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Goshia [24]
3 years ago
15

Solve the inequality 5 - 1/2x great than 30

Mathematics
1 answer:
Mars2501 [29]3 years ago
3 0

Answer:

5-1/2 = 4 1/2 4 1/2= 4.5. and 4.5 is not greater then 30, thus this statement is false

Step-by-step explanation:

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what types of problems can be solved using the greatest common factor what types of problems can be solved using the last common
jonny [76]

Answer:

The greatest common factor is the biggest factor that divides two different numbers. For example, the greatest common factor of 6 and 8 is 2. The least common multiple is the smallest number that two numbers share as a multiple. For example, 12 is a the lowest common multiple of 3 and 4.

Step-by-step explanation:

3 0
4 years ago
2. Chaleah deposited 700$ in a new account that earns 5% simple interest. After 4 years,
Oxana [17]

Answer:

Interest earned: $150.85

Total amount: $850.86

Step-by-step explanation:

Y=700(1.05)^4

Put the right side of the equation into a calculator and get

850.85 that's your total, so to get the amount earned subtract the original 700 from it and get 150.85

5 0
3 years ago
Question number 7 I need help on
Burka [1]

Answer:

Hi please mark me as brainliest

Step-by-step explanation:

It's A

6 0
3 years ago
If it is 4:40 now, what time was it 5 hours and 15 minutes ago?
mamaluj [8]
The answer is that it would be 11:25
3 0
3 years ago
Read 2 more answers
Attached as picture. Please read fully
Troyanec [42]

a. The velocity t = v = Ce_{n} (\frac{mo}{mo - kt} ) - gt

b. v60 = 7164

<h3>How to solve for the velocity</h3>

mdv/dt = ck - mg

dv/dt = ck/m - mg/m

= ck/m - g

dv = (\frac{ck}{Mo-Kt} -g)dv

Integrate the two sides of the equation to get

v -\frac{ck}{k} e_{n} (Mo- kt)-gt+c

v = Ce_{n} (\frac{mo}{mo - kt} ) - gt

b. fuel accounts for 55% of the mass

So final mass after fuel is burned out is = 0.45

c=2500

g=9.8

t=60

v = -2500ln0.45 - 9.8 x 60

= 7752 - 588

= 7164

<h3>Complete question</h3>

A rocket, fired from rest at time t = 0, has an initial mass of m0 (including its fuel). Assuming that the fuel is consumed at a constant rate k, the mass m of the rocket, while fuel is being burned, will be given by m0 - kt. It can be shown that if air resistance is neglected and the fuel gases are expelled at a constant speed c relative to the rocket, then the velocity of the rocket will satisfy the equation where g is the acceleration due to gravity.

dv dt m =ck - mg

(a) Find v(t) keeping in mind that the mass m is a function of t.

v(t) =

m/sec

(b) Suppose that the fuel accounts for 55% of the initial mass of the rocket and that all of the fuel is consumed at 60 s. Find the velocity of the rocket in meters per second at the instant the fuel is exhausted. [Note: Take g = 9.8 m/s² and c = 2500 m/s.]

v(60) =

m/sec [Round to nearest whole number]

Raed more on velocity here

brainly.com/question/25749514

#SPJ1

8 0
2 years ago
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