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lapo4ka [179]
3 years ago
9

Which Trig. Function do you use to find X?

Mathematics
2 answers:
zvonat [6]3 years ago
7 0

Answer:

Cosine

Step-by-step explanation:

Cosine = \frac{adjacent}{hypothenuse}

Cos 70 = \frac{x}{8}

0.633 = \frac{x}{8}

0.633 x 8 = \frac{x}{8} x 8

5.064 = x

MAXImum [283]3 years ago
4 0

Answer:

Cosine.

Step-by-step explanation:

Hypotenuse : Opposite the right angle

Opposite : opposite the angle

Adjacent : The only side left

Soh (sin) Cah (Cos) Toa (Tan)

Hypotenuse and adjacent are the only ones that have a side on them

so h and a. Cah Cosine Adjacent and hypotenuse. Cosine

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valkas [14]
Because you know that both the rectangular prism and the rectangular prism have the same volume, you can divide 588 cm^3 by 12cm to find the area of the base and get 49 cm^2. The base of the rectangular prism is x^2 so to solve just find the square rootof 49 cm^2 to get the side lengths (7 cm).

Hope this helped
7 0
3 years ago
the home of bill burton is assessed at $91,000. the tax rate is 18.5 mills. what is the property tax on bills home?
IrinaVladis [17]

Answer:

  $1683.50

Step-by-step explanation:

You are expected to know that a "mill" is one thousandth of a dollar. In this context, it is the amount of tax on one dollar of assessed valuation. So, the tax amount is found by multiplying the valuation by 18.5/1000:

  tax = 0.0185 · $91,000 = $1683.50

6 0
3 years ago
How to find the vertex calculus 2What is the vertex, focus and directrix of x^2 = 6y
son4ous [18]

Solution:

Given:

x^2=6y

Part A:

The vertex of an up-down facing parabola of the form;

\begin{gathered} y=ax^2+bx+c \\ is \\ x_v=-\frac{b}{2a} \end{gathered}

Rewriting the equation given;

\begin{gathered} 6y=x^2 \\ y=\frac{1}{6}x^2 \\  \\ \text{Hence,} \\ a=\frac{1}{6} \\ b=0 \\ c=0 \\  \\ \text{Hence,} \\ x_v=-\frac{b}{2a} \\ x_v=-\frac{0}{2(\frac{1}{6})} \\ x_v=0 \\  \\ _{} \\ \text{Substituting the value of x into y,} \\ y=\frac{1}{6}x^2 \\ y_v=\frac{1}{6}(0^2) \\ y_v=0 \\  \\ \text{Hence, the vertex is;} \\ (x_v,y_v)=(h,k)=(0,0) \end{gathered}

Therefore, the vertex is (0,0)

Part B:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

\begin{gathered} 4p(y-k)=(x-h)^2 \\  \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}

Rewriting the equation in standard form,

\begin{gathered} x^2=6y \\ 6y=x^2 \\ 4(\frac{3}{2})(y-k)=(x-h)^2 \\ \text{putting (h,k)=(0,0)} \\ 4(\frac{3}{2})(y-0)=(x-0)^2 \\ Comparing\text{to the standard form;} \\ p=\frac{3}{2} \end{gathered}

Since the parabola is symmetric around the y-axis, the focus is a distance p from the center (0,0)

Hence,

\begin{gathered} Focus\text{ is;} \\ (0,0+p) \\ =(0,0+\frac{3}{2}) \\ =(0,\frac{3}{2}) \end{gathered}

Therefore, the focus is;

(0,\frac{3}{2})

Part C:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

\begin{gathered} 4p(y-k)=(x-h)^2 \\  \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}

Rewriting the equation in standard form,

\begin{gathered} x^2=6y \\ 6y=x^2 \\ 4(\frac{3}{2})(y-k)=(x-h)^2 \\ \text{putting (h,k)=(0,0)} \\ 4(\frac{3}{2})(y-0)=(x-0)^2 \\ Comparing\text{to the standard form;} \\ p=\frac{3}{2} \end{gathered}

Since the parabola is symmetric around the y-axis, the directrix is a line parallel to the x-axis at a distance p from the center (0,0).

Hence,

\begin{gathered} Directrix\text{ is;} \\ y=0-p \\ y=0-\frac{3}{2} \\ y=-\frac{3}{2} \end{gathered}

Therefore, the directrix is;

y=-\frac{3}{2}

3 0
1 year ago
Find the circumference of the circle in terms of pi? [30]
son4ous [18]

Answer:

60π

Step-by-step explanation:

If the circle has radius of 30 units, substitute r=30 into the formula C = 2πr.

C = 2π(30)

C = 60π

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