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1 year ago

Ped

Mathematics

1 answer:

Stells [14]1 year ago

6 0

So the correct statement is:

" The GCF of the numbers in each term in the expression is 2"

<h3></h3><h3>How to rewrite the expression as a product?</h3>

Here we have the expression:

12d - 26c

To rewrite ti as a product, we need to find the greatest common factor between 12 and 26.

The decomposition of these two numbers is:

12 = 2*2*3

26 = 2*13

The greatest common factor between the two numbers is 2, then we can rewrite:

12d - 26c = 2*(6d - 13c)

So the correct statement is:

" The GCF of the numbers in each term in the expression is 2"

If you want to learn more about greatest common factors:

brainly.com/question/219464

#SPJ1

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2 years ago

Natalie has three apples she slices each Apple into eighths how many 1/8 apple slices does she have

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2 years ago

Read 2 more answers

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

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2 years ago