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Elden [556K]
3 years ago
8

Suppose your club is selling candles to raise money. It cost $100 to rent a booth from which to sell the candles. If the candles

cost your club $1 and are sold for $5 each, how many candles must be sold to equal your expenses?
Mathematics
2 answers:
Pani-rosa [81]3 years ago
7 0
Let x = # candles sold

100 + x = 5x
4x = 100
  x = 25

answer
25 candles must be sold to equal expenses
77julia77 [94]3 years ago
4 0
-jaydonw2255

Equation:
Expenses=Earnings
100+X=5x
4x=100
x=25(numbers of candles you need to sell)
Simple answer-25
Hope this helped
-Gabrielle
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What is an equation of the line that passes through the points (-5, -6) and (5, 6)?
anyanavicka [17]

Answer:

It would be -5 + -6 = 65

Step-by-step explanation:

Add it up to together by removing the -

8 0
3 years ago
Choose the solution(s) of the following system of equations x2 + y2 = 6 x2 – y = 6
ElenaW [278]
<h2>Answer</h2>

x=\sqrt{6}, y=0\\ x=-\sqrt{6} , y=0\\x=\sqrt{5} , y=-1\\x=-\sqrt{5} , y=-1

Or as ordered pairs: (\sqrt{6} ,0),(-\sqrt{6} ,0),(\sqrt{5} ,-1),(-\sqrt{5} ,-1)

<h2>Explanation</h2>

Lets solve our system of equations  step by step

x^2+y^2=6 equation (1)

x^2-y=6 equation (2)

1. Solve for x^2 in equation (2)

x^2-y=6

x^2=6+y equation (3)

2. Replace equation (3) in equation (1) and solve for y

x^2+y^2=6

6+y+y^2=6

y^2+y=0

y(y+1)=0

y=0 or y=-1

3. Replace the values of y in equation (3) and solve for x

- For y=0

x^2=6+y

x^2=6

x=\sqrt{6} or x=-\sqrt{6}

- For y=-1

x^2=6+y

x^2=6-1

x^2=5

x=\sqrt{5} or x=-\sqrt{5}

So, the solutions of our system of equation are:

x=\sqrt{6}, y=0\\ x=-\sqrt{6} , y=0\\x=\sqrt{5} , y=-1\\x=-\sqrt{5} , y=-1

6 0
3 years ago
Read 2 more answers
Maggie graphed the image of a 90 counterclockwise rotation about vertex A of . Coordinates B and C of are (2, 6) and (4, 3) and
Lemur [1.5K]

Answer:

A(2,2)

Step-by-step explanation:

Let the vertex A has coordinates (x_A,y_A)

Vectors AB and AB' are perpendicular, then

\overrightarrow {AB}=(2-x_A,6-y_A)\\ \\\overrightarrow {AB'}=(-2-x_A,2-y_A)\\ \\\overrightarrow {AB}\perp\overrightarrow {AB'}\Rightarrow \overrightarrow {AB}\cdot \overrightarrow {AB'}=0\Rightarrow (2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0

Vectors AC and AC' are perpendicular, then

\overrightarrow {AC}=(4-x_A,3-y_A)\\ \\\overrightarrow {AC'}=(1-x_A,4-y_A)\\ \\\overrightarrow {AC}\perp\overrightarrow {AC'}\Rightarrow \overrightarrow {AC}\cdot \overrightarrow {AC'}=0\Rightarrow (4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0

Now, solve the system of two equations:

\left\{\begin{array}{l}(2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0\\ \\(4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0\end{array}\right.\\ \\\left\{\begin{array}{l}-4-2x_A+2x_A+x_A^2+12-6y_A-2y_A+y^2_A=0\\ \\4-4x_A-x_A+x_A^2+12-3y_A-4y_A+y_A^2=0\end{array}\right.\\ \\\left\{\begin{array}{l}x_A^2+y_A^2-8y_A+8=0\\ \\x_A^2+y_A^2-5x_A-7y_A+16=0\end{array}\right.

Subtract these two equations:

5x_A-y_A-8=0\Rightarrow y_A=5x_A-8

Substitute it into the first equation:

x_A^2+(5x_A-8)^2-8(5x_A-8)+8=0\\ \\x_A^2+25x_A^2-80x_A+64-40x_A+64+8=0\\ \\26x_A^2-120x_A+136=0\\ \\13x_A^2-60x_A+68=0\\ \\D=(-60)^2-4\cdot 13\cdot 68=3600-3536=64\\ \\x_{A_{1,2}}=\dfrac{60\pm8}{2\cdot 13}=\dfrac{34}{13},2

Then

y_{A_{1,2}}=5\cdot \dfrac{34}{13}-8 \text{ or } 5\cdot 2-8\\ \\=\dfrac{66}{13}\text{ or } 2

Rotation by 90° counterclockwise about A(2,2) gives image points B' and C' (see attached diagram)

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