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Olegator [25]
3 years ago
12

Brianna has 4 pink bracelets. One third of all her bracelets are pink. How many bracelets does brianna have?

Mathematics
1 answer:
aivan3 [116]3 years ago
8 0
Hey there!

Let's first evaluate this problem, and the keyword we need to use. The problem says that out of the total number of bracelets Brianna has, x, 1/3 of them are pink. The reason the word of is so important throughout math is that it signifies multiplication. For example, 1/2 of 10 is 5 because 1/2 times 10 is 5.

Now, to apply that to this situation, we need to create an equation, a mathematical statement that indicated the equilibrium or equality of two values. Since we have 1/3 of all her bracelets, x is four, we can set up our equation. Here's another good keyword- is. Is signifies an equation, because it means "equal to" mathematically. For example, 5 is equal to 20/4.

We have:

1/3(x) = 4 

(One-third of x, the total amount of bracelets, is four.)

We don't want that fraction there. In order to get rid of it and keep the expression equal, we multiply both sides by 3. We have,

x = 12

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

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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}

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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}

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\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.

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