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horsena [70]
3 years ago
10

Find the length of DE using the distance formula.

Mathematics
2 answers:
liraira [26]3 years ago
4 0

Answer:

Is there a picture included?

Step-by-step explanation:

ryzh [129]3 years ago
3 0

Answer:

The length of DE is 10 units

Step-by-step explanation:

Find the length of DE using the distance formula.

(-4,2) (4,-4)

just did it.

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The answer would be .80 or it may be 1
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Will award Brainlyest!!! Choose the equation that represents a line that passes through points (−3, 2) and (2, 1).
Pachacha [2.7K]

Answer:2. Rate of change, or slope, is the change in y over change in x. Remember that a point is given in (x, y). Also that you need two points to find the change.

If you have two points, (x1, y1) and (x2, y2), the rate of change or slope

In your problem, the given points are (0, -1) and (2, -11)

The "initial value" is the y-intercept - where the graph intersects the y-axis and also where x = 0. The given point (0, -1) has x = 0, so the initial value is -1.

4. Remember the model equation for a graph of a line is

y = mx + b,

where m is slope and b is y-intercept - where the graph intersects the y-axis and also where x = 0.

The x-axis of the graph would be number of weeks, since it is independent.

The y-axis of the graph would be how much money she has.

Since she starts (at 0 weeks) at $25, the first point would be (0, 25). Notice here that it is the y-intercept since x = 0. The value of b in the model equation of a line y = mx + b would be 25.

The rate of change, or slope, is $5/week. This is m in the model equation of a line y = mx + b.

Putting b = 25 and m = 5 into the model equation y = mx + b (remember we use this equation because it works for all lines), the equation would be

y = 5x + 25.

4.

The slope of this will be -0.5degrees/hr, or just -0.5 if it is used as m in the 

y = mx+b equation. The slope is negative because it "decreases".

5. 

Plot each point and connect the dots. Remember the first number corresponds to the x-axis and the second to the y-axis.

For example in the table for x = -2 and y = -6, the point is (-2, -6), and from the origin (0,0) you go left 2 on the x-axis and down 6 to plot the point.

9.

I think you can do this if you can do #3, but now you have to find b by plugging in given values.

y = mx + b

y = 1212x + b

We can plug in the given point (3,1) to find b.

1 = 1212(3) + b

b = -3625

y = 1212 - 3625

10. Don't know which function you are referring to.

The slope is $10/week, or m = 10.

A point is (4, 90) because "After 4 weeks, Roberta has $90."

The y-intercept is 50 because x (weeks) = 0 at that time.

y = mx + b

y = 10x + 50

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klio [65]
Your answer is xy-1<span><span>
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A rectangular sign must have an arae of 43 square yards. its lenght must be 2 yards more than its width. find the dimensions of
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We know A=L•W so set 43 equal to A. Equation should look like this thus far: 43=L•W. Now when the question says more than we know that means adding. Since we don’t know the width let w=width. So know your equation should look like this: 43=2+w•(w). Knowing this equation you should be able to solve for the dimensions of the sign. If you need help on that let me know.
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3 years ago
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
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