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Len [333]
3 years ago
6

Can someone help me out on questions 6, 7, and 8?

Mathematics
1 answer:
PolarNik [594]3 years ago
6 0
For #6 draw a triangle between the points and find the middle by drawing the lines to the middle. There should be triangles within the triangles which finding the hypotenuses should be the same.
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50 points! The campaign manager for one of the five candidates running for office in a city wants to conduct a survey to see how
luda_lava [24]

Answer:

B

Step-by-step explanation:

He chose the names on the voter registration list radomly

4 0
2 years ago
Urgent help needed. Will mark you brainlest when correct
posledela

Question 7: Option (4)

A=(3/8)^2=\frac{9}{64}

Question 8: Option (3)

A=\frac{1}{2}(17)(9)=76.5

Question 9: Option (2)

A=\pi(10)^2 \approx 314

7 0
2 years ago
micheal pays 30 to enter a state fair, plus 4 for each ride. which of the following equations represent his total cost
Shkiper50 [21]

Answer:

I cant see the equation but it should go something along the lines of t = 30 + 4r

Step-by-step explanation:

t = total

r = number of rides

Example: If he went on 4 rides, hen u would imput 4 into r like this

t = 30 + 4 (4)

t = 30 + 16

t = 46

8 0
3 years ago
10 points, please help
weqwewe [10]

Answer:

88.5

Step-by-step explanation:

(90+82+81+2x)/5=86

90+82+81+2x=430

2x=177

x=88.5

3 0
2 years ago
Can someone help me with this? I need to find the points of discontinuity/limits for each of these. I think one point is 4, but
Debora [2.8K]
The answers are shown in the attached image

-------------------------------------------------------------------------

Explanation:

Set the denominator x^4-8x^3+16x^2 equal to zero and solve for x

x^4-8x^3+16x^2 = 0
x^2(x^2-8x+16) = 0
x^2(x-4)^2 = 0
x^2 = 0 or (x-4)^2 = 0
x = 0 or x-4 = 0
x = 0 or x = 4

The x values 0 and 4 make the denominator zero

These x values lead to asymptote discontinuities because the numerator 8x-24 = 8(x-3) has no common factors which cancel with the denominator factors.

There are two vertical asymptotes

Let's see what happens when we plug in a value to the left of x = 0, say x = -1, we'd get
f(x) = (8x-24)/(x^4-8x^3+16x^2)
f(-1) = (8(-1)-24)/((-1)^4-8(-1)^3+16(-1)^2)
f(-1) = -1.28
So as x gets closer and closer to x = 0 from the left side, the f(x) is heading to negative infinity

Now plug in some value to the right of x = 0. I'm going to pick x = 1
f(x) = (8x-24)/(x^4-8x^3+16x^2)
f(1) = (8(1)-24)/((1)^4-8(1)^3+16(1)^2)
f(1) = -1.78 (approximate)
So as x gets closer and closer to x = 0 from the right side, the f(x) is heading to negative infinity

Overall, as x approaches 0 from either the left or right side of x = 0, the y value is heading off to negative infinity

---------------------

Repeat for values to the left and right of x = 4
We can't use x = 1 as it turns out that x = 3 is a root
But we can use something like x = 3.5 to find that...
f(x) = (8x-24)/(x^4-8x^3+16x^2)
f(3.5) = (8(3.5)-24)/((3.5)^4-8(3.5)^3+16(3.5)^2)
f(3.5) = 1.31 approx
So as x gets closer to x = 4 from the left, y is getting closer to positive infinity

Plug in x = 5 to find that
f(x) = (8x-24)/(x^4-8x^3+16x^2)
f(5) = (8(5)-24)/((5)^4-8(5)^3+16(5)^2)
f(5) = 0.64
which has the same behavior as the left side

So overall, as we approach x = 4, the y value is heading off to positive infinity

Again everything is summarized in the image attachment

Note: you could make a table of more values but they would effectively say what has already been said. It would be redundant busy work. However, its always good practice for function evaluation. 

6 0
3 years ago
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