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3 years ago

What is the equation of the line ?

Mathematics

1 answer:

Ronch [10]3 years ago

7 0

Look at the picture.

The point-slope form:

$y-y_1=m(x-x_1)\\\\m=\dfrac{y_2-y_1}{x_2-x_1}$

We have the points (-3, -3) and (2, 0). Substitute:

$m=\dfrac{0-(-3)}{2-(-3)}=\dfrac{3}{5}\\\\y-0=\dfrac{3}{5}(x-2)\qquad|\text{use distributive property}\\\\y=\dfrac{3}{5}x-\dfrac{6}{5}\qquad|\cdot5\\\\5y=3x-6\qquad|-5y\\\\0=3x-5y-6\qquad|+6\\\\3x-5y=6$

Answer:

point-slope form: $y=\dfrac{3}{5}(x-2)$

slope-intercept form: $y=\dfrac{3}{5}x-\dfrac{6}{5}$

standard form: $3x-5y=6$

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3 years ago

Read 2 more answers

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

Nana76 [90]

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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Answer:

Step-by-step explanation:

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<h2>hope it helps you see the attachment for further information ✌✌✌✌✌</h2>

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