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Natali5045456 [20]
4 years ago
10

If Sinx= 3/4, find x (3sf)

Mathematics
1 answer:
Reika [66]4 years ago
3 0
I don't know what (3sf) means

If sine(x) = 3/4
Then you want to know the angle whose sine = 3/4.
This is called the arc sine.
arc sine (3/4) = 48.59 degrees

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Factor the polynomial.
Iteru [2.4K]
3x4+6x3+9x2
Factored: (3x2)(x2+2x+3)

Answer:
C) 3x2(x2+2x+3)

Hope this helps! :)

7 0
3 years ago
Read 2 more answers
Simplify the expression. 2(3y - 7) - 5y(2 - y)
meriva
<span>2(3y - 7) - 5y(2 - y)

2*3y - 2*7 - 5y*2 -5y*(-y)

6y - 14 - 10y + 5y</span>²

5y² + 6y - 10y - 14

<span>5y² -4y - 14</span>
5 0
4 years ago
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Composite volcanoes are known for their beautiful, distinctive cone shape. A composite volcano called Mayon in the Philippines h
Grace [21]

Answer: 77.283km^{2}

Step-by-step explanation:

The formula for the lateral area of a cone L.A is:

L.A=\pi r h

Where:

r=10km is the radius of the circular base of the cone

As we know its diameter d=20km, its radius is r=\frac{d}{2}=10km

h=2.46km is the height of the cone

L.A=\pi (10km)(2.46km)

L.A=77.283km^{2} This is the lateral area of the volcano with the shape of a cone

7 0
4 years ago
A triangular lamina has vertices (0, 0), (0, 1) and (c, 0) for some positive constant c. Assuming constant mass density, show th
a_sh-v [17]

The equation of the line through (0, 1) and (<em>c</em>, 0) is

<em>y</em> - 0 = (0 - 1)/(<em>c</em> - 0) (<em>x</em> - <em>c</em>)   ==>   <em>y</em> = 1 - <em>x</em>/<em>c</em>

Let <em>L</em> denote the given lamina,

<em>L</em> = {(<em>x</em>, <em>y</em>) : 0 ≤ <em>x</em> ≤ <em>c</em> and 0 ≤ <em>y</em> ≤ 1 - <em>x</em>/<em>c</em>}

Then the center of mass of <em>L</em> is the point (\bar x,\bar y) with coordinates given by

\bar x = \dfrac{M_x}m \text{ and } \bar y = \dfrac{M_y}m

where M_x is the first moment of <em>L</em> about the <em>x</em>-axis, M_y is the first moment about the <em>y</em>-axis, and <em>m</em> is the mass of <em>L</em>. We only care about the <em>y</em>-coordinate, of course.

Let <em>ρ</em> be the mass density of <em>L</em>. Then <em>L</em> has a mass of

\displaystyle m = \iint_L \rho \,\mathrm dA = \rho\int_0^c\int_0^{1-\frac xc}\mathrm dy\,\mathrm dx = \frac{\rho c}2

Now we compute the first moment about the <em>y</em>-axis:

\displaystyle M_y = \iint_L x\rho\,\mathrm dA = \rho \int_0^c\int_0^{1-\frac xc}x\,\mathrm dy\,\mathrm dx = \frac{\rho c^2}6

Then

\bar y = \dfrac{M_y}m = \dfrac{\dfrac{\rho c^2}6}{\dfrac{\rho c}2} = \dfrac c3

but this clearly isn't independent of <em>c</em> ...

Maybe the <em>x</em>-coordinate was intended? Because we would have had

\displaystyle M_x = \iint_L y\rho\,\mathrm dA = \rho \int_0^c\int_0^{1-\frac xc}y\,\mathrm dy\,\mathrm dx = \frac{\rho c}6

and we get

\bar x = \dfrac{M_x}m = \dfrac{\dfrac{\rho c}6}{\dfrac{\rho c}2} = \dfrac13

8 0
3 years ago
What is the difference of the polynomials:
Galina-37 [17]

Answer:

(6+4x+8)-4-2x-5

6-4+4x-2x+8-5

2+2x+3.

5 0
3 years ago
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