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Sati [7]
3 years ago
5

What is 2 16/99 as a decimal

Mathematics
1 answer:
Serggg [28]3 years ago
6 0
2 16/99 as decimal should be:
2 + (16/99) = 2 +0.17 = 2.17
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22. A parabola equation is given as y = 3(x - 2)^2 + 5. What are the coordinates of the vertex?
sergiy2304 [10]

Answer:

(2, 5)

Step-by-step explanation:

The x coordinate is the number inside the parentheses, but with the opposite sign, 2

The y coordinate is the number at the end

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76 cm a 52 cm Find the area of the regular pentagon (the answer has one digit after the decimal point)​
alekssr [168]

Answer:

98.8 cm²

Step-by-step explanation:

Hi there!

A=\frac{1}{2} Pa where P is the perimeter of the pentagon and a is the apothem (the line going from the center of the pentagon one edge)

<u>1) Find the perimeter</u>

P=5l where l is the side length

Plug in the side length 7.6 cm

P=5(7.6)\\P=38

Therefore, the perimeter of the pentagon is 38 cm.

<u>2) Find the area</u>

A=\frac{1}{2} Pa

Plug in the perimeter (38) and the apothem (5.2)

A=\frac{1}{2} (38)(5.2)\\A=\frac{1}{2} (38)(5.2)\\A=\frac{1}{2} (197.6)\\A=98.8

Therefore, the area of the pentagon is 98.8 cm².

I hope this helps!

5 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
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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, ∞).

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3 years ago
An equation is shown below.<br> 3.5p = 28
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Answer:

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