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Over [174]
3 years ago
15

What is 6 divied 192 please show work

Mathematics
2 answers:
Eduardwww [97]3 years ago
8 0
Depending on what you are asking, the following will help you:

\frac{6}{192}  = \frac{3}{96}=\frac{1}{32} =\boxed{\bf{0.03125}}\\\\\\\frac{192}{6} =\frac{96}{3} = \boxed{\bf{32}}
suter [353]3 years ago
4 0
The problem presented with the word used as "divied" is very tricky and misleading to the point that it might create an understanding.

So we will approach this is in two ways.

If this was meant to be "divided by" then it would look like this:
6 / 192

If this was meant to be "divided to" then it would look like this:
192 / 6

Let us first solve for the first possible problem:
6 / 192 = ?
0.0312

0.0312 would be the answer.

Let us then solve for the second possible problem:
192 / 6 = ?
32 = ?

32 would be the answer.

So 0.0312 and 32 would be two possible answers to this problem.
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Can someone help please~~!?
stepladder [879]
12x + 12 + 15 + 3x + 18 = 180
15x + 45 = 180
15x = 180 - 45
15x = 135
x = 135/15
x = 9 

m < A = 12x + 12 = 12(9) + 12 = 108 + 12 = 120....A is 120
m < C = 3x + 18 = 3(9) + 18 = 27 + 18 = 45....C is 45

5 0
3 years ago
How does the calculator get 48.8? mean (45, 80, 15, 10, 94)
jok3333 [9.3K]

Answer:

mean= sum of all observations /no.of observation

mean=45+80+15+10+94/5

mean=244/5

mean=48.8

this is how the calculator gets 48.8

5 0
3 years ago
Let S be the solid beneath z = 12xy^2 and above z = 0, over the rectangle [0, 1] × [0, 1]. Find the value of m &gt; 1 so that th
jonny [76]

Answer:

The answer is \sqrt{\frac{6}{5}}

Step-by-step explanation:

To calculate the volumen of the solid we solve the next double integral:

\int\limits^1_0\int\limits^1_0 {12xy^{2} } \, dxdy

Solving:

\int\limits^1_0 {12x} \, dx \int\limits^1_0 {y^{2} } \, dy

[6x^{2} ]{{1} \atop {0}} \right. * [\frac{y^{3}}{3}]{{1} \atop {0}} \right.

Replacing the limits:

6*\frac{1}{3} =2

The plane y=mx divides this volume in two equal parts. So volume of one part is 1.

Since m > 1, hence mx ≤ y ≤ 1, 0 ≤ x ≤ \frac{1}{m}

Solving the double integral with these new limits we have:

\int\limits^\frac{1}{m} _0\int\limits^{1}_{mx} {12xy^{2} } \, dxdy

This part is a little bit tricky so let's solve the integral first for dy:

\int\limits^\frac{1}{m}_0 [{12x \frac{y^{3}}{3}}]{{1} \atop {mx}} \right.\, dx =\int\limits^\frac{1}{m}_0 [{4x y^{3 }]{{1} \atop {mx}} \right.\, dx

Replacing the limits:

\int\limits^\frac{1}{m}_0 {4x(1-(mx)^{3} )\, dx =\int\limits^\frac{1}{m}_0 {4x-4x(m^{3} x^{3} )\, dx =\int\limits^\frac{1}{m}_0 ({4x-4m^{3} x^{4}) \, dx

Solving now for dx:

[{\frac{4x^{2}}{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right. = [{2x^{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right.

Replacing the limits:

\frac{2}{m^{2} }-\frac{4m^{3}\frac{1}{m^{5}}}{5} =\frac{2}{m^{2} }-\frac{4\frac{1}{m^{2}}}{5} \\ \frac{2}{m^{2} }-\frac{4}{5m^{2} }=\frac{10m^{2}-4m^{2} }{5m^{4}} \\ \frac{6m^{2} }{5m^{4}} =\frac{6}{5m^{2}}

As I mentioned before, this volume is equal to 1, hence:

\frac{6}{5m^{2}}=1\\m^{2} =\frac{6}{5} \\m=\sqrt{\frac{6}{5} }

3 0
2 years ago
An $8 item is marked up 75%. What is the cost of the item?
UNO [17]
14 dollars because it’s only 75 % out of. 8
8 0
2 years ago
WILL GIVE BRAINLIEST!!!!
Vesna [10]

(2^8 \times 5^{-5} \times 19^0)^{-2} \times (\dfrac{5^{-2}}{2^3})^4 \times 2^{28} =

= 2^{-16} \times 5^{10} \times 1 \times \dfrac{5^{-8}}{2^{12}} \times 2^{28}

= 2^{-16} \times 5^{10} \times 5^{-8} \times 2^{-12} \times 2^{28}

= 2^{-16} \times 2^{-12} \times 2^{28} \times 5^{10} \times 5^{-8}

= 2^{-16-12+28} \times 5^{10-8}

= 2^{0} \times 5^{2}

= 1 \times 25

= 25

7 0
2 years ago
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