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statuscvo [17]
3 years ago
12

Simplify this expression 5.3x-2.2(9.6x-9.11)

Mathematics
1 answer:
USPshnik [31]3 years ago
6 0

−15.82x + 20.042
calculated it
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7x+3(×+6)=8 .....whut , it makes NO SENSE!!!!!!!
VikaD [51]
Hopefully this helps.

3 0
3 years ago
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If the interest rate on a savings account is 0.018%, approximately how much money do you need to keep in this account for 1 year
Diano4ka-milaya [45]
<span>We are not told how often the interest is compounded, so assuming it is <em /><u><em>compounded yearly</em></u>, you need to keep $9.99 in the account to pay the fee.

<u><em>Explanation: </em></u>
Compound interest follows the formula A=p(1+r)^t,
where:
A is the total amount in the account,
p is the amount of principal,
r is the interest rate as a decimal number,
and t is the number of years.

<u>For our problem: </u>
A = 9.99,
p is unknown,
r = 0.018% = 0.00018,
and t=1.

<u>This gives us: </u>
9.99=p(1+0.00018)^1;
9.99=p(1.00018).

<u>Divide both sides by 1.00018: </u>
9.99=p.</span>
3 0
3 years ago
Read 2 more answers
of all the murder victims in 2010 whose relation to the offender was known, 24.8% were killed by a family member and 53% by an a
ICE Princess25 [194]
The answer is 30,000 and i know cuz 2 go in 5 3 times then add 0
5 0
3 years ago
Solve for x solve for x
alekssr [168]
All the angles added up = 360°

46 + 4x - 2 + 9x + 6 + 8x - 5 = 360
21x + 45 = 360
21x = 315
x = 15

Let me know if you have questions.
7 0
2 years ago
Sam is walking across a bridge and accidentally drops an orange into the river basin below
irina1246 [14]

Answer:

We assume that the orange is dropped at t = 0s.

Once the orange is on the air, the only force acting on it is the gravitational force, then the acceleration of the orange is the gravitational acceleration.

A(t) = -32.17 ft/s^2

Where the negative sign is because this acceleration points downwards.

For the velocity equation, we need to integrate over time, we will get:

V(t) = (-32.17 ft/s^2)*t + V0

Where V0 is the initial vertical velocity of the orange, because the orange is accidentally dropped, this initial velocity is equal to zero.

V(t) =   (-32.17 ft/s^2)*t

For the position equation we need to integrate again, this time we get:

P(t) = (1/2)*(-32.17 ft/s^2)*t^2 + P0

Where P0 is the initial height of the orange, we know that it is 40ft, then the position equation is:

P(t) =  (1/2)*(-32.17 ft/s^2)*t^2 + 40 ft

Now that we know the equation, we can graph it. (you can see the graph below)

Now we also want to find at what time does the orange hit the water.

This happens when:

P(t) = 0 ft =   (1/2)*(-32.17 ft/s^2)*t^2 + 40 ft

We just need to solve that equation for t.

0 ft =   (1/2)*(-32.17 ft/s^2)*t^2 + 40 ft

(1/2)*(32.17 ft/s^2)*t^2 =  40 ft

t^2 = (40ft)/( (1/2)*(32.17 ft/s^2))

t = √(  (40ft)/( (1/2)*(32.17 ft/s^2)) ) = 1.58 s

The orange hits the water 1.58 seconds after it is dropped.

3 0
3 years ago
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