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Degger [83]
3 years ago
14

A painter charges $15.53 per hour, plus an additional amount for the supplies. If he made $157.80 on a job where he worked 5 hou

rs, how much did the supplies cost?
Mathematics
1 answer:
Bingel [31]3 years ago
6 0

Answer:

$80.15

Step-by-step explanation:

We can make an equation to solve this problem, that helps a lot. If h represented hours, we could make the equation 15.53h+s=157.8

Now we know that he worked 5 hours, so we can multiply 15.53 by 5 to simplify the equation.

15.53*5=77.65

So now the equation is 77.65+s=157.8.

Now, we can subtract 77.65 from both sides.

So now, our final equation is s=80.15

It cost $80.15 to buy supplies.

Hope this helps!

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Rewrite the expression in the form x
Serga [27]

Answer:

See below (I hope this helps!)

Step-by-step explanation:

When dividing exponential terms of the same base, we can simply subtract the exponents, therefore:

\frac{x^{6} }{x^{8} } = x^{(6-8)} = x^{-2}

5 0
3 years ago
Can someone please help me with this question?!? I am so confused and I don't know how to answer it.
lbvjy [14]

9514 1404 393

Answer:

  a. f(0) = 1

  b. DNE (does not exist)

  c. DNE

  d. lim = 3

Step-by-step explanation:

The function exists at a point if it is defined there. The function is defined anywhere on the solid line and at solid dots. It is not defined at open circles. So, the function is defined everywhere except (2, 3), which has an open circle.

The open circle at (0, 4) prevents the function from being doubly-defined at x=0, since it is already defined to be 1 at x=0.

This discussion tells you ...

  f(0) = 1

 f(2) does not exist. There is a "hole" in the function definition there.

__

The function has a limit at a point if approaching from the left and approaching from the right have you approaching that same point.

Consider the point (1, 2). The graph is a solid line through that point. Approaching from values less than x=1, we get to the same point (1, 2) as when we approach from values greater than x=1.

Similarly, consider the point (2, 3). Approaching from values of x less than 2, we get to the same point (2, 3) as when we approach from x-values greater than 2. The limit at x=2 is 3. The only difference from the previous case is that the function is not actually defined to be that value there.

__

Now consider what happens at x=0. When we approach from the left, we approach the point (0, 4). When we approach from the right, we approach the point (0, 1). These are different points. Because they are different coming from the left and from the right, we say "the limit as x→0 does not exist."

__

In summary, ...

  a) f(0) = 1

  b) lim x → 0 does not exist

  c) f(2) does not exist

  d) lim x → 2 = 3

_____

<em>Additional comment</em>

The significance of the function not being defined at a point where the limit exists, (2, 3), is that <em>the function is not continuous there</em>. This kind of discontinuity is called "removable", because we could make the function continuous at x=2 by defining f(2) = 3 (that is, "filling the hole").

6 0
2 years ago
During a school event, Noah is selling pizza at the food stand for $1.25 per slice.
Eddi Din [679]

Answer:

1. When he sold 20 slices of pizza at $1.25 each, he made $20

  When he sold 50 slices of pizza at $1.25 each, he made $62.5

2. 96 slices of pizza. You would get that by dividing 120 by 1.25, as he needs to make $120, and each slice costs $1.25

7 0
3 years ago
25 divided by 5<br> IF you get it right I will email you!!!
vlabodo [156]

Answer:

5

Step-by-step explanation:

25/5

5×5=25

5 <---answer

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3 years ago
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16r^2+ 2r -4r 16r^2 - 2r 2r ( 8r -1)
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