of coordinates that the line passes through into the equation. Since the perpendicular bisector passes through the midpoint of the given line, let's find the coordinates of the midpoint.

\boxed{midpoint = ( \frac{x1 + x2}{2} , \frac{y1 + y2}{2})  }

Midpoint of given line

= ( \frac{3  -  9}{2} , \frac{1 - 5}{2} )

= ( \frac{ - 6}{2}  , \frac{ - 4}{2} )

= ( - 3 , - 2)

Substituting (-3, -2) into the equation:

-2= -2(-3) +c

-2= 6 +c

c= -2 -6 <em>(</em><em>-</em><em>6</em><em> </em><em>on both</em><em> </em><em>sides</em><em>)</em>

c= -8

Thus, the equation of the perpendicular bisector is y= -2x -8.

5 0
3 years ago
Mrs. Nicholas baked two batches of cookies. The first batch consisted of 120 gingerbread cookies and the second batch consisted
shusha [124]

Answer:

i think 2/15 of the cookies

Step-by-step explanation:

hope this helps

3 0
1 year ago
Solve the system of equations.<br> – 3y + 5x = 26<br> - 2 - 5x = -16
mylen [45]

Answer:

y=-4,\:x=\frac{14}{5}

Step-by-step explanation:

\begin{bmatrix}-3y+5x=26\\ -2-5x=-16\end{bmatrix}

Isolate x for -2-5x=-16:

x=\frac{14}{5}

\mathrm{Substitute\:}x=\frac{14}{5}

\begin{bmatrix}-3y+5\cdot \frac{14}{5}=26\end{bmatrix}

Simplify

\begin{bmatrix}-3y+14=26\end{bmatrix}

Isolate y for -3y+14=26:

y=-4

\mathrm{The\:solutions\:to\:the\:system\:of\:equations\:are:}

y=-4,\:x=\frac{14}{5}

7 0
2 years ago
A collection of quarters and dimes is worth $14.90. There are 80 coins in all. Find how many of each there are in the collection
sashaice [31]

The trick here is to relate the NUMBER of coins to each other in one equation, and then the VALUE of the coins in another equation. If I have 1 dime, that 1 dime is worth 10 cents. The number of dimes is obviously not equal to the value. Let's call quarters q and dimes d. The number of these 2 types of coins added together is 80 coins. So q + d = 80. Now, we know that quarters are worth .25 and dimes are worth .1, so we express a quarter's worth as .25q; we express a dime's worth as .1d. The value of the coins we have is 14.90. So that equation is .25q + .1d = 14.90. Let's solve the first equation for q. q = 80 - d. We can now use that as a substitution for q into the second equation, giving us an equation with only 1 unknown, d. .25(80-d) + .1d = 14.90. Distributing through the parenthesis we have 20 - .25d + .1d = 14.90. Combining like terms gives us - .15d = - 5.1. We will divide both sides by - .15 to get that the number of dimes is 34. If we had a total of 80 coins, then the number of quarters is 80 - 34, which is 46. 46 quarters and 34 dimes

7 0
3 years ago
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