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2 years ago

Find x and y intercepts y=(2x+10)(x-2)

Mathematics

1 answer:

slega [8]2 years ago

8 0

Step-by-step explanation:

$y = (2x + 10)(x - 2)$

$y = 2x(x - 2) + 10(x - 2)$

$y = 2x ^{2} - 4x + 10x - 20$

$y = 2x ^{2} + 6x - 20$

$to \: find \: x \: intercept \: y = 0$

$0 = 2x ^{2} + 6x - 20$

$x = 2 \: or \: - 5$

$to \: find \: y \: intercept \: x = 0$

$y = 2(0)^{2} + 6(0) - 20$

$y = - 20$

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three bags of sweets weigh 27/4 kg. two of them have the same weight and the third bag is heavier than each of the bags of equal

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I'm here buddy,

so, let's take the value of the two bags with equal weight as x.

= x + x + (x + $\frac{6}{5}$ ) = $\frac{27}{4}$

= 3x + $\frac{6}{5}$ = $\frac{27}{4}$

= 3x = $\frac{27}{4}$ - $\frac{6}{5}$

( let's take the LCM of 4 and 5 = 20

= 3x = $\frac{135}{20}$ - $\frac{24}{20}$

= 3x = $\frac{111}{20}$

= x = $\frac{111}{20}$ ÷ $\frac{3}{1}$ = $\frac{111}{20}$ × $\frac{1}{3}$ = $\frac{37}{20}$

So, the weight of the equal bags are $\frac{37}{20}$ and the weight of the third bag ( heavy one ) is $\frac{37}{20}$ + $\frac{6}{5}$ = $\frac{37}{20}$ + $\frac{24}{20}$ = $\frac{61}{20}$

1st bag = $\frac{37}{20}$ kg

2nd bag = $\frac{37}{20}$ kg

3rd bag = $\frac{61}{20}$ kg

Hope it helps...

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3 years ago

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