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3 years ago

Write an equation of the line that passes through (-2,3) and is parallel to the line y = 2x – 4.

Mathematics

1 answer:

allochka39001 [22]3 years ago

3 0

Answer:

The parallel line should be y = 2x + 7

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A fashion designer makes and sells hats. The material for each hat cost $5.50. The Hats sell for $12.50 each. The designer spend

labwork [276]

The designer must sell 200 hats for breaking even.

Step-by-step explanation:

Given,

Cost for material for each hat = $5.50

Selling price of each hat = $12.50

Cost spend on advertising = $1400

Let,

x represents the number of hats.

Revenue = 12.50x

Costs = 5.50x+1400

For breaking even,

Revenue = Cost

$12.50x=5.50x+1400\\12.50x-5.50x=1400\\7x=1400$

Dividing both sides by 7

$\frac{7x}{7}=\frac{1400}{7}\\x=200$

The designer must sell 200 hats for breaking even.

Keywords: addition, division

Learn more about division at:

#LearnwithBrainly

7 0

3 years ago

Mrs. Claus makes a

Slav-nsk [51]

Answer:

14 or 11

Step-by-step explanation:

because 14 is more than 11 and the question asks how many people it feeds not how many people can it feed out the people there

6 0

3 years ago

Read 2 more answers

What is the slope of a line that is perpendicular to 3x+2y=14

djyliett [7]

Answer:

2/3

Step-by-step explanation:

7 0

3 years ago

If someone can help me out :D!

Nataliya [291]

Hello

Answer:

28.378

Explanation:

8 0

2 years ago

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Try to sketch by hand the curve of intersection of the parabolic cylinder y = x2 and the top half of the ellipsoid x2 + 7y2 + 7z

vovikov84 [41]

Plug $y=x^2$ into the equation of the ellipsoid:

$x^2+7(x^2)^2+7z^2=49$

Complete the square:

$7x^4+x^2=7\left(x^4+\dfrac{x^2}7+\dfrac1{14^2}-\dfrac1{14^2}\right)=7\left(x^2+\dfrac1{14}\right)^2+\dfrac1{28}$

Then the intersection is such that

$7\left(x^2+\dfrac1{14}\right)^2+7z^2=\dfrac{1371}{28}$

$\left(x^2+\dfrac1{14}\right)^2+z^2=\dfrac{1371}{196}$

which resembles the equation of a circle, and suggests a parameterization is polar-like coordinates. Let

$x(t)^2+\dfrac1{14}=\sqrt{\dfrac{1371}{196}}\cos t\implies x(t)=\pm\sqrt{\sqrt{\dfrac{1371}{196}}\cos t-\dfrac1{14}}$

$y(t)=x(t)^2=\sqrt{\dfrac{1371}{196}}\cos t-\dfrac1{14}$

$z=\sqrt{\dfrac{1371}{196}}\sin t$

(Attached is a plot of the two surfaces and the intersection; red for the positive root $x(t)$ , blue for the negative)

4 0

3 years ago