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3241004551 [841]
2 years ago
10

=" - \frac{5}{8} + \frac{1}{4} " alt=" - \frac{5}{8} + \frac{1}{4} " align="absmiddle" class="latex-formula">
​
Mathematics
1 answer:
bagirrra123 [75]2 years ago
4 0

Answer:

\large\boxed{-\dfrac{5}{8}+\dfrac{1}{4}=-\dfrac{3}{8}}

Step-by-step explanation:

-\dfrac{5}{8}+\dfrac{1}{4}=\dfrac{1}{4}+\left(-\dfrac{5}{8}\right)=\dfrac{1}{4}-\dfrac{5}{8}\\\\\text{find LCD:}\\\\8=2\cdot4,\ \text{therefore. LCD(4, 8) = 8}\\\\\dfrac{1}{4}=\dfrac{1\cdot2}{4\cdot2}=\dfrac{2}{8}\\\\\text{continuing}\\\\=\dfrac{2}{8}-\dfrac{5}{8}=\dfrac{2-5}{8}=-\dfrac{3}{8}

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3 years ago
A, b, c and d are positive integers, such that a+b+ ab = 76, c+d+ cd = 54. Find (a+b+c+d)·a·b·c·d.
lutik1710 [3]

Notice that

(1 + <em>x</em>)(1 + <em>y</em>) = 1 + <em>x</em> + <em>y</em> + <em>x y</em>

So we can add 1 to both sides of both equations, and we use the property above to get

<em>a</em> + <em>b</em> + <em>a b</em> = 76  ==>  (1 + <em>a</em>)(1 + <em>b</em>) = 77

and

<em>c</em> + <em>d</em> + <em>c d</em> = 54  ==>  (1 + <em>c</em>)(1 + <em>d</em>) = 55

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(or the other way around, since the given relations are symmetric)

and

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<em>d</em> + 1 = 11  ==>  <em>d</em> = 10

Now substitute these values into the desired quantity:

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