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lara31 [8.8K]
3 years ago
15

Janie receives an allowance of $3 per week.

Mathematics
2 answers:
inn [45]3 years ago
8 0
Janie receives an allowance of $3 per week.
In addition, she can earn $2 for each chore she does.
This week, she wants to earn enough money to buy a CD for $13.
Janie can do fractions of chores.

Write an inequality to determine the number of chores Janie must do this week to earn enough money to buy a CD.
13 <= 2c + 3 is the inequality.

Hope this helps!
soldier1979 [14.2K]3 years ago
7 0
13 = 2c + 3 is the inequality for one week
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Use long division to find the quotient below (30x^3+4x^2-150) / (6x-10)
Anni [7]

Answer:

The answer is

A.5x^2+9x+15

Step-by-step explanation:

kindly find attached the solving for proper understanding and solution flow.

Given Data

the divisor=  6x-10

dividend= 30x^3+4x^2-150

firstly for us to perform the division we need to re write the dividend and include the missing coefficient of x

dividend =30x^3+4x^2+0x-150

3 0
4 years ago
What is the solution of StartRoot x + 12 EndRoot = x?
Ber [7]

Answer:

x = 4      or x = -3

Step-by-step explanation:

\sqrt{x+12}=x

Take square both the sides,

(\sqrt{x+12})^{2}=x^{2}\\\\x+12=x^{2}

x² - x - 12 = 0

x² - 4x + 3x - 3*4 =0

x*(x - 4) + 3*(x - 4)= 0

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4 0
3 years ago
Read 2 more answers
A triangle has vertices at (1,10), (-5,2), and (7,2). What is its orthocenter. Show your work.
kondor19780726 [428]

Answer:

Question: What is the orthocenter of a triangle with the vertices (-1,2) (5,2) and (2,1)?

The coordinates of point A are (-1,2), point B are (5,2), and point C are (2,1).

The orthocent is the intersection of the three altitudes. An altitude goes from a vertex and is perpendicular to the line containing the opposite side.

In the coordinate plane the equations of the altitudes can be found and then a system of equations can be solved.

Altitude 1. From point C perpendicular to the line containing side AB.

Slope of line AB is 0 (horizontal line), a vertical line is perpendicular to a horizontal line. Thus, the equation of altitude 1 is  x=2 .

Altitude 2. From point B perpendicular to the line containing side AC.

Slope of line AC is  −13 , the slope of a line perpendicular to line AC is 3. The equation of altitude 2 is  y=3x−13  

Altitude 3. From point A perpendicular to the line containing side BC.

Slope of line BC is  13 , the slope of a line perpendicular to line BC is  −3 . The equation of altitude 3 is  y=−3x−1  

The orthocenter is the point where all three altitudes intersect.

x=2  

y=3x−13  

y=−3x−1  

Use substitution to solve the first two equations  y=3(2)−13=−7  

The orthocenter is the point  (2,−7)  

we did not need the third equation, but we can use it as a check, plug the coordinates into the third equation:

−7=−3(2)−1  

−7=−6−1  

−7=−7  it works.

3 0
3 years ago
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