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makkiz [27]
2 years ago
14

The circumference of the yellow circle is about for 2.5

Mathematics
2 answers:
ikadub [295]2 years ago
6 0
What are you asking to solve?
RSB [31]2 years ago
4 0

Answer:

what are you trying to solve?

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The domain of a function is always equal to which one of the following options?
Ivanshal [37]

Answer:

all possible input values to the function

Step-by-step explanation:

The domain is the input to the function and the range is the output of the function

3 0
2 years ago
Read 2 more answers
Please select the best answer.
Amiraneli [1.4K]

ANSWER

C. The left end goes up; the right end goes down.

EXPLANATION

The given function is

f(x) =  - 2( {x - 2)}^{5}

We analyze the end behavior of this function using the leading term.

The leading term of this function is:

- 2 {x}^{5}

Since the degree(5) is odd and the leading coefficient (-2) is negative, the graph rises on the left and falls on the right.

In other words, the left end of the graph goes up and the right end goes down.

The correct option is C.

5 0
3 years ago
Read 2 more answers
There are 40 grams of sugar in a 12 oz can of Dr Pepper. If your max intake of sugar permitted for the day is 25 grams, how much
LenKa [72]
You would be able to drink 7.5oz of the can while you would only intake 25grams of sugar
7 0
3 years ago
Read 2 more answers
2 1⁄2 ÷ 1 3⁄6 please help me with the RIGHT answer.
ycow [4]

Answer:

1\frac{2}{3}

Step-by-step explanation:

We want to simplify:

2 \frac{1}{2}  \div 1 \frac{3}{6}

This is the same as:

2 \frac{1}{2}  \div 1 \frac{1}{2}

Now let us convert the mixed numbers to improper fractions.

\frac{5}{2}  \div  \frac{3}{2}

We multiply by the reciprocal of the second fraction to get:

\frac{5}{2}  \times  \frac{2}{3}

Cancel out the common factors to obtain:

\frac{5}{3}  = 1 \frac{2}{3}

5 0
2 years ago
Use properties to find sum or product of 4 + (6+21)
Maurinko [17]
You need to use  pemdas so first you do inside the parenthesis 6+21=27 then you add 4 27+4=31 that is your answer



4 0
3 years ago
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