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3 years ago

PLEASE HELP ME! 20 points and brainlist.

Mathematics

2 answers:

spin [16.1K]3 years ago

7 0

Answer:

Greatest common factor

Step-by-step explanation:

You need the upside down birthday cake method to find prime numbers and gcf

motikmotik3 years ago

4 0

Answer:

greatest common factor

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Is (2,17) a one solution, no solution, infinite solution,or two solution

WARRIOR [948]

Answer:

one solution

Step-by-step explanation:

because it's a gksbn smksidihdbd jaiaigsvs j

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3 years ago

Can i pleaseee get help:(

AfilCa [17]

Answer:

Step-by-step explanation:

P=5/25=1/5=0.20

8 0

3 years ago

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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

Can a pentagon have angles that measure 95°, 67°, 105°, 95°, and 177°?

pochemuha

Answer:

No, it cannot have angles that measure 95°, 67°, 105°, 95°, and 177°.

Step-by-step explanation:

S = (n-2) * 180

A pentagon has 5 sides, so substitute n for sides (n = 5)

s = 5-2) * 180 = 540°

95° + 67° + 105° + 95° + 177° = 539°

Hope that Helps!

7 0

4 years ago

Find the value of sin 0 in each of the following figures.

alexdok [17]

Answer:

sinθ = $\frac{8}{10}$

Step-by-step explanation:

sinθ = $\frac{opposite}{hypotenuse}$ = $\frac{BC}{AC}$ = $\frac{8}{10}$

5 0

3 years ago

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