All categories

3 years ago

HELPPP From 8 A.M. to noon, the temperature change was -16F. Find the average change per hour

1 answer:

3 0

-16F temperature change divided by 4 hour difference from 8 to 12/noon.
-16 divided by 4 = -4
The answer is -4 degrees Fahrenheit.

You might be interested in

Given the functions below, find g(5) + h(2).<br> g(x) = 2x - 5<br> h(x) = 4x + 5

Ksju [112]

G(5) = 2(5)-5=5
h(2)=4(2)+5=13

So, g(5)+h(2)=13+5=18.

4 0

2 years ago

Read 2 more answers

In 2002, there were 972 students enrolled at Oakview High School. Since then, the number of students has increased by 1.5% each

Rudiy27

Answer:

$N(t) = 972(1.015)^{t}$

Growth function.

The number of students enrolled in 2014 is 1162.

Step-by-step explanation:

The number of students in the school in t years after 2002 can be modeled by the following function:

$N(t) = N(0)(1+r)^{t}$

In which N(0) is the number of students in 2002 and r is the rate of change.

If 1+r>1, the function is a growth function.

If 1-r<1, the function is a decay function.

In 2002, there were 972 students enrolled at Oakview High School.

This means that $N(0) = 972$

Since then, the number of students has increased by 1.5% each year.

Increase, so r is positive. This means that $r = 0.015$

Then

$N(t) = N(0)(1+r)^{t}$

$N(t) = 972(1+0.015)^{t}$

$N(t) = 972(1.015)^{t}$

Growth function.

Find the number of students enrolled in 2014.

2014 is 2014-2002 = 12 years after 2002, so this is N(12).

$N(t) = 972(1.015)^{t}$

$N(12) = 972(1.015)^{12}$

$N(12) = 1162$

The number of students enrolled in 2014 is 1162.

7 0

3 years ago

dante has been putting one peeny in his bank everyday Now he has 161 pennies Hor how many weeks has d

nikklg [1K]

All you need to do is divide 161 by 7 and get 23 weeks

6 0

3 years ago

Nick goes to the store and buys 22 notebooks for $1.50 each. Then he buys 8 pens for $0.55 each. How much money did he spend in

rosijanka [135]

Answer:

$33 for the notebooks and $4.40 for the pens

Step-by-step explanation:

if u would multiply 33 by 1.50 you would get $33

if u would multiply 8 by 0.55 you would get $4.40

8 0

3 years ago

Find the area between y = 8 sin ( x ) y=8sin⁡(x) and y = 8 cos ( x ) y=8cos⁡(x) over the interval [ 0 , π ] . [0,π]. (Use decima

Marina86 [1]

Answer:

0.416 au

Step-by-step explanation:

Let y1=8sin(x) and y2=8cos(x), we must find the area between y1 and y2

$\int\limits^\pi _0{(8cos(x)-8sin(x))} \, dx = 8\int\limits^\pi _0{(cos(x)-sin(x))} \, dx =\\8(sin(x)+cos(x)) evaluated(0-\pi )=\\8(sin(\pi )-sin(0))+8(cos(\pi )-cos(0))=\\8(0.054-0)+8(0.998-1)=8(0.054)+8(-0.002)=0.432-0.016=0.416$

3 0

3 years ago