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Reptile [31]
4 years ago
10

Which of the following is a composite number?

Mathematics
1 answer:
Anuta_ua [19.1K]4 years ago
7 0

B. 91 is composite because it can be divided by 7 and 13.

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F(9) if f (x)=5x-16, what is the answer to the math problem
Anastasy [175]
F(9)
sub 9 forx
f(9)=5(9)-16
f(9)=45-16
f(9)=29

f(9)=29
3 0
3 years ago
A triangle has sides with lengths of 3 feet, 6 feet, and 8 feet is it a right triangle?
Zielflug [23.3K]

Answer:

Solution,\\Longest~side(h) = 8ft.\\Perpendicular(p) = 3ft.\\Base(b) = 6ft.\\Now, \\h^2 = (8)^2 = 64sq.ft.\\p^2+b^2 = 3^2+6^2 = 9+36 = 45sq.ft.\\Since, h^2\neq p^2+b^2, ~the~triangle~is~not~a~right~triangle.

8 0
3 years ago
What is the remainder R when the polynomial p(x) is divided by (x - 2)? Is (x - 2) a factor of p(x)?
puteri [66]

Answer:

The answer can be calculated by doing the following steps;

Step-by-step explanation:

5 0
3 years ago
Explain how to use the vertex and the value of “A” to determine the range of an absolute value function. PLEASE HELP!!
Thepotemich [5.8K]

Answer:

First, a absolute value function is something like:

y = f(x) = IxI

remember how this work:

if x ≥ 0, IxI = x

if x ≤ 0, IxI = -x

Notice that I0I = 0.

And the range of this function is all the possible values of y.

For example for the parent function IxI, the range will be all the positive reals and the zero.

First, if A is the value of the vertex of the absolute function, then we know that A is the maximum or the minimum value of the function.

Now, if the arms of the graph open up, then we know that A is the minimum of the function, and the range will be:

y ≥ A

Or all the real values equal to or larger than A.

if the arms of the graph open downwards, then A is the maximum of the function, and we have that the range is:

y ≤ A

Or "All the real values equal to or smaller than A"

5 0
3 years ago
The volume of 10,000 drops of liquid is a 1 fluid ounce. what is the volume of 1000 drops?
Paraphin [41]
1/10 is the answer if you divide 10,000 from 1,000
10000 \div 1000
3 0
3 years ago
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