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s344n2d4d5 [400]
3 years ago
14

Kala has scored 21,22,20,20 and 17 point in her five basketball game so far. How many points does she need to score in her next

game so that her average (mean) is 21 points per game?
Mathematics
1 answer:
Oliga [24]3 years ago
8 0
Step 1: Add up all the scores.
Step 2: count the number of  scores.
step 3: divide the sum of the scores by the number of scores themselves.
step 4: The answer is 26 points.
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Determine the vertex of the function f(x) = 3(x − 2)2 + 1.
anastassius [24]

The vertex point of the function f(x) = 3(x - 2)² + 1 is (2 , 1) ⇒ answer A

Step-by-step explanation:

Any quadratic function represented graphically by a parabola

1. If the coefficient of x² is positive, then the parabola open upward

  and its vertex is a minimum point

2. If the coefficient of x² is negative, then the parabola open

   downward and its vertex is a maximum point

3. The standard form of the quadratic function is: f(x) = ax² + bx + c

    where a, b , c are constants

4. The vertex form of the quadratic function is: f(x) = a(x - h)² + k,

     where h , k are the coordinates of its vertex point

∵ The function f(x) = 3(x - 2)² + 1

∵ The f(x) = a(x - h)² + k in the vertex form

∴ a = 3 , h = 2 , k = 1

∵ h , k are the coordinates of the vertex point

∴ The coordinates of the vertex point are (2 , 1)

The vertex point of the function f(x) = 3(x - 2)² + 1 is (2 , 1)

Learn more:

You can learn more about quadratic function in brainly.com/question/1357167

#LearnwithBrainly

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Step-by-step explanation:

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Step-by-step explanation:

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This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive an
Lapatulllka [165]

Answer:

Step-by-step explanation:

Given that there is a function of x,

f(x) = 2sin x + 2cos x,0\leq x\leq 2\pi

Let us find first and second derivative for f(x)

f'(x) = 2cosx -2sinx\\f"(x) = -2sinx-2cosx

When f'(x) =0 we have tanx = 1 and hence

a) f'(x) >0 for I and III quadrant

Hence increasing in (0, \pi/2) U(\pi,3\pi/2)\\

and decreasing in (\pi/2, \pi)U(3\pi/2,2\pi)

x=\frac{\pi}{4}, \frac{3\pi}{4}

f"(\pi/4)

Hence f has a maxima at x = pi/4 and minima at x = 3pi/4

b) Maximum value = 2sin \pi/4+2cos \pi/4 =2\sqrt{2}

Minimum value = 2sin 3\pi/4+2cos 3\pi/4 =-2\sqrt{2}

c)

f"(x) =0 gives tanx =-1

x= 3\pi/4, 7\pi/4

are points of inflection.

concave up in (3pi/4,7pi/4)

and concave down in (0,3pi/4)U(7pi/4,2pi)

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