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Anettt [7]
2 years ago
8

A nick on the edge of a CD rotates to (−6, 5) during one song when represented graphically.

Mathematics
2 answers:
Allisa [31]2 years ago
6 0
By definition, the sine of an angle is given by:
 sin \alpha =  \frac{C.O}{h}
 Where,
 C.O: opposite leg
 h: hypotenuse
 The hypotenuse is given by:
 h =  \sqrt{(-6)^2+(5)^2}
 h =  \sqrt{(61}
 So, replacing values we have:
 sin \alpha = \frac{5}{\sqrt{61}}
 Rewriting we have:
 sin \alpha = \frac{5\sqrt{61}}{61}
 Answer:
 
the sine value of this function is:
 
A)5 times square root 61 over 61
anygoal [31]2 years ago
6 0
"5 times square root 61 over 61" is the one among the following choices given in the question that <span>is the sine value of this function. The correct option among all the options that are given in the question is the first option or option "A".  hope that this is the answer that has come to your great help.</span>
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Which of the following graphs could be the graph of the function f(x) = x4 + x3 – x2 – x?
vitfil [10]

<u> Solution-</u>

The given function is,

f(x) = x^4+ x^3-x^2-x

     = x^3(x+1)- x(x+1)

     = (x+1)(x^3-x)

     = (x+1)x(x^2-1)

     = (x+1)x(x+1)(x-1)

     = {(x+1)}^2x(x-1)

Therefore, at x=0, -1, 1 , f(x) will be 0 . Hence, 0, -1 ,1 are the x-intercepts.

Plotting the graph on desmos, the graph will be as in the attachment.

7 0
2 years ago
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If 2x+9&lt;32 then x could be
Y_Kistochka [10]

Answer:

x < 11.5

Step-by-step explanation:

2x + 9 < 32

(2x + 9) - 9  < 32 - 9

2x < 23

2x/2 < 23/2

x < 11.5

3 0
2 years ago
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Six identical square pyramids can fill the same volume as a cube with the same base. If the height of the cube is h units, what
Rus_ich [418]

Answer:

I believe "The height of each pyramid is one-half h units"

4 0
2 years ago
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Using the U- Substitution u=sqrt(2x), integral form 2-8 dx/ sqrt(2x) + 1 is equivalent to ...
AleksAgata [21]
We will use u-substitute:u= \sqrt{2x} , \frac{du}{dx}= \frac{1}{ \sqrt{2x} }= \frac{1}{u}Then for substitution:dx=u du. and integral becomes:\int { \frac{u}{u+1} } \, du = \int { \frac{u+1-1}{u+1} } \, du= \int{1} \, du- \int { \frac{1}{u-1} } \, dx=u-ln(u+1)=\sqrt{2x}-ln( \sqrt{2x}+1). Now we will change the values of limits: \sqrt{16}-ln( \sqrt{16}+1)-( \sqrt{4}-ln( \sqrt{4}+1))=4-ln(5)-2+ln(3)=2+ln(0,6)=2-0.51=1.49

8 0
2 years ago
There is a number between 100 and 115, such that if you find the sum of the cubes of the digits, subtract this sum from 99, divi
balandron [24]
114

1^3 + 1^3 + 4^3 = 1 + 1 + 64 = 66
99 - 66 = 33
33/3 = 11
1 + 1 = 2

I am not sure of the exact way to solve this....I just did it by trial and error
7 0
1 year ago
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