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marishachu [46]
2 years ago
9

4. CM Manufacturing has provided the following unit costs pertaining to a component they manufacture

Mathematics
1 answer:
TEA [102]2 years ago
3 0

uhh sorry i dont know

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To construct a line parallel to a given line m to a point not on m, you need to know how to construct_______angles?
Lelu [443]
It would be either C or A, sorry I cant help anymore than that
6 0
3 years ago
Read 2 more answers
The step by step and final answer
Mumz [18]
32. (a) For an even function, f(x) = f(-x). Given f(5) = 3, we know f(-5) = 3.
Therefore (-5, 3) is also on the graph.

For an odd function, f(-x) = -f(x). Given f(5) = 3, we know f(-5) = -3.
Therefore (-5, -3) is also on the graph.

33. f(-x) = -f(x). The function is odd.
34. f(-x) = x/(x-1) ≠ -f(x) ≠ f(x). The function is neither even nor odd.
35. f(-x) = f(x). The function is even.

4 0
3 years ago
If A=3x^2+5x-6 and B= -2x^2-6x=7, then A-B equals
Kryger [21]
Yup ur right les goo
6 0
2 years ago
8x2 − 5 + 7x4 − 9x − x5
Veseljchak [2.6K]

Answer:

=-\left(x-1\right)\left(x^4-6x^3-6x^2-14x-5\right)

Step-by-step explanation:

Factor out comon Term -1

=-\left(x^5-7x^4-8x^2+9x+5\right)

Factor x^5-7x^4-8x^2+9x+5:\quad \left(x-1\right)\left(x^4-6x^3-6x^2-14x-5\right)

x^5-7x^4-8x^2+9x+5

Use the rational root theorem

a_0=5,\:\quad a_n=1

The dividers of a_{0}: 1, 5, The dividers of a_n: 1

Therefore, check the following rational numbers: ±\frac{1,\:5}{1}

\frac{1}{1} is a root of the expression, so factor out x-1

=\left(x-1\right)\frac{x^5-7x^4-8x^2+9x+5}{x-1}

\frac{x^5-7x^4-8x^2+9x+5}{x-1}=x^4-6x^3-6x^2-14x-5

=x^4-6x^3-6x^2-14x-5

=-\left(x-1\right)\left(x^4-6x^3-6x^2-14x-5\right)

8 0
3 years ago
Which expression is equivalent to
lutik1710 [3]

Answer:

\frac{\sqrt[4]{3x^2} }{2y}

Step-by-step explanation:

We can simplify the expression under the root first.

Remember to use  \frac{a^x}{a^y}=a^{x-y}

Thus, we have:

\sqrt[4]{\frac{24x^{6}y}{128x^{4}y^{5}}} \\=\sqrt[4]{\frac{3x^{2}}{16y^{4}}}

We know 4th root can be written as "to the power 1/4th". Then we can use the property  (ab)^{x}=a^x b^x

<em>So we have:</em>

<em>\sqrt[4]{\frac{3x^{2}}{16y^{4}}} \\=(\frac{3x^{2}}{16y^{4}})^{\frac{1}{4}}\\=\frac{3^{\frac{1}{4}}x^{\frac{1}{2}}}{2y}\\=\frac{\sqrt[4]{3x^2} }{2y}</em>

<em />

<em>Option D is right.</em>

8 0
3 years ago
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