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dangina [55]
3 years ago
11

Find the explicit formula for the given sequence.

Mathematics
1 answer:
Art [367]3 years ago
4 0

Answer:

Option a.) an = -4(-3)^(n-1)

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A parabola can be drawn given a focus of
Volgvan

Answer:

y=-\frac{1}{4}(x-2)^{2}+9

Step-by-step explanation:

Any point on a given parabola is equidistant from focus and directrix.

Given:

Focus of the parabola is at (2,8).

Directrix of the parabola is y=10.

Let (x,y) be any point on the parabola. Then, from the definition of a parabola,

Distance of (x,y) from focus = Distance of (x,y) from directrix.

Therefore,

\sqrt{(x-2)^{2}+(y-8)^{2}}=|y-10|

Squaring both sides, we get

(x-2)^{2}+(y-8)^{2}=(y-10)^{2}\\(x-2)^{2}=(y-10)^{2}-(y-8)^{2}\\(x-2)^{2}=(y-10+y-8)(y-10-(y-8))...............[\because a^{2}-b^{2}=(a+b)(a-b)]\\(x-2)^{2}=(2y-18)(y-10-y+8)\\(x-2)^{2}=2(y-9)(-2)\\(x-2)^{2}=-4(y-9)\\y-9=-\frac{1}{4}(x-2)^{2}\\y=-\frac{1}{4}(x-2)^{2}+9

Hence, the equation of the parabola is y=-\frac{1}{4}(x-2)^{2}+9.

4 0
3 years ago
Plzzzzz teach this.....<br>I wiil Mark as brainlist answer if correct​
GuDViN [60]
Omg don’t open that file it’s takes all your information
7 0
3 years ago
What value of b makes the trinomial below a perfect square x^2-bx+100
Aloiza [94]
1. Perfect square trinomials, are 2nd degree polynomials, of the form          a x^{2} +bx+c so that a \neq 0, b \neq 0, c \neq 0, which can be written as perfect squares.

2. For example (x+1) ^{2} = x^{2} +2x+1&#10; &#10;(3x-1)^{2}= (3x)^{2}-2(3x)+(-1) ^{2}= 9x^{2}-6x+1    &#10;

3. Thus x^{2} +2x+1, 9x^{2}-6x+1 are perfect square trinomials.

4. x^{2} -bx+100= x^{2} -bx+ 10^{2}= (x+10)^{2}   or  (x-10)^{2}

5. In the first case -b=20, so b=-20. In the second case, -b=-20, so b=20.

6. b∈{-20, 20}
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3 years ago
Find the midpoint (8, -6) (6,5)​
andrey2020 [161]
I think it’s (7, -0.5)
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2 years ago
Three sister shared $3.60 equally. how much did each sister get?
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Each sister got $1.20
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