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Sergeu [11.5K]
3 years ago
6

What is 1/3 divided by 2

Mathematics
1 answer:
Sunny_sXe [5.5K]3 years ago
8 0

Answer:

0.16666666666 so just put 0.16

Step-by-step explanation:

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2. Write 7x2 - 4 + 6x3 - 4x - x4 in standard form.
Hunter-Best [27]

Answer:

16-5x

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
After John worked at a job for 10 years, his salary doubled. If he started at $x, hls salary after 10 years is​
Sholpan [36]

Answer:

$2x

Step-by-step explanation:

  1. x × 2 = 2x

I hope this helps!

4 0
3 years ago
Evaluate expression 8s-5t+s2;s=4,t=1
jeka57 [31]
8(4)-5(1)+2(4)
Multiply
32-5+8
Solve from left to right
28+8
36

The answer is 36
4 0
2 years ago
I WILL MARK BRAINIEST<br> I got one correct there should be about one more that’s also right
Lesechka [4]

So essentially an adjacent angle is when two angles have a common side and a common vertex.

So knowing that we can determine from the picture all the adjacent angles:

D) <2 and <3 are adjacent angles

E) <1 and <3 are adjacent angles

F) <1 and <5 are adjacent angles

I hope this helps.

7 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
2 years ago
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