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mojhsa [17]
3 years ago
13

Sketch the graph of each function

Mathematics
2 answers:
IrinaVladis [17]3 years ago
8 0
<h2>Answer:</h2>

The graph is a sin wave given in the image

jenyasd209 [6]3 years ago
4 0

Answer:

See diagram

Step-by-step explanation:

First, consider the graph of the parent function y=\cos x. This function has the period 2\pi.

Rewrite the function y=2(1+\cos (x-\frac{\pi}{2})) as y=2+2\cos (x-\frac{\pi}{2}). The graph of this function has the same period of 2\pi, is twice streched and translated \frac{\pi}{2} units to the right and 2 units up graph of the parent function (see diagram).

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At the school carnival the sixth graders are making directional arrows each arrow is to be painted red how much area needs to be
Anna35 [415]

We can divide arrow in two sections:

A 1 ( rectangle ) = 4 x 5 = 20 sq in.

A 2 ( triangle ) = ( 6 x 3 ) / 2 = 9 sq in.

A = A 1 + A 2 = 20 + 9 = 29 sq in

Answer: B )29 sq in.

7 0
4 years ago
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Describe the steps you would take to solve for the variable x in the following
adoni [48]

Answer:

x =  19

Step-by-step explanation:

1. add 31 on both sides (you wanna get the "x" by itself)

ex: 2x - 31 = 7

          +31   +31

2. the 31s cancel each other out so now you add 31 + 7

2x       =  7

           +31

___________

             38

3. divide 2 on both sides so the 2s cancel each other out and only the x is left

ex: 2x / 2 = x         38 ( the 38 stands alone)

4. now divide 38/2 and you have your answer

X= 19

5 0
3 years ago
The cost of producing x soccer balls in thousands of dollars is represented by h(x) = 5x + 6. The revenue is represented by k(x)
Korolek [52]
(9x-2) - (5x+6) =

4x-8 , this is your profit function
8 0
3 years ago
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You throw a ball up and its height h can be tracked using the equation h=2x^2-12x+20.
postnew [5]

<em><u>This problem seems to be wrong because no minimum point was found and no point of landing exists</u></em>

Answer:

1) There is no maximum height

2) The ball will never land

Step-by-step explanation:

<u>Derivatives</u>

Sometimes we need to find the maximum or minimum value of a function in a given interval. The derivative is a very handy tool for this task. We only have to compute the first derivative f' and have it equal to 0. That will give us the critical points.

Then, compute the second derivative f'' and evaluate the critical points in there. The criteria establish that

If f''(a) is positive, then x=a is a minimum

If f''(a) is negative, then x=a is a maximum

1)

The function provided in the question is

h(x)=2x^2-12x+20

Let's find the first derivative

h'(x)=4x-12

solving h'=0:

4x-12=0

x=3

Computing h''

h''(x)=4

It means that no matter the value of x, the second derivative is always positive, so x=3 is a minimum. The function doesn't have a local maximum or the ball will never reach a maximum height

2)

To find when will the ball land, we set h=0

2x^2-12x+20=0

Simplifying by 2

x^2-6x+10=0

Completing squares

x^2-6x+9+10-9=0

Factoring and rearranging

(x-3)^2=-1

There is no real value of x to solve the above equation, so the ball will never land.

This problem seems to be wrong because no minimum point was found and no point of landing exists

3 0
3 years ago
What two numbers plus 35 equal 180
Andrews [41]
X^2+35=180?
Subtract 35 on each side

x^2=145
Take the square root

x= -sqrt(145)
+sqrt 145...

I hope this helps!!
4 0
3 years ago
Read 2 more answers
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