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Klio2033 [76]
2 years ago
6

Can someone help me with this

Mathematics
1 answer:
Art [367]2 years ago
5 0

Ans57 , 58 , 59

Step-by-step explanation:

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Simplify completely 12 times x to the third power minus 4 times x to the 2nd power plus 8 times x all over negative 2 times x.
Alenkinab [10]
The answer is: -6x^2+2x-4
5 0
3 years ago
Question was deleted
Nikolay [14]

Answer:

uh okay I'm just stumbling upon questions like this

7 0
3 years ago
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Two integers, a and b, have a product of 36. What is the least possible sum of a and b?
Minchanka [31]
<h3>Answer:  12</h3>

==============================================================

Explanation:

The two integers multiply to 36, so,

ab = 36

which solves to

a = 36/b

Then we want to add the numbers such that we get the smallest possible result.

a+b = (36/b)+b

So we want (36/b)+b to be as small as possible.

Let's say we replace b with x and we consider this function

f(x) = (36/x) + x

The goal is to find when f(x) is smallest, ie, we want to minimize the function.

If we were to graph out the function, we get the curve shown below.

To make things easier, we'll only focus on positive values of x.

The lowest part of the curve is what we're after. Using the "minimum" function/feature on the graphing calculator, we would then find the lowest point occurs at (6,12). This point is considered a local minimum because it's the lowest point in that given neighborhood of x values.

So the input x = 6 leads to the smallest output f(x) = 12.

This in turn means b = 6 is going to pair with a = 36/b = 36/6 = 6.

In short, a = 6 and b = 6.

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

As a check,

a*b = 6*6 = 36

a+b = 6+6 = 12

We can make a table of various values to help confirm that 12 is the smallest sum.

Side note: If you're not allowed to use a graphing calculator, then you'll need to use calculus.

8 0
3 years ago
AABC has vertices at A(1, -9), B(8,0), and C(9,-8).
Rainbow [258]

Check the picture below.

~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ A(\stackrel{x_1}{1}~,~\stackrel{y_1}{-9})\qquad B(\stackrel{x_2}{8}~,~\stackrel{y_2}{0}) ~\hfill AB=\sqrt{[ 8- 1]^2 + [ 0- (-9)]^2} \\\\\\ AB=\sqrt{7^2+(0+9)^2}\implies AB=\sqrt{7^2+9^2}\implies \boxed{AB=\sqrt{130}} \\\\[-0.35em] ~\dotfill

B(\stackrel{x_1}{8}~,~\stackrel{y_1}{0})\qquad C(\stackrel{x_2}{9}~,~\stackrel{y_2}{-8}) ~\hfill BC=\sqrt{[ 9- 8]^2 + [ -8- 0]^2} \\\\\\ BC=\sqrt{1^2+(-8)^2}\implies \boxed{BC=\sqrt{65}}

now, we could check for the CA distance, however, we already know that AB ≠ BC, so there's no need.

6 0
3 years ago
A student used his place value chart to show a number. After the teacher instructed him to divide his number by 100,the chart sh
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When you divide by 100 you are essentially moving the decimal of the number two places to the left. In undoing this you would have to move the decimal of the number two places to the right.

28.003 would then turn into 2,800.3

Unfortunately I cannot draw a chart on here but that is the best I can do.

5 0
3 years ago
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