What is this asking? 8x-4y=12

Mathematics

1 answer:

kkurt [141]3 years ago

8 0

Maybe from the equation of a line from y=mx+c
Where c ( intercept) is -12

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Which of the following sets of values correctly completes the T chart for the equation Y =|3x+1|. -5,1,7

Ivahew [28]

Option C:

5, 1, 7

Solution:

Given equation:

y = |3x + 1|

In the chart, x values are given as –2, 0, 2.

To find the value of y for the corresponding x given.

Substitute x = –2 in the equation of y.

y = |3(–2) + 1|

= |–6 + 1|

= |–5| (Absolute value of a negative number is positive)

y = 5

Substitute x = 0 in the equation of y.

y = |3(0) + 1|

= |0 + 1|

= |1|

y = 1

Substitute x = 2 in the equation of y.

y = |3(2) + 1|

= |6 + 1|

= |7| (Absolute value of a negative number is positive)

y = 7

Option C is the correct answer.

The values of y are 5, 1, 7.

The answer image of the table is attached below.

8 0

4 years ago

Integration: How would I get to this answer?

Scrat [10]

Given that $y$ attains a maximum at $x=1$ , it follows that $y'=0$ at that same point. So integrating once gives

$\displaystyle\int\frac{\mathrm d^2y}{\mathrm dx^2}\,\mathrm dx=\int-8x\,\mathrm dx$
$\dfrac{\mathrm dy}{\mathrm dx}=-4x^2+C_1$
$\implies -4(1)^2+C_1=0\implies C_1=4$

and so the first derivative is $y'=-4x^2+4$ .

Integrating again, you get

$\displaystyle\int\frac{\mathrm dy}{\mathrm dx}\,\mathrm dx=\int(-4x^2+4)\,\mathrm dx$
$y=-\dfrac43x^3+4x+C_2$

You know that this curve passes through the point (2, -1), which means when $x=2$ , you have $y=-1$ :

$-1=-\dfrac43(2)^3+4(2)+C_2$
$\implies C_2=\dfrac53$

and so

$y=-\dfrac43x^3+4x+\dfrac53$