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2 years ago

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Which answer is in slope-intercept form for the given equation? 4x - 9y = -36 Thanks

1 answer:

jek_recluse [69]2 years ago

4 0

Answer:

y = 4/9 x + 4

Step-by-step explanation:

Slope intercept form: y = mx = b

So 4x - 9y = -36

9y = 4x+ 36

y = 4/9 x + 4

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Factor 8y^2+17+2. Please

In-s [12.5K]

It is A (8y + 1)(y + 2) because if you multiply it out you get 8y^2 + 16y + 1y + 2 which is equal to 8y^2 + 17y + 2

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2 years ago

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A given line has the equation 10x + 2y = −2.

worty [1.4K]

The equation is $\boxed{ \ y = - 5x + 12 \ or \ y = 12 - 5x} \ }$

<h3>Further explanation </h3>

This case asking the end result in the form of a slope-intercept.

<u>Step-1: find out the gradient. </u>

10x + 2y = -2

We isolate the y variable on the left side. Subtract both sides by 10x, we get:

2y = - 10x - 2

Divide both sides by two

y = -5x -1

The slope-intercept form is $\boxed{ \ y = mx + c \ }$ , with the coefficient m as a gradient. Therefore, the gradient is m = -5.

If you want a shortcut to find a gradient from the standard form, implement this:

$\boxed{ \ ax + by = k \rightarrow m = - \frac{a}{b} \ }$

10x + 2y = −2 ⇒ a = 10, b = 2

$\boxed{m = - \frac{10}{2} \rightarrow m = -5}$

<u>Step-2:</u> the conditions of the two parallel lines

The gradient of parallel lines is the same $\boxed{ \ m_1 = m_2 \ }$ . So $\boxed{m_1 = m_2 = -5}.$

<u>Final step:</u> figure out the equation, in slope-intercept form, of the parallel line to the given line and passes through the point (0, 12)

We use the point-slope form.

$\boxed{ \ \boxed{ \ y - y_1 = m(x - x_1)} \ }$

Given that

m = -5
(x₁, y₁) = (0, 12)

y - 12 = - 5(x - 0)

y - 12 = - 5x

After adding both sides by 12, the results is $\boxed{ \ y = - 5x + 12 \ or \ y = 12 - 5x} \ }$

<u>Alternative steps </u>

Substitutes m = -5 and (0, 12) to slope-intercept form $\boxed{ \ y = mx + c \ }$

12 = -5(0) + c

Constant c is 12 then arrange the slope-intercept form.

Similar results as above, i.e. $\boxed{ \ y = - 5x + 12 \ or \ y = 12 - 5x} \ }$

<u>Note: </u>

$\boxed{Standard \ form: ax + by = c, with \ a > 0}$

$\boxed{Point-slope \ form: y - y_1 = m(x - x_1)}$

$\boxed{Slope-intercept \ form: y = mx + k}$

<h3>Learn more </h3>

A similar problem brainly.com/question/10704388
Investigate the relationship between two lines brainly.com/question/3238013
Write the line equation from the graph brainly.com/question/2564656

Keywords: given line, the equation, slope-intercept form, standard form, point-slope, gradien, parallel, perpendicular, passes, through the point, constant

7 0

3 years ago

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Which of the following describes the transformations of g (x) = negative (2) Superscript x + 4 Baseline minus 2 from the parent

o-na [289]

Answer: First option.

Step-by-step explanation:

Below are some transformations for a function $f(x)$ :

1. If $f(x)+k$ , the function is shifted "k" units up.

2 If $f(x)-k$ , the function is shifted "k" units down.

3. If $f(x-k)$ , the function is shifted "k" units right.

4. If $f(x+k)$ , the function is shifted "k" units left.

5. If $-f(x)$ , the function is reflected over the x-axis.

6. If $f(-x)$ , the function is reflected over the y-axis.

Then, given the parent function $f(x)$ :

$f(x)=2^x$

And knowing that the the other function is:

$g(x)=-2^{(x+4)}-2$

You can identify that the function $g(x)$ is obtained by:

- Shifting the function $f(x)$ 4 units left.

- Reflecting the function $f(x)$ 4 over the x-axis.

- Shifting the function $f(x)$ 2 units down.

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2 years ago

Can you help my sister with fractions while I'm studying. I need to review on my exams

Travka [436]

Answer:

4 + $\frac{1}{2}$ = 4 $\frac{1}{2}$

Step-by-step explanation:

We have 4 whole circles and half a circle so 4 + $\frac{1}{2}$ = 4 $\frac{1}{2}$

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1 year ago

Find a closed-form solution to the integral equation y(x) = 3 + Z x e dt ty(t) , x > 0. In other words, express y(x) as a fun

MrMuchimi

Answer:

$y{x} = \sqrt{7+2Inx}$

Step-by-step explanation:

$y(x)= 3 + \int\limits^x_e {dx}/ \, ty(t) , x>0}$

Let say; By y(x)= y(e)

we have;

$y(e)= 3 + \int\limits^e_e {dt}/ \, ty= 3+0$

Using Fundamental Theorem of Calculus and differentiating by Lebiniz Rule:

$y^{1} (x) = 0 + 1/ xy$

$y^{1} = 1/xy$

dy/dx = 1/xy

$\int\limits {y} \, dxy = \int\limits \, dx/x$

$y^{2}/2 Inx + C$

RECALL: y(e) = 3

$(3)^{2} / 2 = In (e) + C$

$\frac{9}{2} =In(e)+C$

$\frac{9}{2} - 1 = C$

$\frac{7}{2} = C$

$y^{2} / 2 = In x +C$

$y^{2} / 2 = In x +7/2$

MULTIPLYING BOTH SIDE BY 2 , TO ELIMINATE THE DENOMINATOR, WE HAVE;

$y^{2} = {7+2Inx}$

$y{x} = \sqrt{7+2Inx}$

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3 years ago