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3 years ago

Can you help me whith number two?

Mathematics

2 answers:

trapecia [35]3 years ago

7 0

The answer is C

Just divide by 100 and multiply by whatever the percent is in that case.

Hope this helps!

Otrada [13]3 years ago

5 0

14.47 is the closest to the answer. To get the answer use 72.34×.20 getting 14.468 or 14.47

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Is 13t an algebraic expression like does it mean 13 Times T or is it just 13 and a number representing t

spin [16.1K]

Answer:

This is an algebraic expression

Step-by-step explanation:

U won't ever use t to mean multiplication u will use a . Or × or a letter next to a number to symbolise mutiplication

6 0

3 years ago

Seven students chose integers: -16, 12, -13, -6, 5, 6 and 1 Order the numbers from least to greatest. When all numbers are order

Lorico [155]

Answer:-16<-13<-6<1<5<6<12

-16

Step-by-step explanation:

8 0

3 years ago

Why is volume not a dose?

Ket [755]

Answer:

concentration

Step-by-step explanation:

Because concentration is a factor to how much of a volume to give

4 0

2 years ago

Wes bought a conference table for $960. what is it worth after depreciating at a rate of 12% per year for 4 years?

murzikaleks [220]

Answer:

$594.03

Step-by-step explanation:

Using the exponential function;

A = Pe^-rt

Principal = $960

Rate r = 12% = 0.12

Time t = 4years

Substitute

A = 960e^-(0.12*)*4

A = 960e^-0.48

A = 960(0.61878)

A = 594.03

HEnce the worth after 4years will be $594.03

4 0

3 years ago

The earth rotates once per day about an axis passing through the north and south poles, an axis that is perpendicular to the pla

forsale [732]

Answer:

Speed at Equator = 463.97 meters per second

Centripetal Acceleration at Equator = $3.37*10^{-2}$ meters per second squared

Speed at 30 degrees north of equator = 401.79 meters per second

Centripetal Acceleration at 30 degrees north of equator = $2.92*10^{-2}$ meters per second squared

Step-by-step explanation:

The formula is:

$v=\frac{2 \pi R}{T}$

Where

v is speed

R is radius

T is time

and another formula for centripetal acceleration:

$a_c=\frac{4 \pi^{2} R}{T^2}$

Now,

at equator, the radius is radius of earth (given), time in seconds is

T = 24 * 60 * 60 = 86,400

THus,

$v_E=\frac{2 \pi (6.38*10^{6}}{86,400}=463.97$

Speed at Equator = 463.97 meters per second

Centripetal Acceleration:

$a_{cE}=\frac{v_E^2}{R_E}=\frac{463.97}{6.38*10^{6}}=3.37*10^{-2}$

Centripetal Acceleration at Equator = $3.37*10^{-2}$ meters per second squared

At 30.0° north of the equator:

$R_N=R_E Cos (30)= (6.38*10^6)Cos(30)=5.53*10^6$

Now,

Speed = $v_{30N}=\frac{2 \pi (5.53*10^6)}{86,400}=401.79$

Speed at 30 degrees north of equator = 401.79 meters per second

Centripetal Acceleration:

$a_{30N}=\frac{v_E^2}{R_E}=\frac{401.79}{5.53*10^6}=2.92*10^{-2}$

Centripetal Acceleration at 30 degrees north of equator = $2.92*10^{-2}$ meters per second squared

4 0

3 years ago