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g100num [7]
3 years ago
11

The graph of y=x^2 was transformed to Y= -(x+3)^2 +5. Describe all transformations applied to the original function and determin

e by how many units if necessary
Mathematics
1 answer:
mina [271]3 years ago
3 0

Hey there!

1. The graph was flipped vertically and now opens downward.

We know this because there is a negative sign in front of the equation:

y= \boxed- (x+3)^2+5

2. The graph was moved 3 units to the left.

We know this because there is 3 added to x inside of the parenthesis. Remember, inside (parenthesis) is opposite and acts on x, outside (parenthesis) is same and acts on y.

y=-(x\boxed{+3})^2+5

3. The graph was moved 5 units up.

We know this because there is 5 added at the end of the equation. It acts on y because it is outside of the parenthesis.

y=-(x+3)^2\boxed{+5}

Hope this helps!

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Answer:

\mathbf{y(t) = t + \dfrac{7}{2}t^2 - \dfrac{2}{3}t^3+ ...}

Step-by-step explanation:

THe interpretation of the given question is as follows:

y'' + ky +  ry³ = A cos ωt

Let k = 4, r = 3, A = 7 and ω = 8

The objective is to find the first three non zero terms in the Taylor polynomial approximation to the solution with initial values y(0) = 0 ; y' (0) = 1

SO;

y'' + ky " ry³ = A cos ωt

where;

k = 4, r = 3, A = 7 and ω = 8

y(0) = 0 ; y' (0) = 1

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y'' = - 4y - 3y³ + 7 cos 8t     ---- (1)

∴

y'' (0) = -4y(0) - 3y³(0) + 7 cos (0)

y'' (0) = - 4 × 0 - 3 × 0 + 7

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Differentiating equation (1) with respect to  t ; we have:

y''' = - 4y' - 9y² × y¹ - 56 sin 8t

y''' (0) = -4y'(0) - 9y²(0)× y¹ (0) - 56 sin (0)

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Thus; we have :

y(0) = 0  ; y'(0) = 1  ; y'' (0) = 7 ; y'''(0) = -4

Therefore; the Taylor polynomial approximation to the first three nonzero terms is :

y(t) = y(0) + y'(0) t + y''(0) \dfrac{t^2}{2!} + y'''(0) \dfrac{t^3}{3!}+...

y(t) = 0 +  t + 7 \dfrac{t^2}{2!} + \dfrac{-4}{3!} {t^3}+ ...

\mathbf{y(t) = t + \dfrac{7}{2}t^2 - \dfrac{2}{3}t^3+ ...}

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4 years ago
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CD = BA = √2*2+1*1 = √5

Sum is 2√10 + 2√5 ≈ 10.8. So answer A.
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