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sladkih [1.3K]
3 years ago
10

Find cos(sin-1(3/5))

Mathematics
1 answer:
melamori03 [73]3 years ago
5 0
Cos(sin⁻¹(³/₅))
cos(36.87)
0.79
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48. A map is drawn to a scale of 1cm to represent 50km. If the actual distance between two villages is 480km, what is the distan
lana66690 [7]

Set up a ratio:

1/50 = x/480

Cross multiply:

50x = 480

Divide both sides by 50:

X = 9.6

The distance on the map is 9.6 cm

5 0
3 years ago
From a piece of tin in the shape of a square 6 inches on a side, the largest possible circle is cut out. What is the ratio of th
wel

Answer:

\sf \dfrac{1}{4} \pi \quad or \quad \dfrac{7}{9}

Step-by-step explanation:

The <u>width</u> of a square is its <u>side length</u>.

The <u>width</u> of a circle is its <u>diameter</u>.

Therefore, the largest possible circle that can be cut out from a square is a circle whose <u>diameter</u> is <u>equal in length</u> to the <u>side length</u> of the square.

<u>Formulas</u>

\sf \textsf{Area of a square}=s^2 \quad \textsf{(where s is the side length)}

\sf \textsf{Area of a circle}=\pi r^2 \quad \textsf{(where r is the radius)}

\sf \textsf{Radius of a circle}=\dfrac{1}{2}d \quad \textsf{(where d is the diameter)}

If the diameter is equal to the side length of the square, then:
\implies \sf r=\dfrac{1}{2}s

Therefore:

\begin{aligned}\implies \sf Area\:of\:circle & = \sf \pi \left(\dfrac{s}{2}\right)^2\\& = \sf \pi \left(\dfrac{s^2}{4}\right)\\& = \sf \dfrac{1}{4}\pi s^2 \end{aligned}

So the ratio of the area of the circle to the original square is:

\begin{aligned}\textsf{area of circle} & :\textsf{area of square}\\\sf \dfrac{1}{4}\pi s^2 & : \sf s^2\\\sf \dfrac{1}{4}\pi & : 1\end{aligned}

Given:

  • side length (s) = 6 in
  • radius (r) = 6 ÷ 2 = 3 in

\implies \sf \textsf{Area of square}=6^2=36\:in^2

\implies \sf \textsf{Area of circle}=\pi \cdot 3^2=28\:in^2\:\:(nearest\:whole\:number)

Ratio of circle to square:

\implies \dfrac{28}{36}=\dfrac{7}{9}

5 0
2 years ago
A scale drawing of a school bus has a scale of 1/2 inch to 5feet. If the length of the school bus is 4 1/2 inches on the scale d
Sedaia [141]
Answer: actual length = 45 ft

Explanation:

Scale drawn:

1/2 inch = 5ft

Meaning that if the drawing scale is 0.5 inch then the actual bus would be 5ft.

If the length of the drawn school bus is 4 1/2 inches. We can write:

1/2 = 5
4 1/2 = x

=> x = (5 x 4 1/2)/1/2
Or x = (5 x 4.5)/0.5
x = 22.5/0.5
x = 45

Therefore the actual length of the bus is 45 ft

3 0
3 years ago
<img src="https://tex.z-dn.net/?f=%5Cfrac%7B3%2F7%7D%7B7%5Csqrt%7B10%7D%2F2%20%7D" id="TexFormula1" title="\frac{3/7}{7\sqrt{10}
VashaNatasha [74]

Multiply the numerator and denominator by 7×2 = 14 to eliminate the denominators of those fractions:

\dfrac{\dfrac37}{\dfrac{7\sqrt{10}}2}\times\dfrac{14}{14}=\dfrac{3\times2}{7\sqrt{10}\times7}=\dfrac6{49\sqrt{10}}

Rationalize the denominator by multiplying both numerator and denominator by √10:

\dfrac6{49\sqrt{10}}\times\dfrac{\sqrt{10}}{\sqrt{10}}=\dfrac{6\sqrt{10}}{49(\sqrt{10})^2}=\dfrac{6\sqrt{10}}{49\times10}=\dfrac{6\sqrt{10}}{490}

Lastly, cancel the common factor of 2 in both the numerator and denominator (which comes from 6 = 2×3 and 490 = 2×245):

\dfrac{6\sqrt{10}}{490}=\dfrac{3\sqrt{10}}{245}}

8 0
2 years ago
Which choice is equivalent to the expression below when y2 0?<br> √y^2 + √16y^3 – 4y√y
DanielleElmas [232]

Answer:

Option C.

Step-by-step explanation:

We start with the expression:

\sqrt{y^3}  + \sqrt{16*y^3} - 4*y\sqrt{y}

where y > 0. (this allow us to have y inside a square root, so we don't mess with complex numbers)

We want to find the equivalent expression to this one.

Here, we can do the next two simplifications:

\sqrt{16*y^3} = \sqrt{16} \sqrt{y^3} = 4*\sqrt{y^3}

And:

y*\sqrt{y} = \sqrt{y^2} *\sqrt{y} = \sqrt{y^2*y} = \sqrt{y^3}

If we apply these two to our initial expression, we can rewrite it as:

\sqrt{y^3}  + \sqrt{16*y^3} - 4*y\sqrt{y}

\sqrt{y^3}  + 4*\sqrt{y^3} - 4\sqrt{y^3} = \sqrt{y^3}

Here we can use the second simplification again, to rewrite:

\sqrt{y^3} = y*\sqrt{y}

So, concluding, we have:

\sqrt{y^3}  + \sqrt{16*y^3} - 4*y\sqrt{y} = y*\sqrt{y}

Then the correct option is C.

8 0
3 years ago
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