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3 years ago

What is the rate of change for the numbers 3,5,7,9

Mathematics

2 answers:

JulsSmile [24]3 years ago

6 0

Answer:

Plus 2 (+2)

Step-by-step explanation:

The rate of change is always a constant change from the given first term to the next one. In this case, you are adding 2 each time:

3 + 2 = 5

5 + 2 = 7

7 + 2 = 9

etc.

Luden [163]3 years ago

3 0

Change means difference in the state or condition of a thing and here the thing is number. The change made within them must be following some certain rules or patterns.

They can be uniform and here it is uniform for sure. The terms are increasing by 2 or we can say that for getting th next number, we need to add 2 with the previous.

3 + 2 = 5
5 + 2 = 7
7 + 2 = 9

So, the rate of change for the above numbers or number series is +2 or addition of 2 $.$

Carry On Learning !

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A hot air balloon is spotted simultaneously by a man and a woman 3 ½ miles apart. If the angle of elevation from the man to the

V125BC [204]

See the picture attached to better understand the problem

we know that
in the right triangle ABD
tan 27°=BD/AD-----> BD=(3.5-x)*tan 27°------> equation 1

in the right triangle BDC
tan 41°=BD/DC-----> BD=x*tan 41°-----> equation 2

equate equation 1 and equation 2
(3.5-x)*tan 27°=x*tan 41°----> 3.5*tan 27-x*tan 27=x*tan 41
x*[tan 27+tan 41]=3.5*tan 27
x=3.5*tan 27/[tan 27+tan 41]-----> x=1.29 miles

BD=x*tan 41-----> BD=1.29*tan 41°----> BD=1.124 miles

convert miles to ft
1 mile-------> 5280 ft
1.124 miles----> x
x=1.124*5280-----> x=5934.72 ft

the answer is
<span>the altitude of the balloon in feet is 5934.72</span>

4 0

2 years ago

A house cost $120,000 when it was purchased. The value of the house increases by 10% each year. Find the rate of growth each mon

olga_2 [115]

FIRST MODEL:

Well the model for the value of the house is:

$V={ \left( \frac { 11 }{ 10 } \right) }^{ t }\cdot 120000$

V = Value

t = Years passed {t≥0}

-----------------------

When t=0, V=120000

When t=1, V=132000

When t=2, V=145200

etc... etc...

---------------------------

Now, this model is actually curved so there is no constant rate of growth each month. We can only calculate what the rate of growth is at a particular time. If we want to find out the rate of growth at a particular time, we must differentiate the formula (model) above.

--------------------------

$V={ \left( \frac { 11 }{ 10 } \right) }^{ t }\cdot 120000\\ \\ \ln { V=\ln { \left( { \left( \frac { 11 }{ 10 } \right) }^{ t }\cdot 120000 \right) } }$

$\\ \\ \ln { V=\ln { \left( { \left( \frac { 11 }{ 10 } \right) }^{ t } \right) } } +\ln { \left( 120000 \right) } \\ \\ \ln { V=t\ln { \left( \frac { 11 }{ 10 } \right) } } +\ln { \left( 120000 \right) }$

$\\ \\ \frac { 1 }{ V } \cdot \frac { dV }{ dt } =\ln { \left( \frac { 11 }{ 10 } \right) } \\ \\ V\cdot \frac { 1 }{ V } \cdot \frac { dV }{ dt } =\ln { \left( \frac { 11 }{ 10 } \right) } \cdot V$

$\\ \\ \therefore \quad \frac { dV }{ dt } =\ln { \left( \frac { 11 }{ 10 } \right) } \cdot { \left( \frac { 11 }{ 10 } \right) }^{ t }\cdot 120000$

Plug any value of (t) that is greater than 0 into the formula above to find out how quickly the investment is growing. If you want to find out how quickly the investment was growing after 1 month had passed, transform t into 1/12.

The rate of growth is being measured in years, not months. So when t=1/12, the rate of growth turns out to be 11528.42 per annum.

SECOND MODEL (What you are ultimately looking for):

$V={ \left( \frac { 11 }{ 10 } \right) }^{ \frac { t }{ 12 } }\cdot 120000$

V = Value of house

t = months that have gone by {t≥0}

Formula above differentiated:

$V={ \left( \frac { 11 }{ 10 } \right) }^{ \frac { t }{ 12 } }\cdot 120000\\ \\ \ln { V } =\ln { \left( { \left( \frac { 11 }{ 10 } \right) }^{ \frac { t }{ 12 } }\cdot 120000 \right) }$

$\\ \\ \ln { V=\ln { \left( { \left( \frac { 11 }{ 10 } \right) }^{ \frac { t }{ 12 } } \right) } } +\ln { \left( 120000 \right) }$

$\\ \\ \ln { V=\frac { t }{ 12 } } \ln { \left( \frac { 11 }{ 10 } \right) } +\ln { \left( 120000 \right) }$

$\\ \\ \frac { 1 }{ V } \cdot \frac { dV }{ dt } =\frac { 1 }{ 12 } \ln { \left( \frac { 11 }{ 10 } \right) }$

$\\ \\ V\cdot \frac { 1 }{ V } \cdot \frac { dV }{ dt } =\frac { 1 }{ 12 } \ln { \left( \frac { 11 }{ 10 } \right) } \cdot V$

$\\ \\ \therefore \quad \frac { dV }{ dt } =\frac { 1 }{ 12 } \ln { \left( \frac { 11 }{ 10 } \right) } \cdot { \left( \frac { 11 }{ 10 } \right) }^{ \frac { t }{ 12 } }\cdot 120000$

When t=1, dV/dt = 960.70 (2dp)

dV/dt in this case will measure the rate of growth monthly. As more money is accumulated, this rate of growth will rise. The rate of growth is constantly increasing as the graph of V is actually a curve. You can only find out the rate at which the house value is growing monthly at a particular time.

6 0

2 years ago

Of Composite Shapes P5

ycow [4]

It is 114, because you subtract 6 from 15. So, the bigger thing is 90, add the smaller one onto it, 6 times 4 equals 24, 90+24=114.

8 0

3 years ago

Trig Project

Flauer [41]

Answer:

bump

Step-by-step explanation:

bump

4 0

1 year ago

What are the solutions of the equation x4 + 95x2 – 500 = 0?

Ugo [173]

X4+(95x2)-500=0
x4+(190)-500=0
x4-310=0
x4=310
x=4√(310)
x=4.196

8 0

3 years ago