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pishuonlain [190]
3 years ago
5

He earns a fixed weekly salary of $300 plus a fee of $5 for each car he parks. His potential earnings for a week are shown in th

e graph. At what point does he begin to earn more from fees than from his fixed salary?
Mathematics
1 answer:
tamaranim1 [39]3 years ago
3 0
I think what you are looking for is 5x=300 therefore x=60 so at 60cars he would start to make more.
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Please answer 4 - 9 will mark brainlyist
miv72 [106K]

There are 180 degrees in a triangle

4. 38+38= 76 degrees

180-76=104

Isosceles, Obtuse

5. 65+ x+x=65

Well we know one of the x

It is 90 degrees because it is a right angle

65+90=155

180-155=25

Scalene, Right triangle

6. 71+45= 116

180-116=64

7.  Scalene, acute

8. 126+22=148

180-148=32

Scalene, Obtuse

9. 50+90=140

180-140=40

Right angle,  Scalene

Please let me know if i got anything wrong

8 0
3 years ago
Simplify.<br> (-8V2)(+1/2V32)
erastovalidia [21]

Hello!!

\rule{300}{5}

\huge\blue{{\fbox{\tt{ANSWER:-\:}}}}

-32

\huge\red{{\underline{\overline{\textbf{MY\:EXPLANATION~!}}}}}

Apply rule:  (-a) = -a

(-8\sqrt{2}) = - 8\sqrt{2}

= -8\sqrt{2}~\frac{1}{2}\sqrt{32}

\sqrt{2}~\sqrt{32} =\sqrt{64}

= -8\sqrt{64}~\frac{1}2}

\sqrt{64} = 8

= -8 · 8 · \frac{1}{2}

Multiply the numbers: -~8 · 8= -64

=-64 · \frac{1}{2} = -32

= -32

Hope It Helps. . .

#LearnWithBrainly

\huge\blue{{\fbox{\tt{ANSWER:-\:}}}}

\bold{Jace\;} ^-^

6 0
3 years ago
(3x-5)/(x-5)&gt;=0 <br> How do I get the answers for this problem?
Diano4ka-milaya [45]
\frac{3x-5}{x-5} > 0 

First, note that x is undefined at 5. / x ≠ 5
Second, replace the inequality sign with an equal sign so that we can solve it like a normal equation. / Your problem should look like: \frac{3x-5}{x-5} = 0
Third, multiply both sides by x - 5. / Your problem should look like: 3x - 5 = 0
Forth, add 5 to both sides. / Your problem should look like: 3x = 5
Fifth, divide both sides by 3. / Your problem should look like: x = \frac{5}{3}
Sixth, from the values of x above, we have these 3 intervals to test: 
x < \frac{5}{3}
\frac{5}{3} < x < 5
x > 5
Seventh, pick a test point for each interval. 

1. For the interval x < \frac{5}{3} :
Let's pick x - 0. Then, \frac{3x0-5}{0-5} > 0
After simplifying, we get 1 > 0 which is true.
Keep this interval.

2. For the interval \frac{5}{3} < x < 5:
Let's pick x = 2. Then, \frac{3x2-5}{2-5} > 0
After simplifying, we get -0.3333 > 0, which is false.
Drop this interval.

3. For the interval x > 5:
Let's pick x = 6. Then, \frac{3x6-5}{6-5} > 0
After simplifying, we get 13 > 0, which is ture.
Keep this interval.
Eighth, therefore, x < \frac{5}{3} and x > 5

Answer: x < \frac{5}{3} and x > 5

3 0
4 years ago
Represent 42 divided by 6 = 7 using subtraction
GuDViN [60]

Answer:

42 - 6 - 6 - 6 - 6 - 6 - 6 - 6 = 0

or

42 - 7 - 7 - 7 - 7 - 7 - 7 = 0

5 0
3 years ago
Use the formula to find the value of $400 invested in 4%
Anna [14]

Answer:

Some part of the question is missing , you are requested to kindly recheck it once. There must be some time provided in the problem

Step-by-step explanation:

3 0
3 years ago
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