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navik [9.2K]
3 years ago
15

OK GUYS PLEASE HELP: USE THE INFORMATION PROVIDED TO WRITE THE VERTEX FORM EQUATION OF EACH PARABOLA Y=-12X^2+192X-769 EXPLAIN P

LEASE THIS IS A GRADE
2)SOLVE EACH EQUATION BY COMPLETING THE SQUARE : N^2+16N-74=-3
3)SOLVE EACH EQUATION BY FACTORING: 6N^2+48N+90=0
Mathematics
1 answer:
levacccp [35]3 years ago
7 0
The vertex for the first one is: (8,-1)
the answer for the second one is: n = -8 + with a - at the bottom square root of 151
the answer for the third question is: n = (-3,-5) 

hope i helped please thank, rate, and give me best anyswer if you're on a desktop thank you :)
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Taya2010 [7]
The answer is 46°.

The measure of an inscribed angle is equal to (1/2) the measure of the intercepted arc. 

That means that, since ADC is 23 degrees, doubling that gives you 46 degrees, the measure of the intercepted arc.

Central angles are equal to the measure of the intercepted arc, which in this case is the same arc we just calculated.

Therefore it's 46.
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What is the answer to f(x)=x^2-10x+4 f(-2)
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Answer:

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3 0
3 years ago
F(x) = x2 − x − ln(x) (a) find the interval on which f is increasing. (enter your answer using interval notation.) find the inte
Mekhanik [1.2K]

Answer:

(a) Decreasing on (0, 1) and increasing on (1, ∞)

(b) Local minimum at (1, 0)

(c) No inflection point; concave up on (0, ∞)

Step-by-step explanation:

ƒ(x) = x² - x – lnx

(a) Intervals in which ƒ(x) is increasing and decreasing.

Step 1. Find the zeros of the first derivative of the function

ƒ'(x) = 2x – 1 - 1/x = 0

           2x² - x  -1 = 0

     ( x - 1) (2x + 1) = 0

         x = 1 or x = -½

We reject the negative root, because the argument of lnx cannot be negative.

There is one zero at (1, 0). This is your critical point.

Step 2. Apply the first derivative test.

Test all intervals to the left and to the right of the critical value to determine if the derivative is positive or negative.

(1) x = ½

ƒ'(½) = 2(½) - 1 - 1/(½) = 1 - 1 - 2 = -1

ƒ'(x) < 0 so the function is decreasing on (0, 1).

(2) x = 2

ƒ'(0) = 2(2) -1 – 1/2 = 4 - 1 – ½  = ⁵/₂

ƒ'(x) > 0 so the function is increasing on (1, ∞).

(b) Local extremum

ƒ(x) is decreasing when x < 1 and increasing when x >1.

Thus, (1, 0) is a local minimum, and ƒ(x) = 0 when x = 1.

(c) Inflection point

(1) Set the second derivative equal to zero

ƒ''(x) = 2 + 2/x² = 0

             x² + 2 = 0

                   x² = -2

There is no inflection point.

(2). Concavity

Apply the second derivative test on either side of the extremum.

\begin{array}{lccc}\text{Test} & x < 1 & x = 1 & x > 1\\\text{Sign of f''} & + & 0 & +\\\text{Concavity} & \text{up} & &\text{up}\\\end{array}

The function is concave up on (0, ∞).

6 0
3 years ago
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