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3 years ago

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A barrel had 20 liters of water. Water leaked out from the barrel at a constant rate. After 24 hours, 14 liters of water was lef

t in the tank. How many liters of water leaked out per hour? 0.25 0.50 1.42 3.14

1 answer:

SpyIntel [72]3 years ago

7 0

6/24 is .25

So water leaked out at 0.25 liters per hour

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there are quite a bit of them but i just need someone to convert these fractions to decimals....thanks<3

Mariana [72]

Answer:

1/9=0.111111111 2/9=0.222222222 3/9=0.333333333 4/9=0.444444444

5/9=0.555555556 6/9=0.666666667 7/9=0.777777778 8/9=0.888888889

Step-by-step explanation:

sorry i dont have time to do them all but heres all of the nines

4 0

2 years ago

Solve for x. 2=4x-6

77julia77 [94]

Answer:

x = 2

Step-by-step explanation:

Rewrite the equation as 4x - 6 = 2

Move all terms not containing x to the right side of the equation:

Add 6 to both sides of the equation.

4x = 2 + 6

Add 2 and 6.

4x = 8

Divide each term by 4 and simplify.

Divide each term in 4x = 8 by 4.

$\frac{4x}{4} =\frac{8}{4}$

Reduce the expression by cancelling the common factors.

$x=\frac{8}{4}$

Divide 8 by 4.

x = 2

8 0

2 years ago

Can 4/3 be simplified ?

Novosadov [1.4K]

Answer: yes, 4/3 can be simplified

Step-by-step explanation:

Since it is an improper fraction, it can be simplified.

three goes into four once, And there is a remainder of one. so the new fraction is: 1 1/3

6 0

3 years ago

Read 2 more answers

In the circus, a lion has been trained to jump through a ring of fire that has a diameter of 8 feet. What is the ring's area?

liberstina [14]

Answer:

50.265

Step-by-step explanation:

radius is 4 then you put it in A=Pi r^2

5 0

2 years ago

Find the values of m and b that make the following function differentiable. the piecewise function f of x equals x cubed when x

aalyn [17]

$f(x)=\begin{cases}x^3&\text{for }x\le1\\mx+b&\text{for }x>1\end{cases}$

$f$ must be continuous in order to be differentiable, so we need to have

$\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1^+}f(x)=f(1)$

By its definition, $f(1)=1^3=1$ , and

$\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1}x^3=1$

$\displaystyle\lim_{x\to1^+}f(x)=\lim_{x\to1}(mx+b)=m+b$

so that $\boxed{m+b=1}$ .

We want the derivative to exist at $x=1$ , which requires that we pick an appropriate value for $f'(1)$ so that $f'(x)$ is also continuous. At the moment, we know

$f'(x)=\begin{cases}3x^2&\text{for }x1\end{cases}{/tex]so we need to pick [tex]f'(1)$ such that

$\displaystyle\lim_{x\to1^-}f'(x)=\lim_{x\to1^+}f'(x)$

We have

$\displaystyle\lim_{x\to1^-}f'(x)=\lim_{x\to1}3x^2=3$

$\displaystyle\lim_{x\to1^+}f'(x)=\lim_{x\to1}m=m$

so that $\boxed{m=3}$ (which means we need to pick $f'(1)=3$ ) and so $m+b=1\implies\boxed{b=-2}$

5 0

3 years ago