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grin007 [14]
3 years ago
9

Show 65% as a decimal and a fraction. Show your work

Mathematics
1 answer:
xxTIMURxx [149]3 years ago
5 0
65=.65
65 is 65 pieces out of 100 so 65/100
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When dividing 878 by 31 a student finds a quotient of 28 with a remainder of 11 check the students work and use the check to fin
sveticcg [70]
878 ÷ 31 = 28 with a remainder of 10
7 0
3 years ago
Find the missing dimension of each triangle described height 14 inches area 245 inches Square
Amiraneli [1.4K]
Given:
Triangle 
height 14 inches
area 245 inches square

Formula in finding the area of a triangle is:

Area = (height * base) / 2

The base is missing, so we need to compute its value using the given figures.

245 = (14 * b) / 2
245 * 2 = 14b
490 = 14b
490/14 = b
35 = b

The base is 35 inches.
4 0
3 years ago
If the speed limit of a highway is 60 mph, which car exceeded the speed limit? Assume all the cars were traveling at a constant
JulsSmile [24]
There was only one car that went over the speed limit and it's car A (183). the other cars were all under and how i found this out was take 60 mph x hours n that's the miles they should of drove n that many hours if they are above it the miles will be higher.
7 0
3 years ago
Read 2 more answers
1. Which of the following is a perfect square number? A. 24 B. 125 C. 121 D. 8 20​
zepelin [54]

Answer:

125 is the square of 25 I hope

6 0
2 years ago
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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