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3 years ago

5

Find the solution to Y=-x^2+3 for X=-3,0, and 3

2 answers:

Alecsey [184]3 years ago

8 0

For this case we must evaluate the following quadratic equation, $y = -x ^ 2 + 3,$ for $x = -3$ , $x = 0$ and $x = 3$

For $x = -3$ :

$y = - (- 3) ^ 2 + 3\\y = -9 + 3 =\\y = -6$

For $x = 0$ :

$y = - (0) ^ 2 + 3\\y = 0 + 3\\y = 3$

For $x = 3$ :

$y = - (3) ^ 2 + 3\\y = -9 + 3\\y = -6$

Answer:

$(-3, -6)\\(0,3)\\(3, -6)$

BabaBlast [244]3 years ago

7 0

Answer:

3-,3,0,3

Step-by-step explanation:

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3 years ago

Read 2 more answers

3=b+3<br><br> A. 0<br> B. 1<br> C. 9<br> D. 6

jok3333 [9.3K]

Your Answer Is...

A) 0.

Why?

This is correct because if 3 = b + 3, then ? + 3 = 3. 0 + 3 = 3, making the answer A) 0.

≈≈≈≈ Glad I Could Help, And Good Luck! ≈≈≈≈

My Name: AnonymousGiantsFan

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3 years ago

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riadik2000 [5.3K]

Let

$S_n = \displaystyle \sum_{k=0}^n r^k = 1 + r + r^2 + \cdots + r^n$

where we assume |r| < 1. Multiplying on both sides by r gives

$r S_n = \displaystyle \sum_{k=0}^n r^{k+1} = r + r^2 + r^3 + \cdots + r^{n+1}$

and subtracting this from $S_n$ gives

$(1 - r) S_n = 1 - r^{n+1} \implies S_n = \dfrac{1 - r^{n+1}}{1 - r}$

As n → ∞, the exponential term will converge to 0, and the partial sums $S_n$ will converge to

$\displaystyle \lim_{n\to\infty} S_n = \dfrac1{1-r}$

Now, we're given

$a + ar + ar^2 + \cdots = 15 \implies 1 + r + r^2 + \cdots = \dfrac{15}a$

$a^2 + a^2r^2 + a^2r^4 + \cdots = 150 \implies 1 + r^2 + r^4 + \cdots = \dfrac{150}{a^2}$

We must have |r| < 1 since both sums converge, so

$\dfrac{15}a = \dfrac1{1-r}$

$\dfrac{150}{a^2} = \dfrac1{1-r^2}$

Solving for r by substitution, we have

$\dfrac{15}a = \dfrac1{1-r} \implies a = 15(1-r)$

$\dfrac{150}{225(1-r)^2} = \dfrac1{1-r^2}$

Recalling the difference of squares identity, we have

$\dfrac2{3(1-r)^2} = \dfrac1{(1-r)(1+r)}$

We've already confirmed r ≠ 1, so we can simplify this to

$\dfrac2{3(1-r)} = \dfrac1{1+r} \implies \dfrac{1-r}{1+r} = \dfrac23 \implies r = \dfrac15$

It follows that

$\dfrac a{1-r} = \dfrac a{1-\frac15} = 15 \implies a = 12$

and so the sum we want is

$ar^3 + ar^4 + ar^6 + \cdots = 15 - a - ar - ar^2 = \boxed{\dfrac3{25}}$

which doesn't appear to be either of the given answer choices. Are you sure there isn't a typo somewhere?

7 0

2 years ago