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dem82 [27]
3 years ago
13

Oliver is determining the number of cats in his math class. Oliver is collecting what kind of data?

Mathematics
2 answers:
Burka [1]3 years ago
6 0

Answer:

quantitative data

Step-by-step explanation:

lianna [129]3 years ago
3 0
Oliver is collecting Quantitative data , is that a choice?
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Dora buys one package each of 1- pound, 2- pound, and 4-pound packages if ground beef to make hamburgers. How many 1/4-pound ham
Alisiya [41]
82. Dora bought one package of each 1-pound pork, 2-pound pork and 4-pound pork. Thus, she got a total of: => 1 pound + 2 pounds + 4 pounds = 7 pounds of pork. Question: How many ¼ pound of hamburger she can make then with this given number of pork in pounds. => ¼ = 1 / 4 = .25 Now, let’s divide 7 pounds by .25 pounds => 7 / .25 = 28 Thus, She can make 28 hamburgers in all.
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3 years ago
Write a polynomial equation of least degree with roots 2, -1 and 4.
Rasek [7]
A polynomial with roots a and b is (x - a)(x - b).

(x - 2)(x - (-1))(x - 4) = 0

(x - 2)(x + 1)(x - 4) = 0

has roots 2, -1, and 4.
3 0
2 years ago
48 divided by blank =6
poizon [28]

Answer:

i think ANSWER IS 8

Step-by-step explanation:

48÷8 = 6

6 0
2 years ago
Read 2 more answers
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
Q. Which graph is linear?
Sophie [7]

Answer:

d trust me

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
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