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3 years ago

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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What base side length does the rectangular prism need to have the same volume??????

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,