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3 years ago

11/8+(-11/10)= Plz, I don't know the answer.

Mathematics

2 answers:

hammer [34]3 years ago

6 0

Answer:

11/40

Step-by-step explanation:

If your answering in fraction and not in decimal this is how you do it:

1. Get rid of the brackets surrounding -11/10. You do this by multiplying the positive sign (+) by the negative sign (-) in front of 11. Now the rule says that multiplying a positive with a negative always gives you a negative therefore your fraction in the brackets will remain negative while removing the plus sign. Now your problem should look like this: 11/8-11/10.

2. Finding the common multiple of the denominators. You can't just minus the fractions unless the denominators are the same. To do this you first have to find the lowest common multiple (lcm) of denominators 8 and 10. You can list 5 multiples for each to find this or more depending on how many multiples you need to reach the lcm.

8= 8,16,24,32,40,48,etc.

10=10,20,30,40,50,etc.

The lcm for 8 and 10 is 40.

3. Change the fractions to substitute the new value of the denominators in. To do this you have to multiply the fractions. For example, multiply 11/8 by 5/5 to get 55/40(11×5/8×5). Do the same for the second fraction: 11/10×4/4(11×4/10×4) which gives 44/40.

4. Now finally just minus which is just 55/40-44/40 and your answer will be 11/40. Minus the numerators only not the denominators.

I'm not really good at explaining but I hope this helped

sukhopar [10]3 years ago

5 0

0.275 is the answer

-11/10=-1.1

11/8=1.375

1.375 + (-1.1) = 0.275

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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:

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

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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:

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

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

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