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Elenna [48]
3 years ago
8

Your teacher says he will give you 2200 seconds to complete the practice quiz, how long is this in hours? Express your answer ro

unded to the nearest hundredth of an hour.
Mathematics
1 answer:
Rama09 [41]3 years ago
7 0
0.611 hour

hope this helps!
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What is the value of x in the equation shown below<br> 0.75x + 3 = 6(x - 3)
EleoNora [17]

Answer: x=4

Step-by-step explanation: 0.75x+3=6(x−3)

0.75x+3=(6)(x)+(6)(−3)

0.75x+3=6x+−18

0.75x+3=6x−18

0.75x+3−6x=6x−18−6x

−5.25x+3−3=−18−3

−5.25x /−5.25 =− 21 /−5.25

x=4

7 0
3 years ago
Find a linear equation in your life?
Crank

My life is not linear. Time is not constant. My life (and time included) is just a series of times that do not run parallel to each other. Instead, they drift further or closer as a function of velocity and curvature of space time.

5 0
3 years ago
Read 2 more answers
Factor completely 2x^2 − 2x − 40. 2(x − 5)(x 4) (2x − 10)(x 4) (x − 5)(2x 8) 2(x − 4)(x 5)
Zina [86]
2x^2 - 2x - 40 = 2x^2 -10x + 8x - 40 = 2x(x - 5) + 8(x - 5)

= (x-5)(2x+8) = 2(x-5)(x+4)
3 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
Joe works on an assembly line that fills boxes with various toy trinkets. Every sixth box gets a red ball and every fourt box ge
Genrish500 [490]
Every 12th box gets a whistle and a ball because 12 is the lowest common factor of 6 and 4.
6 0
3 years ago
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