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tia_tia [17]
3 years ago
7

Is 8.4 a rational number

Mathematics
2 answers:
Murljashka [212]3 years ago
6 0
Yep.

All rational numbers are terminating or repeating decimals (this one is terminating because it ends). This number can also be written as a ratio of two integers (84/10).
Valentin [98]3 years ago
4 0
I think 8.4 is a rational number
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A glass of water was seven- eighths of a liter. With a 6 liter jug, how many glasses could you pour? Do you have to multiply div
ivolga24 [154]

Let us suppose, we pour x glass of water.

The given jug is of 6 liters. And the glass of water was seven- eighths of a liter. Hence, we have the below equation

\frac{7}{8} x= 6

Now, in order to solve for x, we will divide 6 with 7/8. So, we have to divide.

x=\frac{6}{7/8} \\
\\
x=\frac{48}{7} \\
\\
x=6.9

Hence, we can pour 7 glass of water. But the 7th glass is not full of water. It is partially full.

5 0
3 years ago
Jerome recently moved to a new city and bought a home. The assessment rate is 51%. The property tax rate is $53.26 per $1000. Wh
mrs_skeptik [129]
Given:
assessment rate = 51%
tax rate = 53.26 per 1,000

53.26 / 1000 = 0.05326
0.05326 x 100% = 5.326%

Effective tax rate = assessment rate * tax rate
ETR = 51% * 5.326%
ETR = 2.72%



8 0
3 years ago
1. (5pts) Find the derivatives of the function using the definition of derivative.
andreyandreev [35.5K]

2.8.1

f(x) = \dfrac4{\sqrt{3-x}}

By definition of the derivative,

f'(x) = \displaystyle \lim_{h\to0} \frac{f(x+h)-f(x)}{h}

We have

f(x+h) = \dfrac4{\sqrt{3-(x+h)}}

and

f(x+h)-f(x) = \dfrac4{\sqrt{3-(x+h)}} - \dfrac4{\sqrt{3-x}}

Combine these fractions into one with a common denominator:

f(x+h)-f(x) = \dfrac{4\sqrt{3-x} - 4\sqrt{3-(x+h)}}{\sqrt{3-x}\sqrt{3-(x+h)}}

Rationalize the numerator by multiplying uniformly by the conjugate of the numerator, and simplify the result:

f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x} - 4\sqrt{3-(x+h)}\right)\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x}\right)^2 - \left(4\sqrt{3-(x+h)}\right)^2}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16(3-x) - 16(3-(x+h))}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16h}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}

Now divide this by <em>h</em> and take the limit as <em>h</em> approaches 0 :

\dfrac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ \displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-x}\left(4\sqrt{3-x} + 4\sqrt{3-x}\right)} \\\\ \implies f'(x) = \dfrac{16}{4\left(\sqrt{3-x}\right)^3} = \boxed{\dfrac4{(3-x)^{3/2}}}

3.1.1.

f(x) = 4x^5 - \dfrac1{4x^2} + \sqrt[3]{x} - \pi^2 + 10e^3

Differentiate one term at a time:

• power rule

\left(4x^5\right)' = 4\left(x^5\right)' = 4\cdot5x^4 = 20x^4

\left(\dfrac1{4x^2}\right)' = \dfrac14\left(x^{-2}\right)' = \dfrac14\cdot-2x^{-3} = -\dfrac1{2x^3}

\left(\sqrt[3]{x}\right)' = \left(x^{1/3}\right)' = \dfrac13 x^{-2/3} = \dfrac1{3x^{2/3}}

The last two terms are constant, so their derivatives are both zero.

So you end up with

f'(x) = \boxed{20x^4 + \dfrac1{2x^3} + \dfrac1{3x^{2/3}}}

8 0
2 years ago
7) Find the perimeter of triangle ABC. Round all answers to the nearest tenth.[6 points)
dsp73

Answer: 14.6

Step-by-step explanation:

I made a square around the triangle which I then counted the squares, found the Pythagorean theorem, and then added the missing sides together

6 0
3 years ago
Need Help on this assignment ASAP!
allsm [11]
The answer is y=(1/2)x -5

You are given the slope and a core donate pair so you plug them in to the equation. 2=x. -4=y. And (1/2)= m

Then work out like a one step equation

7 0
3 years ago
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