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2 years ago

What is the simplest form of the ratio 11 : 16? A. 1 : 6 B. 5 : 8 C. 11 : 16 D. 22 : 32

Mathematics

2 answers:

Volgvan2 years ago

4 0

Answer:

Step-by-step explanation:

c because 11 and 16 have no GCF other than 1

Maslowich2 years ago

4 0

11:16 may not be simplified, so 11:16 is the simplest form.

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1) In the year 2010, Shaniya invested $1 in the stock market. In the year 2022, her $1 was worth $2. In the year 2034, it was wo

vovikov84 [41]

Answer:

$1,073,741,824

Step-by-step explanation:

Every 12 years, the price increases by double. So, first we would need to find the difference of years between 2370 and 2046.

2370-2046=324

Divide this number by 12 to see how many times the number is doubled.

324/12=27

Continue this pattern of "every 12 years, the price increases by 2x" 27 times.

Ex: 2058-->$16

Ex: 2070-->$32

Ex: 2082-->$64

etc....

3 0

2 years ago

Which is the slope of for (-5,9) and (-5,-5).

Drupady [299]

C) Undefined

formula: y-y/x-x

9--5= 9+5=14

-5--5= -5+5= 0

14/0

you can't divide by 0 so it's undefined

7 0

3 years ago

Read 2 more answers

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 > 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

A bottle filled with juice weighs 8 pounds. One morning, George drank 2/3 of the juice in the filled bottle. The bottle and the

Marysya12 [62]

The remaining weight of the empty bottle is 0.5 because each third of the bottle weighs 2.5 pounds and of the remaining juice and the bottle was 3 pounds 2.5 subtracted from 3 is 0.5

8 0

3 years ago

The price to ship a box varies directly as the weight of the box in pounds. A box that weighs 10 pounds costs $10.90 to ship.

xenn [34]

Well we can say 10.90/10 = the cost per pound to ship the item
we can infer this from the above text so
10.90/10 = 1.09.
So 1.09 * 37 = 40.33
$40.33
or 40 dollars and 33 cents to ship the second box with a weight of 37lb (pounds).

3 0

3 years ago