U + 4/5 = 2 and 1/3
Subtract 4/5 from each side of the equation:
U = (2 and 1/3) - (4/5)
Now it's just problem in plain old adding and subtracting fractions,
just like the ones you've done many times before.
First let's change (2 and 1/3) to a fraction: 2 and 1/3 = 7/3
So you have to find the value of (7/3) - (4/5) .
In order to add or subtract fractions, they need to have a common denominator.
The least common multiple of 3 and 5 is 15, so that's a good choice.
7/3 = 35/15
4/5 = 12/15
Now the problem is: (35/15) - (12/15).
That's 23/15 . . . . . the same thing as <u>1 and 8/15</u> .
That's the value of ' U '. What an ugly number !
Answer:
Its 7.333333333333333333 (3 forever)
First you isolate y onto one side of the equation
y/9= 44/54 --> y = 9 x 44/54
Multiply the right side of the equation
9 x 44/54 = 396/54
Now divide to get the whole answer
396/54 = 7.3333333333 (3 forever)
y = 7.333333
Answer:
2.52
Step-by-step explanation:
21*.12=
2.52
Answer:
A
Step-by-step explanation:
The position of an atom moving inside a cathode ray tube is given by the function:
Where f(t) is in meters and <em>t</em> is in seconds.
And we want to determine its instantaneous velocity at <em>t</em> = 2.5 seconds.
The velocity function is the derivative of the position function. Thus, find the derivative of the function:
<em />
<em />
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Then the instantaneous velocity at <em>t</em> = 2.5 will be:
Our answer is A.
Answer:
i) 
\frac{3}{5} + (- \frac{1}{2}) = \frac{6}{10} - \frac{5}{10} = \frac{1}{10}
ii)
4^{3} + 8^{2} + \sqrt{9}
iii) ![(\frac{4}{5})^{2}. \sqrt[3]{8} \leqx^{3} - 3x + 6 \leq \sqrt{\frac{1}{3}} \Rightarrow \hspace{0.2cm}](https://tex.z-dn.net/?f=%28%5Cfrac%7B4%7D%7B5%7D%29%5E%7B2%7D.%20%5Csqrt%5B3%5D%7B8%7D%20%5Cleqx%5E%7B3%7D%20-%203x%20%2B%206%20%5Cleq%20%5Csqrt%7B%5Cfrac%7B1%7D%7B3%7D%7D%20%5CRightarrow%20%5Chspace%7B0.2cm%7D)
(\frac{4}{5})^{2}. \sqrt[3]{8} \leqx^{3} - 3x + 6 \leq \sqrt{\frac{1}{3}}
Step-by-step explanation:
i) \frac{3}{5} + (- \frac{1}{2}) = \frac{6}{10} - \frac{5}{10} = \frac{1}{10}
ii) 4^{3} + 8^{2} + \sqrt{9}
iii) (\frac{4}{5})^{2}. \sqrt[3]{8} \leqx^{3} - 3x + 6 \leq \sqrt{\frac{1}{3}}
