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3 years ago

© Find the LCM of 220 and 88.

Mathematics

1 answer:

Morgarella [4.7K]3 years ago

7 0

Answer:

440

Step-by-step explanation:

multiple of 220

220, 440

multiple of 88

88, 176, 264, 352, 440

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Round 51.78 to the greatest number.

soldier1979 [14.2K]

Answer:

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3 years ago

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I need help with this math problem!

kherson [118]

Answer: (50, 70)

The answer is (50, 70).

Step-by-step explanation:

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2 years ago

What is the volume of the composite figure? Explain your work. A complete answer should include how you broke up the figure, whi

Sphinxa [80]

Answer:

$15,000\:\mathrm{mm^3}$

Step-by-step explanation:

The composite figure consists of a square prism and a trapezoidal prism. By adding the volume of each, we obtain the volume of the composite figure.

The volume of the square prism is given by $V=s^2\cdot h$ , where $s$ is the base length and $h$ is the height. Substituting given values, we have: $V=14^2\cdot 30=196\cdot 30=5,880\:\mathrm{mm^3}$

The volume of a trapezoidal prism is given by $V=\frac{b_1+b_2}{2}\cdot l\cdot h$ , where $b_1$ and $b_2$ are bases of the trapezoid, $l$ is the length of the height of the trapezoid and $h$ is the height. This may look very confusing, but to break it down, we're finding the area of the trapezoid (base) and multiplying it by the height. The area of a trapezoid is given by the average of the bases ( $\frac{b_1+b_2}{2}$ ) multiplied by the trapezoid's height ( $l$ ).

Substituting given values, we get:

$V=\frac{14+24}{2}\cdot (30-14)\cdot 30,\\V=19\cdot 16\cdot 30=9,120\:\mathrm{mm^3}}$

Therefore, the total volume of the composite figure is $5,880+9,120=\boxed{15,000\:\mathrm{mm^3}}$ (ah, perfect)

Alternatively, we can break the figure into a larger square prism and a triangular prism to verify the same answer:

$V=30^2\cdot 14+\frac{1}{2}\cdot10\cdot 16\cdot 30=\boxed{15,000\:\mathrm{mm^3}}\checkmark$

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2 years ago

Find the distance between the points:

lesantik [10]

To calculate distance between two points we use the distance formula sqrt((x2−x1)^2+(y2−y1)^2).

To start, we find the square of the distance between x1 and x2 and y1 and y2. The distance between x1 and x2, or 1 and 3, is 2. The distance between y1 and y2, or 3 and -4, is 7.

Now we square 2 and 7 and add them together to get 4 + 49 = 53.

The last thing we do to find the distance is take the square root of 53. 53 is not a perfect square and is also a prime number so our answer in simplest form is still sqrt53.<span />

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3 years ago

Simplify the expression. tan(sin^−1 x)

Blizzard [7]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2799412

_______________

Let $\mathsf{\theta=sin^{-1}(x)\qquad\qquad-\dfrac{\pi}{2}\ \textless \ \theta\ \textless \ \dfrac{\pi}{2}.}$

(that is the range of the inverse sine function).

So,

$\mathsf{sin\,\theta=sin\!\left[sin^{-1}(x)\right]}\\\\ \mathsf{sin\,\theta=x\qquad\quad(i)}$

Square both sides:

$\mathsf{sin^2\,\theta=x^2\qquad\qquad(but~sin^2\,\theta=1-cos^2\,\theta)}\\\\ \mathsf{1-cos^2\,\theta=x^2}\\\\ \mathsf{1-x^2=cos^2\,\theta}\\\\ \mathsf{cos^2\,\theta=1-x^2}$

Since $\mathsf{-\,\dfrac{\pi}{2}\ \textless \ \theta\ \textless \ \dfrac{\pi}{2},}$ then $\mathsf{cos\,\theta}$ is positive. So take the positive square root and you get

$\mathsf{cos\,\theta=\sqrt{1-x^2}\qquad\quad(ii)}$

Then,

$\mathsf{tan\,\theta=\dfrac{sin\,\theta}{cos\,\theta}}\\\\\\ \mathsf{tan\,\theta=\dfrac{x}{\sqrt{1-x^2}}}\\\\\\\\ \therefore~~\mathsf{tan\!\left[sin^{-1}(x)\right]=\dfrac{x}{\sqrt{1-x^2}}\qquad\qquad -1\ \textless \ x\ \textless \ 1.}$

I hope this helps. =)

Tags: <em>inverse trigonometric function sin tan arcsin trigonometry</em>

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3 years ago

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