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Anestetic [448]
3 years ago
10

PLEASE HELP What is the slope of this line? a) -2 b) -1/2c) 1/2d) 2 ​

Mathematics
2 answers:
anyanavicka [17]3 years ago
5 0

Answer:

C: 1/2

Step-by-step explanation:

Use the slope formula

IgorLugansk [536]3 years ago
3 0

Answer:1/2

Step-by-step explanation:

1__1/_2__6

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Find the slope of the line going through the points (11,5) and (20,5)
Pani-rosa [81]
So remember the formula for working out the slope:

(y2 - y1) ÷ (x2 - x1)

We already know two points:

(11, 5) and (20, 5) (Remember points are like (x, y)


Therefore y2 = 5 and y1 = 5 and x2 = 20 and x1 = 11

Substitute these into the formula from the start:

(y2 - y1) ÷ (x2 - x1)
(5 - 5) ÷ (20 - 11)
0 ÷ 9

And we can determine the slope is equal to 0 as 0 ÷ 9 = 0
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2 years ago
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zysi [14]
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3 years ago
Find the total cost if tools plus company makes 210 drills
Serggg [28]
How unspecific? What's the cost of one
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3 years ago
Need help with problems 16 and 17, please? I'll do brainiac
natali 33 [55]

Answer: 16) y=-0.5x+3 and 17) y=2x-3

Step-by-step explanation:

16: The slope-intercept form of a line is y=mx+b, where m is the slope, and b is the y-value at the y-intercept.

Since y=3 at x=0, the y-intercept is (0,3), and b=3.

Slope=\frac{y_{2} -y_{1} }{x_{2}-x_{1}  } where (x_{1} ,y_{1}) and (x_{2} ,y_{2}) are two points on the line. Choose (0,3) and (2,2):

Slope=\frac{2-3}{2-0}=-0.5

Plug m=-0.5 and b=3 into y=mx+b:

y=-0.5x+3

17: The slope-intercept form of a line is y=mx+b, where m is the slope, and b is the y-value at the y-intercept.

Because y=-3 at x=0, the y-intercept is (0,-3), and b=-3.

Slope=\frac{y_{2} -y_{1} }{x_{2}-x_{1}  } where (x_{1} ,y_{1}) and (x_{2} ,y_{2}) are two points on the line. Choose (0,-3) and (1,-1):

Slope=\frac{-3-(-1)}{0-1}=\frac{-2}{-1}=2

Plug m=2 and b=-3 into y=mx+b:

y=2x-3

5 0
2 years ago
Find gradient <br><br>xe^y + 4 ln y = x² at (1, 1)​
cricket20 [7]

xe^y+4\ln y=x^2

Differentiate both sides with respect to <em>x</em>, assuming <em>y</em> = <em>y</em>(<em>x</em>).

\dfrac{\mathrm d(xe^y+4\ln y)}{\mathrm dx}=\dfrac{\mathrm d(x^2)}{\mathrm dx}

\dfrac{\mathrm d(xe^y)}{\mathrm dx}+\dfrac{\mathrm d(4\ln y)}{\mathrm dx}=2x

\dfrac{\mathrm d(x)}{\mathrm dx}e^y+x\dfrac{\mathrm d(e^y)}{\mathrm dx}+\dfrac4y\dfrac{\mathrm dy}{\mathrm dx}=2x

e^y+xe^y\dfrac{\mathrm dy}{\mathrm dx}+\dfrac4y\dfrac{\mathrm dy}{\mathrm dx}=2x

Solve for d<em>y</em>/d<em>x</em> :

e^y+\left(xe^y+\dfrac4y\right)\dfrac{\mathrm dy}{\mathrm dx}=2x

\left(xe^y+\dfrac4y\right)\dfrac{\mathrm dy}{\mathrm dx}=2x-e^y

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2x-e^y}{xe^y+\frac4y}

If <em>y</em> ≠ 0, we can write

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2xy-ye^y}{xye^y+4}

At the point (1, 1), the derivative is

\dfrac{\mathrm dy}{\mathrm dx}\bigg|_{x=1,y=1}=\boxed{\dfrac{2-e}{e+4}}

4 0
3 years ago
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