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3 years ago

What is 6.63 rounded to the nearest tenth

Mathematics

2 answers:

sertanlavr [38]3 years ago

4 0

Google will helped you

konstantin123 [22]3 years ago

4 0

6.6
because 3 is less than 5 so you don't do anything

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Name the line segments

ratelena [41]

Answer:

EF, FG, GH, EG, FH, EH

Step-by-step explanation:

These are all the ones under def. line segment.

See the attachment. It will be helpful.

6 0

3 years ago

PLEASE HELP ASAP!!!!

Vesna [10]

Answer:

I think the second one, the one you already put

Step-by-step explanation:

8 0

3 years ago

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If we assume the population of Grand Rapids is growing at a rate of approximately 4% per decade, we can model the population fun

saw5 [17]

Answer:

The average rate of change of the population on the intervals [ 5 , 10 ] , [ 5 , 9 ] , [ 5 , 8 ] , [ 5 , 7 ] , and [ 5 , 6 ] are 734.504, 733.06, 731.62, 730.185 and 728.75 respectively.

Step-by-step explanation:

The given function is

$P(t)=181843(1.04)^{(\frac{t}{10})}$

where, P(t) is population after t years.

At t=5,

$P(5)=181843(1.04)^{(\frac{5}{10})}=185444.20$

At t=6,

$P(6)=181843(1.04)^{(\frac{6}{10})}=186172.95$

At t=7,

$P(7)=181843(1.04)^{(\frac{7}{10})}=186904.57$

At t=8,

$P(8)=181843(1.04)^{(\frac{8}{10})}=187639.06$

At t=9,

$P(9)=181843(1.04)^{(\frac{9}{10})}=188376.44$

At t=10,

$P(10)=181843(1.04)^{(\frac{10}{10})}=189116.72$

The rate of change of P(t) on the interval $[x_1,x_2]$ is

$m=\frac{P(x_2)-P(x_1)}{x_2-x_1}$

Using the above formula, the average rate of change of the population on the intervals [ 5 , 10 ] is

$m=\frac{P(10)-P(5)}{10-5}=\frac{189116.72-185444.20}{5}=734.504$

The average rate of change of the population on the intervals [ 5 , 9 ] is

$m=\frac{P(9)-P(5)}{9-5}=\frac{188376.44-185444.20}{4}=733.06$

The average rate of change of the population on the intervals [ 5 , 8 ] is

$m=\frac{P(8)-P(5)}{8-5}=\frac{187639.06-185444.20}{3}=731.62$

The average rate of change of the population on the intervals [ 5 , 7 ] is

$m=\frac{P(7)-P(5)}{7-5}=\frac{186904.57-185444.20}{2}=730.185$

The average rate of change of the population on the intervals [ 5 , 6 ] is

$m=\frac{P(6)-P(5)}{6-5}=\frac{186172.95-185444.20}{1}=728.75$

Therefore the average rate of change of the population on the intervals [ 5 , 10 ] , [ 5 , 9 ] , [ 5 , 8 ] , [ 5 , 7 ] , and [ 5 , 6 ] are 734.504, 733.06, 731.62, 730.185 and 728.75 respectively.

7 0

4 years ago

A scientist is working with 0.7m of gold wire. How long is the wire in millimeters?

Kaylis [27]

Answer:

700mm

Step-by-step explanation:

1 metre = 100 centimetres

1 centimetre = 10 millimeters

1 metre = 100 centimetres = 1000 millimeters

0.7 metre = 70 centimetres = 700 millimeters

3 0

3 years ago

A worker is paid from Noon to 3 PM. They get a 15-minute break during their paid time.

jeka57 [31]

Answer:

8.3%

Step-by-step explanation:

A worker is paid from noon to 3 pm. So the worker works for 3 hours.

To find the percentage, we need to have the same units.

To do this, convert 3 hours to minutes and get 180 minutes.

We can now write a fraction (break minutes/total work time) to find the percent of the number of minutes in their break as a part of their whole paid shift.

This fraction is 15/180.

We can now simplify this fraction to 1/12.

Then convert into decimal form and get 0.0833.

Finally convert this decimal into a percentage by multiplying by 100 and get 8.3%

4 0

2 years ago

Read 2 more answers