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Black_prince [1.1K]
3 years ago
13

Can someone show me how to do this ?

Mathematics
1 answer:
Savatey [412]3 years ago
7 0
The point slope formula is y - y1 = m(x - x1)

the value x1 is the x coordinate of a point
the value y1 is the y coordinate of a point
the value m is the slope, which is 3

therefore, x1 = -2 & y1 = 1

Then, all you do is plug in the points:

y - 1 = 3(x + 2)

NOTE:
It is 3(x + 2) instead of 3(x - 2) because the formula is (x - x1). If you plug x1 in, it is (x - (-2)). 2 negatives cancel out to be a positive, so it is (x + 2)

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The students want to make care packages for unhoused people for the winter season. They would like to put 5 boxes of tissues int
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Answer:

2250

Step-by-step explanation:

From my understanding you have 450 packs to make and you need to put 5 tissue boxes in each one so we would just simply multiply 450 packages by 5 tissue boxes each package and get 2250

5 0
3 years ago
Which is the best estimate for the area of the circle shown below?
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Answer:

c) 380 m²

Step-by-step explanation:

A = π·r²

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3 0
3 years ago
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Please give the answer
dybincka [34]

Answer:

When N is an even number, use the first equation in the matrix. When N is negative use the second equation.

Step-by-step explanation:

n = 6 | 6/2 = 3

n=5 | 3(5)+1 = 16

5 0
3 years ago
Write and equation of the translated or rotated graph in general form (picture below)
WINSTONCH [101]

Answer:

The answer is hyperbola; (x')² - (y')² - 16 = 0 ⇒ answer (a)

Step-by-step explanation:

* At first lets talk about the general form of the conic equation

- Ax² + Bxy + Cy²  + Dx + Ey + F = 0

∵ B² - 4AC < 0 , if a conic exists, it will be either a circle or an ellipse.

∵ B² - 4AC = 0 , if a conic exists, it will be a parabola.

∵ B² - 4AC > 0 , if a conic exists, it will be a hyperbola.

* Now we will study our equation:

 xy = -8

∵ A = 0 , B = 1 , C = 0

∴ B² - 4 AC = (1)² - 4(0)(0) = 1 > 0

∴ B² - 4AC > 0

∴ The graph is hyperbola

* The equation xy = -8

∵ We have term xy that means we rotated the graph about

  the origin by angle Ф

∵ Ф = π/4

∴ We rotated the x-axis and the y-axis by angle π/4

* That means the point (x' , y') it was point (x , y)

- Where x' = xcosФ - ysinФ and y' = xsinФ + ycosФ

∴ x' = xcos(π/4) - ysin(π/4) , y' = xsin(π/4) + ycos(π/4)

∴ x' = x/√2 - y/√2 = (x - y)/√2

∴ y' = x/√2 + y/√2 = (x + y)/√2

* Lets substitute x' and y' in the 1st answer

∵ (x')² - (y')² - 16 = 0

∴ (\frac{x-y}{\sqrt{2}})^{2}-(\frac{x+y}{\sqrt{2}})^{2}=

 ( \frac{x^{2}-2xy+y^{2}}{2})-(\frac{x^{2}+2xy+y^{2}}{2})-16=0

* Lets open the bracket

∴ \frac{x^{2}-2xy+y^{2}-x^{2}-2xy-y^{2}}{2}-16=0

* Lets add the like terms

∴ \frac{-4xy}{2}-16=0

* Simplify the fraction

∴ -2xy - 16 = 0

* Divide the equation by -2

∴ xy + 8 = 0

∴ xy = -8 ⇒ our equation

∴ Answer (a) is our answer

∴ The answer is hyperbola; (x')² - (y')² - 16 = 0

* Look at the graph:

- The black is the equation (x')² - (y')² - 16 = 0

- The purple is the equation xy = -8

- The red line is x'

- The blue line is y'

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Answer:

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Step-by-step explanation:

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