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Fittoniya [83]
3 years ago
7

Jose earns 954 per hour at his job he worked 18 hours last week calculate Jose's pay before taxes​

Mathematics
1 answer:
PSYCHO15rus [73]3 years ago
6 0

Answer:

$171.72

Step-by-step explanation:

$9.54 x 18

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5x - 2 equivalent to to  2x−2+3x

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If a is an odd integer and b is an even integer, which of the following makes an odd integer?
Serga [27]

9514 1404 393

Answer:

  D.) a+2b

Step-by-step explanation:

The integers 'a' and 'b' can be any, so you can choose a couple and evaluate these expressions to see what you get. For example, we can let a=1 and b=0. For these values, the offered expressions evaluate to ...

  A) 3(0) = 0 . . . even

  B) 1 +3 = 4 . . . even

  C) 2(1+0) = 2 . . . even

  D) 1 +2(0) = 1 . . . odd

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<em>Additional comment</em>

These rules apply to even/odd:

  • odd × odd = odd
  • odd × even = even
  • even × even = even
  • odd + odd = even
  • odd + even = odd
  • even + even = even

Then A is (odd)(even) = even; B is (odd)+(odd) = even; C is (even)(whatever) = even; D = (odd)+(even) = odd.

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3 years ago
PLEASE HELP!!!
Morgarella [4.7K]

Answer:

A

Step-by-step explanation:

mid point of AC=((2m+2p)/2,(2n+2r)/2)=(m+p,n+r)

4 0
4 years ago
Describe the behavior of the function ppp around its vertical asymptote at x=-2x=−2x, equals, minus, 2. ​
insens350 [35]

Answer:

x->-2^{-}, p(x)->-\infty and as x->-2^{+}, p(x)->-\infty

Step-by-step explanation:

Given

p(x) = \frac{x^2-2x-3}{x+2} -- Missing from the question

Required

The behavior of the function around its vertical asymptote at x = -2

p(x) = \frac{x^2-2x-3}{x+2}

Expand the numerator

p(x) = \frac{x^2 + x -3x - 3}{x+2}

Factorize

p(x) = \frac{x(x + 1) -3(x + 1)}{x+2}

Factor out x + 1

p(x) = \frac{(x -3)(x + 1)}{x+2}

We test the function using values close to -2 (one value will be less than -2 while the other will be greater than -2)

We are only interested in the sign of the result

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

As x approaches -2 implies that:

x -> -2^{-} Say x = -3

p(x) = \frac{(x -3)(x + 1)}{x+2}

p(-3) = \frac{(-3-3)(-3+1)}{-3+2} = \frac{-6 * -2}{-1} = \frac{+12}{-1} = -12

We have a negative value (-12); This will be called negative infinity

This implies that as x approaches -2, p(x) approaches negative infinity

x->-2^{-}, p(x)->-\infty

Take note of the superscript of 2 (this implies that, we approach 2 from a value less than 2)

As x leaves -2 implies that: x>-2

Say x = -2.1

p(-2.1) = \frac{(-2.1-3)(-2.1+1)}{-2.1+2} = \frac{-5.1 * -1.1}{-0.1} = \frac{+5.61}{-0.1} = -56.1

We have a negative value (-56.1); This will be called negative infinity

This implies that as x leaves -2, p(x) approaches negative infinity

x->-2^{+}, p(x)->-\infty

So, the behavior is:

x->-2^{-}, p(x)->-\infty and as x->-2^{+}, p(x)->-\infty

6 0
3 years ago
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