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Luden [163]
3 years ago
14

Image attached Please help asap

Mathematics
1 answer:
bearhunter [10]3 years ago
6 0
First question: 2x = 50
Explanation:
“2 times as many” in this situation means that the pages justin read (50) is twice as Colin’s. Meaning that to find the number of pages Colin’s read, we will have to divide 50 by 2.
2x = 50 is an equivalent representation of that.

2nd question:
(Choose these correct statements):
• 9h represents how much money Sofia earns babysitting for h hours.
• Sofia babysat for 3 hours.
• If Sofia babysits for 5 hours, she will earn $45

Have a nice day (brainliest please)
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Answer:

I can't understand this that well could you take a picture of the question

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Can some explain steps​
Mashutka [201]

Answer:

r = 3.

Step-by-step explanation:

16 = 10 + √(3r + 27)

√(3r + 27) = 6

Square both sides:

3r + 27 = 36

3r = 36 - 27 = 9

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Left side of the equation = 16

Right side = 10 + √(9 + 27)

                  = 10 + √36 = 16

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2 years ago
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The ratio of boys to girls in a group is 7:1. If there are 66 more boys than girls, work out how many girls there are
nevsk [136]

Answer:11

Step-by-step explanation:

Boys : girls=7:1

Sum of ratio=7+1=8

Let them number of girls be y

Then they number of boys=y+66

Total number of pupils=y+y+66

Total number of pupils=2y+66

Number of girls=(girls ratio)/(sum of ratio) x (total number of pupils)

y=1/8 x (2y+66)

Cross multiply

y x 8=2y + 66

8y=2y + 66

Collect like terms

8y-2y=66

6y=66

Divide both sides by 6

6y/6=66/6

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The number of girls is 11

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2 years ago
You complete 40% of your homework problems before dinner. what fraction of the problems did you complete before dinner?
Rzqust [24]

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4 0
3 years ago
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Consider the following function.
Kryger [21]

Answer:

See below

Step-by-step explanation:

I assume the function is f(x)=1+\frac{5}{x}-\frac{4}{x^2}

A) The vertical asymptotes are located where the denominator is equal to 0. Therefore, x=0 is the only vertical asymptote.

B) Set the first derivative equal to 0 and solve:

f(x)=1+\frac{5}{x}-\frac{4}{x^2}

f'(x)=-\frac{5}{x^2}+\frac{8}{x^3}

0=-\frac{5}{x^2}+\frac{8}{x^3}

0=-5x+8

5x=8

x=\frac{8}{5}

Now we test where the function is increasing and decreasing on each side. I will use 2 and 1 to test this:

f'(2)=-\frac{5}{2^2}+\frac{8}{2^3}=-\frac{5}{4}+\frac{8}{8}=-\frac{5}{4}+1=-\frac{1}{4}

f'(1)=-\frac{5}{1^2}+\frac{8}{1^3}=-\frac{5}{1}+\frac{8}{1}=-5+8=3

Therefore, the function increases on the interval (0,\frac{8}{5}) and decreases on the interval (-\infty,0),(\frac{8}{5},\infty).

C) Since we determined that the slope is 0 when x=\frac{8}{5} from the first derivative, plugging it into the original function tells us where the extrema are. Therefore, f(\frac{8}{5})=1+\frac{5}{\frac{8}{5}}-\frac{4}{\frac{8}{5}^2 }=\frac{41}{16}, meaning there's an extreme at the point (\frac{8}{5},\frac{41}{16}), but is it a maximum or minimum? To answer that, we will plug in x=\frac{8}{5} into the second derivative which is f''(x)=\frac{10}{x^3}-\frac{24}{x^4}. If f''(x)>0, then it's a minimum. If f''(x), then it's a maximum. If f''(x)=0, the test fails. So, f''(\frac{8}{5})=\frac{10}{\frac{8}{5}^3}-\frac{24}{\frac{8}{5}^4}=-\frac{625}{512}, which means (\frac{8}{5},\frac{41}{16}) is a local maximum.

D) Now set the second derivative equal to 0 and solve:

f''(x)=\frac{10}{x^3}-\frac{24}{x^4}

0=\frac{10}{x^3}-\frac{24}{x^4}

0=10x-24

-10x=-24

x=\frac{24}{10}

x=\frac{12}{5}

We then test where f''(x) is negative or positive by plugging in test values. I will use -1 and 3 to test this:

f''(-1)=\frac{10}{(-1)^3}-\frac{24}{(-1)^4}=-34, so the function is concave down on the interval (-\infty,0)\cup(0,\frac{12}{5})

f''(3)=\frac{10}{3^3}-\frac{24}{3^4}=\frac{2}{27}>0, so the function is concave up on the interval (\frac{12}{5},\infty)

The inflection point is where concavity changes, which can be determined by plugging in x=\frac{12}{5} into the original function, which would be f(\frac{12}{5})=1+\frac{5}{\frac{12}{5}}+\frac{4}{\frac{12}{5}^2 }=\frac{43}{18}, or (\frac{12}{5},\frac{43}{18}).

E) See attached graph

5 0
3 years ago
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