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baherus [9]
2 years ago
12

Can someone pls help me!!!

Mathematics
1 answer:
Annette [7]2 years ago
8 0

Answer:

10 i believe

Step-by-step explanation:

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B is the midpoint of AC and E' is the midpoint of BD. If A(-9,-4), C(-1, 6), and E(-4,-3), find the coordinates of D.
Ipatiy [6.2K]
The coordinates for D are (-4, -7)

First we must locate point B as it is vital to finding the midpoint of BD. To do this, we take the average of the endpoints AC since B is its midpoint. 

x values = -9 + 1 = -8
Then divide by 2 for the average -8/2 = -4

y values = -4 + 6 = 2
Then divide by 2 for the average 2/2 = 1

Therefore B must be (-4, 1)

Now we know the values of E must be the average of B and D. So we can write equations for each coordinate since we know they are averages. 

x - values = (Bx + Dx)/2 = Ex
(-4 + Dx)/2 = -4 ---> multiply both sides by 2
-4 + Dx = -8 ---> add -4 to both sides
Dx = -4

y - values = (By + Dy)/2 = Ey
(1 + Dy)/2 = -3 ---> multiply both sides by 2
1 + Dy = -6 ---> subtract 1 from both side
Dy = -7

So the coordinates for D must be (-4, -7)
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3 years ago
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Help please this is urgent
Dennis_Churaev [7]

Answer:

2

Step-by-step explanation:

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3 years ago
How does a change in demand relate to a demand curve
marissa [1.9K]
A change in any one of the underlying factors that determine what quantity people are willing to buy at a given price will cause a shift in demand. Graphically, the new demand curve lies either to the right (an increase) or to the left (a decrease) of the original demand curve.
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3 years ago
Suppose Upper F Superscript prime Baseline left-parenthesis x right-parenthesis equals 3 x Superscript 2 Baseline plus 7 and Upp
Sedaia [141]

It looks like you're given

<em>F'(x)</em> = 3<em>x</em>² + 7

and

<em>F</em> (0) = 5

and you're asked to find <em>F(b)</em> for the values of <em>b</em> in the list {0, 0.1, 0.2, 0.5, 2.0}.

The first is done for you, <em>F</em> (0) = 5.

For the remaining <em>b</em>, you can solve for <em>F(x)</em> exactly by using the fundamental theorem of calculus:

F(x)=F(0)+\displaystyle\int_0^x F'(t)\,\mathrm dt

F(x)=5+\displaystyle\int_0^x(3t^2+7)\,\mathrm dt

F(x)=5+(t^3+7t)\bigg|_0^x

F(x)=5+x^3+7x

Then <em>F</em> (0.1) = 5.701, <em>F</em> (0.2) = 6.408, <em>F</em> (0.5) = 8.625, and <em>F</em> (2.0) = 27.

On the other hand, if you're expected to <em>approximate</em> <em>F</em> at the given <em>b</em>, you can use the linear approximation to <em>F(x)</em> around <em>x</em> = 0, which is

<em>F(x)</em> ≈ <em>L(x)</em> = <em>F</em> (0) + <em>F'</em> (0) (<em>x</em> - 0) = 5 + 7<em>x</em>

Then <em>F</em> (0) = 5, <em>F</em> (0.1) ≈ 5.7, <em>F</em> (0.2) ≈ 6.4, <em>F</em> (0.5) ≈ 8.5, and <em>F</em> (2.0) ≈ 19. Notice how the error gets larger the further away <em>b </em>gets from 0.

A <em>better</em> numerical method would be Euler's method. Given <em>F'(x)</em>, we iteratively use the linear approximation at successive points to get closer approximations to the actual values of <em>F(x)</em>.

Let <em>y(x)</em> = <em>F(x)</em>. Starting with <em>x</em>₀ = 0 and <em>y</em>₀ = <em>F(x</em>₀<em>)</em> = 5, we have

<em>x</em>₁ = <em>x</em>₀ + 0.1 = 0.1

<em>y</em>₁ = <em>y</em>₀ + <em>F'(x</em>₀<em>)</em> (<em>x</em>₁ - <em>x</em>₀) = 5 + 7 (0.1 - 0)   →   <em>F</em> (0.1) ≈ 5.7

<em>x</em>₂ = <em>x</em>₁ + 0.1 = 0.2

<em>y</em>₂ = <em>y</em>₁ + <em>F'(x</em>₁<em>)</em> (<em>x</em>₂ - <em>x</em>₁) = 5.7 + 7.03 (0.2 - 0.1)   →   <em>F</em> (0.2) ≈ 6.403

<em>x</em>₃ = <em>x</em>₂ + 0.3 = 0.5

<em>y</em>₃ = <em>y</em>₂ + <em>F'(x</em>₂<em>)</em> (<em>x</em>₃ - <em>x</em>₂) = 6.403 + 7.12 (0.5 - 0.2)   →   <em>F</em> (0.5) ≈ 8.539

<em>x</em>₄ = <em>x</em>₃ + 1.5 = 2.0

<em>y</em>₄ = <em>y</em>₃ + <em>F'(x</em>₃<em>)</em> (<em>x</em>₄ - <em>x</em>₃) = 8.539 + 7.75 (2.0 - 0.5)   →   <em>F</em> (2.0) ≈ 20.164

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Nikki gathered data about the length of time she spent listening to the radio and the number of commercials she heard. She organ
marishachu [46]

Answer:

C. 63

Step-by-step explanation:

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