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anygoal [31]
3 years ago
8

The axis of symmetry for the graph of the function is f(x) = 1/4 x2 +

Mathematics
1 answer:
CaHeK987 [17]3 years ago
3 0

Answer:

Step-by-step explanation:

f(x) = 1/4 x²+bx+10

the derivate is : f'(x) = 1/2 x +b

you have : f'(6)=0

1/2 (6)+b=0

3+b =0

b = -3

so f(x) = 1/4 x²-3x+10.......f(6) =1/4(6)² -3(6)+10 =9-18+10 =1

f(x) = 1/4(x-6)²+1... the vertex form

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I don't know how to answer this question.
laiz [17]

Answer:

- 39/8y

Step-by-step explanation:

First you need to find a common denominator:

There is already a y in both denominators, so you don't have to worry about that. Since only one denominator has an 8, the other needs to have an 8, so you need to multiply the first fraction by 8 → - 4/y * 8 = - 32/8y

From here you can subtract the numerators:

- 32/8y +(- 7/8y) = - 39/8y

Because you are adding a negative to a negative, it makes the number mroe negative


I hope this helps!

3 0
3 years ago
Evaluate the expression without using a calculator: log10^3x-2y
Kruka [31]

Answer:

3x-2y

Step-by-step explanation:

log10^(3x-2y)

We know the base is base 10 since it is not written

log10 10^(3x-2y)

The log10 10 cancels

3x-2y

3 0
3 years ago
Read 2 more answers
1 red marble, 3 green, 4 blue. What is the probability of picking a yellow marble?​
Alexandra [31]

Answer:

zero probability, which means impossible

Step-by-step explanation:

there are no yellow marbles to select from

7 0
3 years ago
I=$310 P=$1,000 t=5 years
PtichkaEL [24]
I = P x r x t
r = I/P x t

r = 310/1000 x 5
r = 0.62
r = 0.62 x 100
r = 62%
4 0
3 years ago
These roots of the polynomial equation x^4-4x^3-2x^2+12x+9=0 are 3,-1,-1. Explain why the fourth root must be a real number. Fin
Alex787 [66]

Roots with imaginary parts always occur in conjugate pairs. Three of the four roots are known and they are all real, which means the fourth root must also be real.

Because we know 3 and -1 (multiplicity 2) are both roots, the last root r is such that we can write

x^4-4x^3-2x^2+12x+9=(x-3)(x+1)^2(x-r)

There are a few ways we can go about finding r, but the easiest way would be to consider only the constant term in the expansion of the right hand side. We don't have to actually compute the expansion, because we know by properties of multiplication that the constant term will be (-3)(1)(1)(-r)=3r.

Meanwhile, on the left hand side, we see the constant term is supposed to be 9, which means we have

3r=9\implies r=3

so the missing root is 3.

Other things we could have tried that spring to mind:

- three rounds of division, dividing the quartic polynomial by (x-3), then by (x+1) twice, and noting that the remainder upon each division should be 0

- rational root theorem

4 0
3 years ago
Read 2 more answers
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