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Dahasolnce [82]
3 years ago
9

Identify the y-intercept (initial value) in the function

Mathematics
1 answer:
Nina [5.8K]3 years ago
8 0

Answer: 13

Step-by-step explanation:

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Simplify your answers to the simplest form with steps
yan [13]
4[2a+b]/c[b+2a] = 4/c.
7 0
3 years ago
Read 2 more answers
N is the centriod of triangle. Find XN if XG = 33​
Maksim231197 [3]

Answer:

22

Step-by-step explanation:

The centroid divides a median in two parts that have this ratio = 1/3 and 2/3

In particular the part between the vertex and the centroid is 2/3 of the median.

So we have:

XN = (33 * 2)/3 = 22

3 0
3 years ago
Find an equation of the set of all points equidistant from the points A(−3, 6, 3) and B(4, 1, −1). Describe the set.
valentinak56 [21]

Answer:

See below

Step-by-step explanation:

I will describe this set in R³. Let P=(x,y,z) be a point equidistant to A and B, that is, the distance from P to A is equal to the distance from P to B.

First, using the usual distance formula, the distance from P to A is equal to d(P,A)=\sqrt{(x-(-3))^2+(y-6)^2+(z-3)^2}=\sqrt{(x+3)^2+(y-6)^2+(z-3)^2}

On the other hand, the distance form P to B is equal to d(P,B)=\sqrt{(x-4)^2+(y-1)^2+(z-(-1))^2}=\sqrt{(x-4)^2+(y-1)^2+(z+1)^2}

P is equidistant from A and B if and only if P satisfies the equation d(P,A)=d(P,B), that is,

\sqrt{(x+3)^2+(y-6)^2+(z-3)^2}=\sqrt{(x-4)^2+(y-1)^2+(z+-1)^2}

Take the square in both sides of this equation to get

(x+3)^2+(y-6)^2+(z-3)^2=(x-4)^2+(y-1)^2+(z+1)^2

(x+3)^2-(x-4)^2+(y-6)^2-(y-1)^2+(z-3)^2-(z+1)^2=0

You can simplify using difference of squares and multiplying like this:

(7)(2x-1)+(-5)(2y-7)+(-4)(2z-2)=0

14x-10y-8z+36=0

which is the equation of a plane.

4 0
4 years ago
Lesson 20 5h grade homework
LUCKY_DIMON [66]
1igjgjyuhr55cysrhgu8rsyggshrborbgxm
5 0
3 years ago
Ive tried over and over and i still cant figure it out some help pls lol
galina1969 [7]
The answer is A because the volume of a cylinder formula is 3.14 time radius squared times the height of the cylinder
6 0
2 years ago
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