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marishachu [46]
3 years ago
5

Given fx) = 17- x2, what is the average rate of change in f(x) over the interval [1, 5]?

Mathematics
2 answers:
Andrei [34K]3 years ago
6 0

Let [a, b] = the interval

The average rate of change can be found by using [f(b) - f(a)]/(b - a).

Let a = 1 and b = 5.

Let R = rate of change

17- x^2

R = [17 - 5^2] - [17 - 1^2]/(5 - 1)

R = [17 - 25] -[17 - 1]/4

R = [(-8) - (16)]/4

R = (-24)/4

R = -6

Done.

vredina [299]3 years ago
6 0

Answer: A. -6

Step-by-step explanation:

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Jessica's heart beats 12 times in 1/6 minute how many times does her heart beat each minute ​
Charra [1.4K]

Answer:

72

Step-by-step explanation:

12 6 times equals 72.

12x6=72

4 0
3 years ago
Find the measures of all four angles if 3·(m∠1+m∠3) = m∠2+m∠4.
mixas84 [53]

Answer:

Step-by-step explanation:

m∠1=m∠3

m∠2=m∠4

3*(m∠1+m∠3)=m∠2+m∠4

3*(m∠1+m∠1)=m∠2+m∠2

3×2 m∠1=2 m∠2

m∠2=3 m∠1

now m∠1+m∠2=180°

m∠1+3 m ∠1=180

4 m∠1=180

m∠1=180/4=45°

m∠3=45°

m∠2=180-m∠1=180-45=135°

m∠4=135°

5 0
3 years ago
(b) In the figure below, m 2AEB= 55° and m AB = 60°. Find m CD.<br><br> m CD =
RUDIKE [14]

Answer:

47 degrees

Explanation:

your welcome

4 0
2 years ago
How do you solve sin (5pi/3) without a calculator?
Savatey [412]

very simple, we use the formula sin(a+b)=sinacosb + sinbcosa and sin(20)=2sinacosa

5pi = 2pi/3+3pi/3,

First, we use sin(a+b)=sinacosb + sinbcosa

sin(5pi/3)=sin(2pi/3+3pi/3)= sin(2pi/3+pi)= sin(2pi/3)cos(pi) +sin(pi)cos(2pi/3)

but we know that sin(pi)= 0, and cos (pi) = -1, so sin(5pi/3)= - sin(2pi/3)

now, use sin(2a)=2sinacosa, sin(5pi/3)= - sin(2pi/3)= -2sin(pi/3)cos(pi/3)

sin<span>(5pi/3)=  -2sin(pi/3)cos(pi/3)</span>

<span>sin(pi/3)= 0.86, cos(pi/3)=0.5, finally we have   </span>sin<span>(5pi/3)=  -0.86 x 0.5= -0.43</span>

5 0
3 years ago
Read 2 more answers
​Evaluate the following expression if y = 4
sleet_krkn [62]
15+5•y/2

15+5•4 / 2

15+20 / 2

15+10

=25
3 0
2 years ago
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