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4 years ago

How can knowing the gcf and lcm help you when you add subtract and multiply fractions?

2 answers:

8 0

If the denominators are different numbers, it is necessary to find a common denominator. Find the Lowest Common Multiple (LCM) of the denominators (also known as the Lowest Common Denominator, or LCD) and convert the fractions to equivalent fractions with the same denominator. Now they have the same denominator, proceed as above.

Example: 1/4 + 1/6 = 3/12 + 2/12 = 5/12 (here LCM=12)

You don't have to do anything to fractions to multiply them, but you may need to reduce one after multiplying. To do that, find the GCF of the numerator and the denominator and divide both of them by it. If the GCF is 1, the fraction is in its simplest form.

Example: 1/4 · 2/5 = 1·2/4·5=2/20=1/10 (here GCF=2)

Maksim231197 [3]4 years ago

4 0

In order to add and subtract fractions, we must be able to find a common denominator.
The LCD, or least common denominator, is the LCM of the denominators. Once we've added, subtracted or multiplied fractions, we often have to simplify the fraction.
In order to do this, we divide the numerator and denominator by their GCF. This means we must be able to find the LCM and GCF in order to work with fractions.

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2 years ago

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How many miles are in 8,216 feet

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Step-by-step explanation:

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4 years ago

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Someone please help me with this lol… have no idea what I’m doing

Sholpan [36]

Given:

$\cos \theta =\dfrac{3}{5}$

$\sin \theta$

To find:

The quadrant of the terminal side of $\theta$ and find the value of $\sin\theta$ .

Solution:

We know that,

In Quadrant I, all trigonometric ratios are positive.

In Quadrant II: Only sin and cosec are positive.

In Quadrant III: Only tan and cot are positive.

In Quadrant IV: Only cos and sec are positive.

It is given that,

$\cos \theta =\dfrac{3}{5}$

$\sin \theta$

Here cos is positive and sine is negative. So, $\theta$ must be lies in Quadrant IV.

We know that,

$\sin^2\theta +\cos^2\theta =1$

$\sin^2\theta=1-\cos^2\theta$

$\sin \theta=\pm \sqrt{1-\cos^2\theta}$

It is only negative because $\theta$ lies in Quadrant IV. So,

$\sin \theta=-\sqrt{1-\cos^2\theta}$

After substituting $\cos \theta =\dfrac{3}{5}$ , we get

$\sin \theta=-\sqrt{1-(\dfrac{3}{5})^2}$

$\sin \theta=-\sqrt{1-\dfrac{9}{25}}$

$\sin \theta=-\sqrt{\dfrac{25-9}{25}}$

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3 years ago

Find the length of diagonal XU in the hexagon below. Round your solution to 2 decimal points

SpyIntel [72]

Answer:

The length of diagonal XU is 6.40 units

Step-by-step explanation:

* Lets explain how to find the distance between two points

- The rule of the distance between two points $(x_{1},y_{1})$

and $(x_{2},y_{2})$ is:

$d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

* Lets solve the problem

-From the attached figure:

- The coordinates of the vertex X are (2 , 2)

- The coordinates of the vertex U are (-2 , 7)

∴ The point $(x_{1},y_{1})$ = (2 , 2)

∴ The point $(x_{2},y_{2})$ = (-2 , 7)

∴ $x_{1}=2,x_{2}=-2$

∴ $y_{1}=2,y_{2}=7$

∵ $d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

∴ $d=\sqrt{(-2-2)^{2}+(7-2)^{2}}=\sqrt{(-4)^{2}+(5)^{2}}=\sqrt{16+25}=\sqrt{41}$

∵ $\sqrt{41}=6.403124$

∴ d = 6.40

* The length of diagonal XU is 6.40 units

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3 years ago

The equation h = -16t 2+ 53x + 8 gives the height h, in feet, of a football as a function of time t, in seconds, after it is kic

Elanso [62]

Answer: The maximum height the football reaches = 51.89 feet.

Step-by-step explanation:

Given: The equation $h = -16t^2+ 53t + 8$ gives the height h, in feet, of a football as a function of time t, in seconds, after it is kicked.

Differentiate the given equation, we get

$h'=-(2)16t+53=-32t+53$ (i)

Put $h'=0$

$\Rightarrow\ -32t+53=0\\\\\Rightarrow\ t=\dfrac{53}{32}\approx1.65625$

Differentiate (i) w.r.t. t, we get

$h''=-32$ < 0 i.e. h is maximum at t= 1.65625

The value of h at t= 1.65625 will be $-16(1.65625)^2+53(1.65625)+8=51.890625\approx51.89\text{ feet}$

Hence, the maximum height the football reaches = 51.89 feet.

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3 years ago