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Vsevolod [243]
3 years ago
9

Helppp btw the number are (0,5) (2,2) (3,1) (4,-1)

Mathematics
1 answer:
boyakko [2]3 years ago
4 0
Your answer is (3,1) because if you graph the line that is the on off the line the other follow through
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Which of the following describes an example of mutually exclusive outcomes A and B for throwing a single 6-sided
swat32
Answer:
A=the outcome is odd, B=the outcome is even
Is mutually exclusive because you can't have a number that is even and odd at the same time.
Hope this helps :)
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2 years ago
Which of the following lines has a negative slope?
Furkat [3]

Answer:

The second option, or the one attached.

Step-by-step explanation:

       This line has a negative slope because the line is going "down the stairs" from the top-left to the bottom-right.

6 0
2 years ago
Read 2 more answers
Please help me with my Algebra homework!!!
balu736 [363]

Answer:

B) No Unique Solutions

Step-by-step explanation:

Given:

z=x+5y-14

-2y=-4\\y={-4}{-2}=2

3x+y-3z=14

now we now the value of y = 2 so we will substitute in equation  z=x+5y-14

z=x+5\times2-14\\z=x+10-14\\z= x-4

Now, We have value of y=2 and value of z=x-4 Substituting in both the value in equation 3x+y-3z=14

3x+2-3(x-4)=14\\3x+2-3x+12=14\\14=14

which means that the equation has Infinite solutions.

7 0
3 years ago
Which of the following is equal to tan(A)?
Scrat [10]

Answer: cot B

Step-by-step explanation: if i am right mark me as brainliest

8 0
3 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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