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3 years ago

50 POINTS AND BRAINLEST Which line is the graph of y=-2x+2

Mathematics

2 answers:

Mariulka [41]3 years ago

8 0

Answer: Line B sir

Step-by-step explanation:

y = -2x+2 has a slope of -2 (means it's going down) and a y-intercept of 2. Line b is the only line that matches this description.

worty [1.4K]3 years ago

6 0

Answer:

line b

Step-by-step explanation:

y = -2x+2 has a slope of -2 (means it's going downhill) and a y-intercept of 2. Line b is the only line that matches this description.

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What is the interquartile range of this data set. 4,5,7,9,10,14,16,24

ehidna [41]

Answer:

The interquartile range is 10.

Step-by-step explanation:

(xu-xl)=10

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3 years ago

Which function has a domain of all real numbers except x=pi/2+-npi?

OLga [1]

It is rather obvious that we will have to deal with a trigonometric function. Let us see... the first hint is that a function is not defined somewhere. cos and sin are defined for all real numbers... what about tan though? Well, as we approach pi/2, tan becomes larger and larger and it is not defined at pi/2. Given that the tan function has a period of pi (cos and sin have a period of 2pi), this function is the right candidate. Since it is periodic and it is not defined at pi/2, it cannot be defined at any x=pi/2+-npi (if it was defined for some n, then it would have the same value at pi/2 due to periodicity). Hence, a function that works is tanx, or 2*tanx or tanx+6.

8 0

3 years ago

Read 2 more answers

Write the equation of the circle below to complete the square. Also determine the area of the circle.

diamong [38]

Answer: $(x-7)^2+(y+2)^2=11^2$ ; $A=121\pi$

Step-by-step explanation:

I have never done a problem like this, so this is my first time attempting it.

The standard form for the equation of a circle is:

$(x-a)^2+(y-b)^2=r^2$

Begin by grouping the 'x' terms and 'y' terms:

$(x^2-14x)+(y^2+4y-68)=0$

Complete the square for the second polynomial:

$(\frac{b}{2} )^2=(\frac{4}{2} )^2=4$

$(x^2-14x)+(y^2+4y+4-4-68)=0$

$(x^2-14x)+(y^2+4y+4)-4-68=0$

$(x^2-14x)+(y^2+4y+4)-72=0$

$(x^2-14x)+(y+2)^2-72=0$

Do the exact same process with the other polynomial:

$(x^2-14x-72)+(y+2)^2=0$

Complete the square:

$(\frac{14}{2} )^2=49$

$(x^2-14x+49-49-72)+(y+2)^2=0$

$(x^2-14x+49)-49-72+(y+2)^2=0$

$(x^2-14x+49)-121+(y+2)^2=0$

$(x-7)^2-121+(y+2)^2=0$

Bring the 121 to the other side to get:

$(x-7)^2+(y+2)^2=121$

$(x-7)^2+(y+2)^2=11^2$

The above expression is the equation of our circle in standard form. It is centered at the coordinates (7, -2) and has a radius of 11 units. The area of a circle if given by:

$A=\pi r^2=121\pi$