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3 years ago

What is 0+9+2 times 8

Mathematics

1 answer:

Anni [7]3 years ago

5 0

Answer:

That would be 88!

Step-by-step explanation:

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What is the image of (9,3) after a dilation by a scale factor of 3 centered at the<br> origin?

sergiy2304 [10]

Answer: (27, 9)

Multiply the original coordinates by the scale factor 3 to get this answer

9*3 = 27

3*3 = 9

This only works if the center of dilation is the origin.

The general rule is $(x,y) \to (kx, ky)$ where k is the scale factor.

7 0

3 years ago

Label the diagram

Tomtit [17]

A) y-axis

B) quad lll

C) x-axis

D) quad IV

Is this the answer you were looking for?

8 0

3 years ago

Lin solved the equation

Alekssandra [29.7K]

Answer: x=1/2

she was wrong, it's supposed to be

Step-by-step explanation: 8(x−3)+7=2x(4−17)

Step 1: Simplify both sides of the equation.

8(x−3)+7=2x(4−17)

(8)(x)+(8)(−3)+7=2x(4−17)(Distribute)

8x+−24+7=−26x

(8x)+(−24+7)=−26x(Combine Like Terms)

8x+−17=−26x

8x−17=−26x

Step 2: Add 26x to both sides.

8x−17+26x=−26x+26x

34x−17=0

Step 3: Add 17 to both sides.

34x−17+17=0+17

34x=17

Step 4: Divide both sides by 34.

34x/34=17/34 x=1/2

3 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

Please i need help on this question im strugling

DerKrebs [107]

Answer:

1: 5

2: 5

3: 1.5

Step-by-step explanation:

6 0

3 years ago