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4 years ago

Please Help Me, Take Your Time.

Mathematics

1 answer:

Doss [256]4 years ago

7 0

<h3>Given</h3>

72 blue, 42 red beads

beads are used to make identical necklaces

(a) the greatest number of necklaces that can be made

(b) the number of each color bead in each necklace

<h3>Solution</h3>

You can write and factor the equation

... necklaces = 72 blue + 42 red

... necklaces = 6(12 blue + 7 red)

where 6 is the greatest common factor (GCF) of 72 and 42.

(a) You can make up to 6 identical necklaces. 6 is the largest common factor of 72 and 42. If you were to try to make more, they could not be identical.

(b) Each necklace can consist of 12 blue and 7 red beads. These are the numbers obtained when the total bead count is divided into 6 equal groups.

_____

There are several ways you can find the GCF of two numbers. For small numbers, it is generally feasible to use your knowledge of multiplication tables and factors to choose the largest common factor of two numbers. You can also use Euclid's algorithm, which is to repeatedly compute

... (largest number) modulo (smallest number)

until the result is zero. The final "smallest number" is the GCF.

Here, that looks like

... 72 mod 42 = 30

... 42 mod 30 = 12

... 30 mod 12 = 6

... 12 mod 6 = 0 . . . . . . . so 6 is the GCF

___

Of course, you know that

... 72 = 2³×3²

... 42 = 2×3×7

so, the largest set of common factors is 2×3 = 6.

___

Your graphing calculator may have a function for computing the greatest common divisor (GCD), too. The TI-84 does, for example.

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Which quadratic equation defines the function that has zeros at − 1/12 and 1/4 ?

Cerrena [4.2K]

$\bf \begin{cases}
x=-\frac{1}{2}\implies &x+\frac{1}{2}=0\\\\
x=\frac{1}{4}\implies &x-\frac{1}{4}=0
\end{cases}
\\\\\\
\left( x+\frac{1}{2} \right)\left( x-\frac{1}{4} \right)=\stackrel{y}{0}\implies \stackrel{FOIL}{x^2+\frac{1}{4}x-\frac{1}{8}}=y$

7 0

4 years ago

A ball is launched upward at 14 m/s from a platform 30 m high.Find the maximum height the ball will reach and how long it will t

BartSMP [9]

Answer:

The ball will reach a maximum height of 39.993 meters after 1.428 seconds.

Step-by-step explanation:

Let suppose that no non-conservative forces acts on the ball during its motion, then we can determine the maximum height reached by the Principle of Energy Conservation, which states that:

$K_{1}+U_{g,1} = K_{2}+U_{g,2}$ (1)

Where:

$K_{1}$ , $K_{2}$ - Initial and final translational kinetic energies, measured in joules.

$U_{g,1}$ , $U_{g,2}$ - Initial and final gravitational potential energies, measured in joules.

By definition of translational kinetic energy and gravitational potential energy we expand and simplify the expression above:

$\frac{1}{2}\cdot m\cdot v_{2}^{2}+m\cdot g\cdot y_{2}= \frac{1}{2}\cdot m\cdot v_{1}^{2}+m\cdot g\cdot y_{1}$ (2)

Where:

$m$ - Mass of the ball, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

$v_{1}$ , $v_{2}$ - Initial and final speed of the ball, measured in meters per second.

$y_{1}$ , $y_{2}$ - Initial and final heights of the ball, measured in meters.

The final height of the ball is determined by the following formula:

$v_{2}^{2}+2\cdot g\cdot y_{2} = v_{1}^{2}+2\cdot g\cdot y_{1}$

$v_{1}^{2}-v_{2}^{2}+2\cdot g \cdot y_{1}=2\cdot g\cdot y_{2}$

$y_{2} = y_{1}+\frac{v_{1}^{2}-v_{2}^{2}}{2\cdot g}$ (3)

If we know that $y_{1} = 30\,m$ , $v_{1} = 14\,\frac{m}{s}$ , $v_{2} = 0\,\frac{m}{s}$ and $g = 9.807\,\frac{m}{s^{2}}$ , the maximum height that the ball will reach is:

$y_{2} = 30\,m + \frac{\left(14\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$y_{2} = 39.993\,m$

The ball will reach a maximum height of 39.993 meters.

Given the absence of non-conservative forces, the ball exhibits a free fall. The time needed for the ball to reach its maximum height is computed from the following kinematic formula:

$t = \frac{v_{2}-v_{1}}{-g}$ (4)

If we know that $v_{1} = 14\,\frac{m}{s}$ , $v_{2} = 0\,\frac{m}{s}$ and $g = 9.807\,\frac{m}{s^{2}}$ , then:

$t = \frac{0\,\frac{m}{s}-14\,\frac{m}{s} }{-9.807\,\frac{m}{s^{2}} }$

$t = 1.428\,s$

The ball will take 1.428 seconds to reach its maximum height.

6 0

3 years ago

What is the value of x, when 10(x + 2) = 5(x + 8)?

olga_2 [115]

Solve for x:
10 (x + 2) = 5 (x + 8)

Expand out terms of the left hand side:
10 x + 20 = 5 (x + 8)

Expand out terms of the right hand side:
10 x + 20 = 5 x + 40

Subtract 5 x from both sides:
(10 x - 5 x) + 20 = (5 x - 5 x) + 40

10 x - 5 x = 5 x:
5 x + 20 = (5 x - 5 x) + 40

5 x - 5 x = 0:
5 x + 20 = 40

Subtract 20 from both sides:
5 x + (20 - 20) = 40 - 20

20 - 20 = 0:
5 x = 40 - 20

40 - 20 = 20:
5 x = 20

Divide both sides of 5 x = 20 by 5:
(5 x)/5 = 20/5

5/5 = 1:
x = 20/5

The gcd of 20 and 5 is 5, so 20/5 = (5×4)/(5×1) = 5/5×4 = 4:

Answer: x = 4

6 0

3 years ago

Read 2 more answers

Find the x-coordinates of any relative extrema and inflection point(s) for the function f(x)= 6x^1/3 + 3x^4/3. You must justify

irina [24]

Answer:

x-coordinates of relative extrema = $\frac{-1}{2}$

x-coordinates of the inflexion points are 0, 1

Step-by-step explanation:

$f(x)=6x^{\frac{1}{3}}+3x^{\frac{4}{3}}$

Differentiate with respect to x

$f'(x)=6\left ( \frac{1}{3} \right )x^{\frac{-2}{3}}+3\left ( \frac{4}{3} \right )x^{\frac{1}{3}}=\frac{2}{x^{\frac{2}{3}}}+4x^{\frac{1}{3}}$

$f'(x)=0\Rightarrow \frac{2}{x^{\frac{2}{3}}}+4x^{\frac{1}{3}}=0\Rightarrow x=\frac{-1}{2}$

Differentiate f'(x) with respect to x

$f''(x)=2\left ( \frac{-2}{3} \right )x^{\frac{-5}{3}}+\frac{4}{3}x^{\frac{-2}{3}}=\frac{-4x^{\frac{2}{3}}+4x^{\frac{5}{3}}}{3x^{\frac{2}{3}}x^{\frac{5}{3}}}\\f''(x)=0\Rightarrow \frac{-4x^{\frac{2}{3}}+4x^{\frac{5}{3}}}{3x^{\frac{2}{3}}x^{\frac{5}{3}}}=0\Rightarrow x=1$