IF $a = 4, b = 8$ $a=4, b=8$ , and $c = 5$ $c=5$ , what is $c \cdot \frac{b c}{4} - (7 - a)$ $c * (bc)/4-(7-a)$ ?

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IF $a = 4, b = 8$ $a=4, b=8$ , and $c = 5$ $c=5$ , what is $c \cdot \frac{b c}{4} - (7 - a)$ $c * (bc)/4-(7-a)$ ?

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Answer 1 · 2018-03-13T03:37:21

$47$ $47$

Explanation:

Given -

$a = 4$ $a=4$
$b = 8$ $b=8$
$c = 5$ $c=5$

$= c . \frac{b c}{4} - (7 - a)$ $=c.(bc)/4-(7-a)$

$= 5 \times \frac{8 \times 5}{4} - (7 - 4)$ $=5xx(8xx5)/4-(7-4)$

$5 \times \frac{40}{4} - (3)$ $5xx40/4-(3)$

$5 \times 10 - 3$ $5xx10-3$

$50 - 3 = 47$ $50-3=47$

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IF $a = 4, b = 8$ $a=4, b=8$ , and $c = 5$ $c=5$ , what is $c \cdot \frac{b c}{4} - (7 - a)$ $c * (bc)/4-(7-a)$ ?

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IF a=4,b=8a=4, b=8, and c=5c=5, what is c⋅bc4−(7−a)c * (bc)/4-(7-a)?

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IF $a = 4, b = 8$ $a=4, b=8$ , and $c = 5$ $c=5$ , what is $c \cdot \frac{b c}{4} - (7 - a)$ $c * (bc)/4-(7-a)$ ?