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telo118 [61]
2 years ago
10

The cube of a whole number is 100 more

Mathematics
1 answer:
Nikitich [7]2 years ago
3 0

Answer:

The number is 0 or 100

Step-by-step explanation:

Let the number be x.

ATQ, x^3=100x^2, x^2(x-100)=0, x=0, 100

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What is 400 in expanded form (Warfare you get brainliest)
MAXImum [283]
Hey there, Lets solve your question together 

Question - <span>What is 400 in Expanded form ? 
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We know that expanded form is a spread out way to write a number by showing the value of each individual digit that contributes<span> to make the full number. 
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Lets use an example to show expanded form

Number - 4,265 

= 4 x 1,000 + 2 x 100 + 6 x 10 + 5 x 1 

Now lets solve for the number 400 in expanded form. 

The number "4" is in the hundreds place 

The number "0" is in the tenths<span> place 
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<span>The number "0" is in the One's place 

</span>Therefore 400 in expanded form is 400+00+0. 
 
<span>          Hope this helps </span>




5 0
3 years ago
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Solve the literal equation for y. 2x-4y=20
Phoenix [80]
2x -4y=20
-4y=-2x+20
Y= 1/2x - 5 that’s the answer
3 0
3 years ago
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50 men working 10 hours daily can complete a job in
Sliva [168]

<h2>Given :-</h2>

  • 50 men do 10 hours work daily complete work in 20 days

  • 80 men do job in 50% time

<h2>To Find :-</h2>

  • Days required

<h2>Solution :-</h2>

  • 1 man may do job in (50 × 20) days =
  • 1 man may do job in 1000 days

Now,

  • 80 man = 1000/80 = 12.5

Now,

  • When hours reduced 50% then we will multiply days by 2
  • Total days = 12.5 × 2 = 25 days.

<h2>Hence</h2>

  • Option C is correct

\begin{gathered} \\ \end{gathered}

8 0
2 years ago
What is the rate of change for y=2x+10
Tanzania [10]

Answer:

its 2  

The general rate of change can be found by using the difference quotient formula. To find the average rate of change over an interval, enter a function with an interval

8 0
3 years ago
The projected rate of increase in enrollment at a new branch of the UT-system is estimated by E ′ (t) = 12000(t + 9)−3/2 where E
nexus9112 [7]

Answer:

The projected enrollment is \lim_{t \to \infty} E(t)=10,000

Step-by-step explanation:

Consider the provided projected rate.

E'(t) = 12000(t + 9)^{\frac{-3}{2}}

Integrate the above function.

E(t) =\int 12000(t + 9)^{\frac{-3}{2}}dt

E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+c

The initial enrollment is 2000, that means at t=0 the value of E(t)=2000.

2000=-\frac{24000}{\left(0+9\right)^{\frac{1}{2}}}+c

2000=-\frac{24000}{3}+c

2000=-8000+c

c=10,000

Therefore, E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000

Now we need to find \lim_{t \to \infty} E(t)

\lim_{t \to \infty} E(t)=-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000

\lim_{t \to \infty} E(t)=10,000

Hence, the projected enrollment is \lim_{t \to \infty} E(t)=10,000

8 0
3 years ago
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