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3 years ago

How to solve? What are the rules in BODMAS

Mathematics

2 answers:

leva [86]3 years ago

8 0

Answer:

18/35

Step-by-step explanation:

1 1/5 ÷ 2 1/3

Change from mixed numbers to improper fractions

1 1/5 = (5*1+1)/5 = 6/5

2 1/3 = (3*2+1)/3 = 7/3

6/5÷ 7/3

Copy dot flip

6/5* 3/7

18/35

Elden [556K]3 years ago

8 0

Answer:

Step-by-step explanation:

I don't say you must have to mark my ans as brainliest but if it has really helped u plz don't forget to thank me my friend...

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Um. i just need help with this! so if anyone knows help me plz <3

kap26 [50]

Answer:

Last one

Step-by-step explanation:

You can simplify the first one or compare both of them if they have the same ratios! Hope this helps <3

6 0

2 years ago

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A sheet of paper has a perimeter of 39 in. if the width of the paper is 8.5 in . what is the length of the sheet of paper

Valentin [98]

If the perimeter is 39 what you have to do is 39 - 8.5 - 8.5 that will give you 22 and divide 22/2 and that gives you 11. So 11 is the length

5 0

3 years ago

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I need help please i will give u brainly

Vinvika [58]

Answer:

6.04 inches.

Step-by-step explanation:

Corbin want's to make a model, in which every 100 feet would be shown by 1 inch.

The monument is 604 feet in height.

Divide 604 with 100:

604/100 = 6.04

The height of Corbin's model would thereby be 6.04 inches.

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3 years ago

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15 points <br> Which graph represents the function g(x) = |x+4|+2?

rewona [7]

Answer:

number 3

Step-by-step explanation:

7 0

3 years ago

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Differentiate with respect to X <br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B%20%5Cfrac%7Bcos2x%7D%7B1%20%2Bsin2x%20%7D%20

Mice21 [21]

Power and chain rule (where the power rule kicks in because $\sqrt x=x^{1/2}$ ):

$\left(\sqrt{\dfrac{\cos(2x)}{1+\sin(2x)}}\right)'=\dfrac1{2\sqrt{\frac{\cos(2x)}{1+\sin(2x)}}}\left(\dfrac{\cos(2x)}{1+\sin(2x)}\right)'$

Simplify the leading term as

$\dfrac1{2\sqrt{\frac{\cos(2x)}{1+\sin(2x)}}}=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}$

Quotient rule:

$\left(\dfrac{\cos(2x)}{1+\sin(2x)}\right)'=\dfrac{(1+\sin(2x))(\cos(2x))'-\cos(2x)(1+\sin(2x))'}{(1+\sin(2x))^2}$

Chain rule:

$(\cos(2x))'=-\sin(2x)(2x)'=-2\sin(2x)$

$(1+\sin(2x))'=\cos(2x)(2x)'=2\cos(2x)$

Put everything together and simplify:

$\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{(1+\sin(2x))(-2\sin(2x))-\cos(2x)(2\cos(2x))}{(1+\sin(2x))^2}$

$=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{-2\sin(2x)-2\sin^2(2x)-2\cos^2(2x)}{(1+\sin(2x))^2}$

$=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{-2\sin(2x)-2}{(1+\sin(2x))^2}$

$=-\dfrac{\sqrt{1+\sin(2x)}}{\sqrt{\cos(2x)}}\dfrac{\sin(2x)+1}{(1+\sin(2x))^2}$

$=-\dfrac{\sqrt{1+\sin(2x)}}{\sqrt{\cos(2x)}}\dfrac1{1+\sin(2x)}$

$=-\dfrac1{\sqrt{\cos(2x)}}\dfrac1{\sqrt{1+\sin(2x)}}$

$=\boxed{-\dfrac1{\sqrt{\cos(2x)(1+\sin(2x))}}}$

5 0

3 years ago