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julia-pushkina [17]
3 years ago
7

Misty took a multiple choice test in science that had 50 questions. If the relationship between the number she got correct and t

he number she got incorrect is 7:3, and she got 9 incorrect, how many did she get correct?
Mathematics
1 answer:
grandymaker [24]3 years ago
6 0
She got 41 right because she only missed 9
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Dan has the same number of nickels and dimes in his piggy bank. If the total amount of those coins is $0.90, how many of each co
SpyIntel [72]

Answer:

6.

Step-by-step explanation:

We can start by defining a couple things.

x = amount of nickels

y = amount of dimes

Since he has the same amount of each, we can say this.

x = y

And because we know it totals 0.90, we can say this.

0.05x + 0.10y = 0.90

then, we substitute and get this.

0.05x + 0.10x = 0.90

Simplify.

0.15x = 0.90

Divide both sides by 0.15.

x = 6

So we have 6 nickels, and they both are the same, so we have 6 dimes.

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3 years ago
Anyone have spare food my children are hungry down there in the basement ):
statuscvo [17]

Answer:

Nope, sorry.

Step-by-step explanation:

6 0
3 years ago
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A toy car can go 3 mph. How long would it take to go 10 miles?
alisha [4.7K]
How long does it take to go 3mph ?
3 0
3 years ago
2+2 pls help quick or ill fail sike awnser 7/3x = 2/3
kompoz [17]

Answer:

21/3 i think

Step-by-step explanation:

7 0
2 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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