What is the median of the set of data 928, 834, 842, 841, 932, 917, 892, 856, 908, 867, 873, 868, 872, 921, and 924?

What is the value of f(-4)

WITCHER [35]

Answer:

7

Step-by-step explanation:

Substitute x = -4 :

f(-4) = 3 - (-4)

= 3 + 4

= 7

4 0

3 years ago

Please answer this, I need help

frutty [35]

Answer:

-$7

Step-by-step explanation:

20 - 15 = 5

5 + 5 = 10

10 - 10 = 0

0 - 7 = -7

8 0

3 years ago

Read 2 more answers

Simplify the expression (7a5b6)4 . Please help

8_murik_8 [283]

Answer:

28a 20b 24

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

joanne cannot decide which of two washing machines to buy. The selling price of each is $660. The first is marked down by 40%.

Alexxx [7]

Answer: The first machine would cost $420

and the second machine would cost $432

you should buy the first machine

Step-by-step explanation:If you get a 30% discount, then you are paying 70% of the selling price.

Machine 1: 600(.70) = 420

Machine 2: A 10% discount means you pay 90%

600(.90) = 540 But now you get a 20% discount on that amount, which means you would pay 80%

So, 540(.80) = 432

6 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:

a)

Speed at Equator = 463.97 meters per second

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

b)

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,

a)

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

b)

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