All categories

2 years ago

When the target number is rounded to the nearest million, the millions digit is 1.

Mathematics

1 answer:

Sladkaya [172]2 years ago

6 0

Answer:

185445

Step-by-step explanation:

am just guessing but plssssssss give me brainliest

You might be interested in

What is the circumference of diameter of 2cm

kicyunya [14]

a circle with a diameter of 2cm has a circumference of 6.28 cm

3 0

2 years ago

<img src="https://tex.z-dn.net/?f=a%20-%207%20%3D%2017" id="TexFormula1" title="a - 7 = 17" alt="a - 7 = 17" align="absmiddle" c

andriy [413]

A would equal 24 because 17+7 is 24

6 0

3 years ago

Is the difderence of 1.48-0.25 less than or greater than one? Explain

luda_lava [24]

Greater than 1.Because 1.25-0.25=1.23.That's it.

6 0

3 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

4 0

3 years ago

Which line is perpendicular?

Serggg [28]

D. You can find a perpendicular line by finding the inverse equations (-(1/)2 x+3

8 0

2 years ago