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Nastasia [14]
3 years ago
5

Please help me with this

Mathematics
2 answers:
vfiekz [6]3 years ago
7 0
This answer is side kk
Nina [5.8K]3 years ago
3 0
The answer to this question is side KL
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castortr0y [4]

Answer:

f(6) = 35 × 6 - 175 = 35

6 0
2 years ago
Lamaj is rides his bike over a piece of gum and continues riding his bike at a constant rate time = 1.25 seconds the game is at
Hitman42 [59]

Lamaj rides his bike over a piece of gum and continues riding his bike at a constant rate. At time = 1.25 seconds, the gum is at a maximum height above the ground and 1 second later the gum is on the ground again.

a. If the diameter of the wheel is 68 cm, write an equation that models the height of the gum in centimeters above the ground at any time, t, in seconds.

b. What is the height of the gum when Lamaj gets to the end of the block at t = 15.6 seconds?

c. When are the first and second times the gum reaches a height of 12 cm?

Answer:

Step-by-step explanation:

a)

We are being told that:

Lamaj rides his bike over a piece of gum and continues riding his bike at a constant rate. This keeps the wheel of his bike in Simple Harmonic Motion and the Trigonometric equation  that models the height of the gum in centimeters above the ground at any time, t, in seconds.  can be written as:

\mathbf {y = 34cos (\pi (t-1.25))+34}

where;

y =  is the height of the gum at a given time (t) seconds

34 = amplitude of the motion

the amplitude of the motion was obtained by finding the middle between the highest and lowest point on the cosine graph.

\mathbf{ \pi} = the period of the graph

1.25 = maximum vertical height stretched by 1.25 m  to the horizontal

b) From the equation derived above;

if we replace t with 1.56 seconds ; we can determine the height of the gum when Lamaj gets to the end of the block .

So;

\mathbf {y = 34cos (\pi (15.6-1.25))+34}

\mathbf {y = 34cos (\pi (14.35))+34}

\mathbf {y = 34cos (45.08)+34}

\mathbf{y = 58.01}

Thus, the  gum is at 58.01 cm from the ground at  t = 15.6 seconds.

c)

When are the first and second times the gum reaches a height of 12 cm

This indicates the position of y; so y = 12 cm

From the same equation from (a); we have :

\mathbf {y = 34 cos(\pi (t-1.25))+34}

\mathbf{12 = 34 cos ( \pi(t-1.25))+34}

\dfrac {12-34}{34} = cos (\pi(t-1.25))

\dfrac {-22}{34} = cos(\pi(t-1.25))

2.27 = (\pi (t-1.25)

t = 2.72 seconds

Similarly, replacing cosine in the above equation with sine; we have:

\mathbf {y = 34 sin (\pi (t-1.25))+34}

\mathbf{12 = 34 sin ( \pi(t-1.25))+34}

\dfrac {12-34}{34} = sin (\pi(t-1.25))

\dfrac {-22}{34} = sin (\pi(t-1.25))

-0.703 = (\pi(t-1.25))

t = 2.527 seconds

Hence, the gum will reach 12 cm first at 2.527 sec and second time at 2.72 sec.

7 0
3 years ago
Given that the diameter of Circle A is 6 cm , and the radius of Circle B is 18 cm , what can be concluded about the two circles?
ELEN [110]

Answer:

The radius of circle B is 6 times greater than the radius of circle A

The area of circle B is 36 times greater than the area of circle A

Step-by-step explanation:

we have

<em>Circle A</em>

D=6\ cm

The radius of circle A is

r=6/2=3\ cm -----> the radius is half the diameter

<em>Circle B</em>

r=18\ cm

Compare the radius of both circles

3\ cm< 18\ cm

18=6(3)

The radius of circle B is six times greater than the radius of circle A

Remember that , if two figures are similar, then the ratio of its areas is equal to the scale factor squared

All circles are similar

In this problem the scale factor is 6

so

6^{2}=36

therefore

The area of circle B is 36 times greater than the area of circle A

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2 years ago
What is the prediction for the outcome when the independent variable is 3? (Round the answer to the nearest hundredth.)
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C bc i am a professor at Yale and it is simplistic............................................................... <span />
6 0
3 years ago
I need help with this
andre [41]

Answer:

  f(x)=\dfrac{(x+3)(x-1)^2}{(x+4)(x+3)(x-2)^2}

Step-by-step explanation:

Each vertical asymptote corresponds to a zero in the denominator. When the function does not change sign from one side of the asymptote to the other, the factor has even degree. The vertical asymptote at x=-4 corresponds to a denominator factor of (x+4). The one at x=2 corresponds to a denominator factor of (x-2)², because the function does not change sign there.

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Each zero corresponds to a numerator factor that is zero at that point. Again, if the sign doesn't change either side of that zero, then the factor has even multiplicity. The zero at x=1 corresponds to a numerator factor of (x-1)².

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Each "hole" in the function corresponds to numerator and denominator factors that are equal and both zero at that point. The hole at x=-3 corresponds to numerator and denominator factors of (x-3).

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Taken altogether, these factors give us the function ...

  \boxed{f(x)=\dfrac{(x+3)(x-1)^2}{(x+4)(x+3)(x-2)^2}}

8 0
3 years ago
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