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2 years ago

PLEASE ANSWER ASAP

Mathematics

1 answer:

gladu [14]2 years ago

6 0

Answer:

Step-by-step explanation:

I'll assume these are the equations:

11x + y = 18

4x + y = 18

-2x - 2y = 18

=======

The easiest approach to graphing, for me, is to rearrange each to y=mx+b (slope-intercept) format:

y = -11x+18 y-intercept of 18 (0,18)
y = -4x + 18 y-intercept of 18 (0,18)
y = - x -9 y-intercept of -9 (0,-9)

It seems to me that all three can be graphed easily from simply reading the slope and y-intercept values. Since these are straight lines, all we need is two points for each line, and one is standing out in plain view, the y-intercept: (0,y-intercept). The second point can be determined by using whatever value of x makes the calculation easy

An example: y = - x -9. Plot (0,-9) for the y intercept, and then calculate one additional point (e.g., for x = - 9). (-9,0) Then connect a straight line between these two points and presto (metric term for magic), a graph. For the first equation, I picked 20 for x. For x = 20, y = -29. That was easy, and we have the second point: (1,-10)

See the attachment for how this was done. The first points are all for the y-intercept (0,[18 or 9]). All of the second points were calculated in my head by using a convenient value for x.

This approach doesn't work well for non-linear equations. There, I find it easier to set up a table of values - a spreadsheet such as Excel is my tool for the calculations.

The attachment also demonstrates the Really Easy Way to graph functions. DESMOS, a free, and excellent, online calculator.

Read more about products at:

brainly.com/question/10873737

#SPJ1

3 0

1 year ago

where would an imaginary line need to be drawn to reflect across an axis of symmetry so that a regular pentagon can carry onto i

irina [24]

Answer:

An imaginary line would need to be drawn at any angle to the center of the side opposite the angle to reflect across an axis of symmetry so that a regular pentagon can carry onto itself. Hope this answer helps.

5 0

2 years ago

BRAINLIEST IF CORRECT

stellarik [79]

The answer is C. f(x) = 0.05

4 0

2 years ago

Read 2 more answers

Find dy/dx if y =x^3+5x+2/x²-1

stiks02 [169]

<u>Differentiate using the Quotient Rule</u> –

$\qquad$ $\pink{\twoheadrightarrow \sf \dfrac{d}{dx} \bigg[\dfrac{f(x)}{g(x)} \bigg]= \dfrac{ g(x)\:\dfrac{d}{dx}\bigg[f(x)\bigg] -f(x)\dfrac{d}{dx}\:\bigg[g(x)\bigg]}{g(x)^2}}\\$

According to the given question, we have –

f(x) = x^3+5x+2
g(x) = x^2-1

Let's solve it!

$\qquad$ $\green{\twoheadrightarrow \bf \dfrac{d}{dx}\bigg[ \dfrac{x^3+5x+2 }{x^2-1}\bigg]} \\$

$\qquad$ $\twoheadrightarrow \sf \dfrac{(x^2-1) \dfrac{d}{dx}(x^3+5x+2) - ( x^3+5x+2) \dfrac{d}{dx}(x^2-1)}{(x^2-1)^2 }\\$

$\qquad$ $\twoheadrightarrow \sf \dfrac{(x^2-1)(3x^2+5) - ( x^3+5x+2) 2x}{(x^2-1)^2 }\\$

$\qquad$ $\pink{\sf \because \dfrac{d}{dx} x^n = nx^{n-1} }\\$

$\qquad$ $\twoheadrightarrow \sf \dfrac{3x^4+5x^2-3x^2-5-(2x^4+10x^2+4x)}{(x^2-1)^2 }\\$

$\qquad$ $\twoheadrightarrow \sf \dfrac{3x^4+5x^2-3x^2-5-2x^4-10x^2-4x}{(x^2-1)^2 }\\$

$\qquad$ $\green{\twoheadrightarrow \bf \dfrac{x^4-8x^2-4x-5}{(x^2-1)^2 }}\\$

$\qquad$ $\pink{\therefore \bf{\green{\underline{\underline{\dfrac{d}{dx} \dfrac{x^3+5x+2 }{x^2-1}} = \dfrac{x^4-8x^2-4x-5}{(x^2-1)^2 }}}}}\\\\$

7 0

2 years ago