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2 years ago

Select the correct answer. The graph of the function is shown f(x)=sin(x). What is the domain of the function? PLATO USERS

Mathematics

1 answer:

pav-90 [236]2 years ago

6 0

Answer:

Domain: (−∞,∞),{x|x∈R}

Step-by-step explanation:

the domain will be Domain: (−∞,∞),{x|x∈R} because

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

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Find the 13th term of the sequence 4, 8, 16, 32, .

topjm [15]

Answer:

The thirteen number will be 16384.

Step-by-step explanation:

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3 years ago

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What is the mode of the data set?

saw5 [17]

Answer: 92 which is choice D

Look at the right side of values for the digit that comes up the most. In this case, it would be '2' in the second row. It shows up 3 times. Those leafs pair up with the stem of 9 to form the full value 92. So we have 92 show up three times which is the most frequent value.

The data set would expand out to

85, 88

92, 92, 92, 98

105, 106, 106, 109

113

126, 127

5 0

3 years ago

if Mr Green drives 100 miles in 2 hours. he continues to drive at the same speed if he doesn't stop how long will it take mr gre

Anton [14]

Answer:

8 hours

Step-by-step explanation:

1 - 50 miles

2 - 100 miles

3 - 150 miles

4 - 200 miles

5 - 250 miles

6 - 300 miles

7 - 350 miles

8 - 400 miles

and so on hope this helps

5 0

2 years ago

To solve the system of equations below, Pedro isolated the variable y in the first equation and then substituted it into the sec

Reil [10]

Answer:

Resulting equation will be 5x²=36.

Step-by-step explanation:

The given equations are 5y = 10x-----(1)

and x²+y²=36------(2)

Now we substitute the value of y from equation (1) into equation (2).

5y = 10x ⇒ y = 2x

Then equation (2) will be x²+(2x)²=36

x²+4x²=36 ⇒ 5x² = 36

So the equation after substitution of the value of y from equation 1 will be 5x² = 36.

8 0

3 years ago

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Match each interval with its corresponding average rate of change for q(x) = (x + 3)2. 1. -6 ≤ x ≤ -4 1 2. -3 ≤ x ≤ 0 -4 3. -6 ≤

MrMuchimi

The average rate of change of a function f(x) in an interval, a < x < b is given by
$\frac{f(b) - f(a)}{b - a}$

Given q(x) = (x + 3)^2

1.) The average rate of change of q(x) in the interval -6 ≤ x ≤ -4 is given by $\frac{q(-4)-q(-6)}{-4-(-6)} = \frac{(-4+3)^2-(-6+3)^2}{-4+6} = \frac{1-9}{2} = \frac{-8}{2} =-4$

2.) The average rate of change of q(x) in the interval -3 ≤ x ≤ 0 is given by $\frac{q(0)-q(-3)}{0-(-3)} = \frac{(0+3)^2-(-3+3)^2}{0+3} = \frac{9-0}{3} = \frac{9}{3} =3$

3.) The average rate of change of q(x) in the interval -6 ≤ x ≤ -3 is given by $\frac{q(-3)-q(-6)}{-3-(-6)} = \frac{(-3+3)^2-(-6+3)^2}{-3+6} = \frac{0-9}{3} = \frac{-9}{3} =-3$

4.) The average rate of change of q(x) in the interval -3 ≤ x ≤ -2 is given by $\frac{q(-2)-q(-3)}{-2-(-3)} = \frac{(-2+3)^2-(-3+3)^2}{-2+3} = \frac{1-0}{1} = \frac{1}{1} =1$

5.) The average rate of change of q(x) in the interval -4 ≤ x ≤ -3 is given by $\frac{q(-3)-q(-4)}{-3-(-4)} = \frac{(-3+3)^2-(-4+3)^2}{-3+4} = \frac{0-1}{1} = \frac{-1}{1} =-1$

6.) The average rate of change of q(x) in the interval -6 ≤ x ≤ 0 is given by $\frac{q(0)-q(-6)}{0-(-6)} = \frac{(0+3)^2-(-6+3)^2}{0+6} = \frac{9-9}{6} = \frac{0}{6} =0$

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3 years ago

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