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Sati [7]
3 years ago
5

How do I add odd number fractions for example 5/9 and 3/5?​​

Mathematics
2 answers:
lilavasa [31]3 years ago
8 0

Answer:

if the denominators are not the same, then you have to use equivalent fractions which do have a common denominator. To do this, you need to find at least common multiple (LCM) of the two denominators. To add fractions with unlike denominators, rename the fractions with common denominator. Then add and simplify

miv72 [106K]3 years ago
7 0

Answer:

\frac{52}{45}  or  1\frac{7}{45}

Step-by-step explanation:

In order to add two fractions with different denominators, you will have to find its least common denominator (LCD):

The LCD of 9 and 5 is 45.

Next, we'll start with 5/9:

Divide 45 by 9, which results in 5. Then, multiply the numerator by 5:\frac{5}{9} = \frac{5*5}{45}  = \frac{25}{45}

The same goes with 3/5:

Divide 45 by the denominator, 5, then multiply 5 by the numerator 3:

\frac{3}{5} = \frac{9*3}{45}  = \frac{27}{45}

Now, we can finally add 25/45 and 27/45 since they have the same denominator. Doing so results in a sum of 52/45:

\frac{25}{45} + \frac{27}{45} = \frac{52}{45}

If you have to transform 52/45 into a proper fraction (which will result in a mixed fraction), then it will be 1\frac{7}{45}.

Please mark my answers as the Brainliest, if you find this explanation helpful :)

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53 cucumbers and 45 carrots

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1. Find the inverse functions of the following<br> a. f (x) = 5x +3
Mrac [35]

Answer:

y = 5x + 3

x = 5y + 3

5y + 3 = x

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Step-by-step explanation:

8 0
3 years ago
F(x) = 5x^3+ 2x^2 - 90x - 36. <br><br> finding all zeros by factoring and explaining the steps
ella [17]

Answer:

f(x)=5x^3-2x^2-90x-36=0

=x^2(5x-2)-18(5x-2)=(x^2-18)(5x-2)=0

x^2-18=0/5x-2=0

x^2=18=x=9√2

5x-2=0

x=2/5

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8 0
3 years ago
Find all solutions of the equation in the interval [0, 2pi).
natali 33 [55]

Answer:

Step-by-step explanation:

Begin by squaring both sides to get rid of the radical. Doing that gives you:

sin^2x=1-cosx

Now use the Pythagorean identity that says

sin^2x =1-cos^2x and make the replacement:

1-cos^2x=1-cosx. Now move everything over to one side of the equals sign and set it equal to 0 so you can factor:

1-cos^2x+cosx-1=0 and then simplify to

cosx-cos^2x=0

Factor out the common cos(x) to get

cosx(1-cosx)=0 and there you have your 2 trig equations:

cos(x) = 0 and 1 - cos(x) = 0

The first one is easy enough to solve. Look on the unit circle and see where, one time around, where the cos of an angle is equal to 0. That occurs at

x=\frac{\pi }{2},\frac{3\pi}{2}

The second equation simplifies to

cos(x) = 1

Again, look to the unit circle and find where the cos of an angle is equal to 1. That occurs at π only.

So, in the end, your 3 solutions are

x=\frac{\pi}{2},\pi,\frac{3\pi}{2}

8 0
3 years ago
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