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WITCHER [35]
3 years ago
11

Jude estimates that his index finger is 3 inches long. He calculates the circumference of one of his bongo drums to be 27 inches

. Suppose Jude used his finger as a referent to measure the circumference of the drum. How many finger lengths along the circumference of the drum would he have counted?
Mathematics
1 answer:
anzhelika [568]3 years ago
7 0

Answer: 9 finger lengths

Step-by-step explanation: Since the circumference is the middle line passing through (usually a circle....) 27/3 equals 9, which basically means that 9 of Jude's index fingers can pass through the circumference of one of his bongo drums. Hope this helps, 15402256.

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Andrews [41]
Eighty-one thousand two hundred seventy-one.
8 0
3 years ago
What are all the roots of the equation x^2+6x-9
frez [133]

Answer:

x = - 3 ± 3\sqrt{2}

Step-by-step explanation:

given the equation x² + 6x - 9 = 0

We can solve for x using the quadratic formula

x = ( - b ± \sqrt{b^2-4ac } ) / 2a

with a = 1, b = 6 and c = - 9

x = ( - 6 ± \sqrt{6^2-(4(1)(-9)} ) / 2

  = ( - 6 ± \sqrt{36+36} ) / 2

  = ( - 6 ± \sqrt{72} ) / 2

  = ( - 6 ± 6\sqrt{2} ) / 2

x = - 3 ± 3\sqrt{2}


5 0
3 years ago
Express the ratio 14 inches to 3 feet as a fraction in simplest form
artcher [175]
The answer is gonna be 14:36
7 0
3 years ago
The position function of a particle in rectilinear motion is given by s(t) = 2t3 – 21t2 + 60t + 3 for t ≥ 0 with t measured in s
Norma-Jean [14]

The positions when the particle reverses direction are:

s(t_1)=55ft\\\\s(t_2)=28ft

The acceleraton of the paticle when reverses direction is:

a(t_1)=-18\frac{ft}{s^{2}}\\ \\a(t_2)=a(5s)=18\frac{ft}{s^{2}}

Why?

To solve the problem, we need to remember that if we derivate the position function, we will get the velocity function, and if we derivate the velocity function, we will get the acceleration function. So, we will need to derivate two times.

Also, when the particle reverses its direction, the velocity is equal to 0.

We are given the following function:

s(t)=2t^{3}-21t^{2}+60t+3

So,

- Derivating to get the velocity function, we have:

v(t)=\frac{ds}{dt}=(2t^{3}-21t^{2}+60t+3)\\\\v(t)=3*2t^{2}-2*21t+60*1+0\\\\v(t)=6t^{2}-42t+60

Now, making the function equal to 0, to find the times when the particle reversed its direction, we have:

v(t)=6t^{2}-42t+60\\\\0=6t^{2}-42t+60\\\\0=t^{2}-7t+10\\(t-5)*(t-2)=0\\\\t_{1}=5s\\t_{2}=2s

We know that the particle reversed its direction two times.

- Derivating the velocity function to find the acceleration function, we have:

a(t)=\frac{dv}{dt}=6t^{2}-42t+60\\\\a(t)=12t-42

Now, substituting the times to calculate the accelerations, we have:

a(t_1)=a(2s)=12*2-42=-18\frac{ft}{s^{2}}\\ \\a(t_2)=a(5s)=12*5-42=18\frac{ft}{s^{2}}

Now, substitutitng the times to calculate the positions, we have:

s(t_1)=2*(2)^{3}-21*(2)^{2}+60*2+3=16-84+120+3=55ft\\\\s(t_2)=2*(5)^{3}-21*(5)^{2}+60*5+3=250-525+300+3=28ft

Have a nice day!

3 0
3 years ago
If f(x) is a linear function, f(−2)=4, and f(1)=−1, find an equation for f(x)
juin [17]

Answer:

y = -\frac{5}{3}x+\frac{2}{3}

Step-by-step explanation:

Given

f(-2) = 4

f(1) = -1

Required

The equation of the function

The given parameters means that:

(x_1,y_1) = (-2,4)

(x_2,y_2) = (1,-1)

Calculate the slope (m)

m = \frac{y_2 -y_1}{x_2 -x_1}

m = \frac{-1-4}{1--2}

m = \frac{-5}{3}

The equation is then calculated using:

y = m(x - x_1) + y_1

This gives:

y =\frac{-5}{3}(x--2)+4

y =\frac{-5}{3}(x+2)+4

Open bracket

y = -\frac{5}{3}x-\frac{10}{3}+4

Take LCM

y = -\frac{5}{3}x+\frac{-10+12}{3}

y = -\frac{5}{3}x+\frac{2}{3}

6 0
3 years ago
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