What is the value of x to the nearest tenth tan 64 degrees = x/7

Find the area of the right triangle. If necessary, round to the nearest tenth

Ad libitum [116K]

Answer:

B. 150 yards

Step-by-step explanation:

Area of a Triangle = .5 * Base * height

So we need the height of the triangle, or the missing side.

We have to use Pythagorean Theorem to solve this:

A^2 + B^2 = C^2

15 ^ 2 + B ^ 2 = 25 ^2

b^2 = 25^2 - 15 ^ 2

b^2 = 625 - 225

b^2 = 400

sqrt (400) = 20

The missing height is 20. Now we plug it into 1/2 * b * H

.5 * 15 * 20

7.5 * 20 = 150

150 Yds

6 0

3 years ago

Combine and simplify the following radical expression ASAP ASAP ASAP ASAP

Montano1993 [528]

Answer:

$\sqrt[3]{3}$

Step-by-step explanation:

Our expression is: $\frac{1}{3} \sqrt[3]{81}$ .

Let's focus on the cube root of 81 first. What's the prime factorisation of 81? It's simply: 3 * 3 * 3 * 3, or $3^3*3$ . Put this in for 81:

$\sqrt[3]{81} =\sqrt[3]{3^3*3}=\sqrt[3]{3^3} *\sqrt[3]{3}$

We know that the cube root of 3 cubed will cancel out to become 3, but the cube root of 3 cannot be further simplified, so we keep that. Our outcome is then:

$\sqrt[3]{3^3} *\sqrt[3]{3}=3\sqrt[3]{3}$

Now, let's multiply this by 1/3, as shown in the original problem:

$\frac{1}{3}* 3\sqrt[3]{3}=\sqrt[3]{3}$

Thus, the answer is $\sqrt[3]{3}$ .

~ an aesthetics lover

6 0

3 years ago

Graph the hyperbola using the transverse axis, vertices, and co-vertices:

Reptile [31]

See attachment for the graph of the hyperbola 12x^2 - 3y^2 - 108 = 0

<h3>How to graph the hyperbola?</h3>

The equation of the hyperbola is given as:

12x^2 - 3y^2 - 108 = 0

Start by calculating the transverse axis

So, we have:

Transverse axis

The vertices of the given hyperbola are (0, 0)

This means that

(h, k) = 0

Where

a = 3 and b = 6

The transverse axis is calculated as:

y = ±b/a(x - h) + k

So, we have:

y = ±6/3(x - 0) + 0

Evaluate the difference and sum

y = ±6/3x

Evaluate the quotient

y = ±2x

This means that the transverse axes are y = 2x and y =-2x

The vertices

In the above section, we have:

The vertices of the given hyperbola are (0, 0)

This means that

(h, k) = 0

The co-vertices

In the above section, we have:

The vertices of the given hyperbola are (0, 0)

This means that

(h, k) = 0

And

a = 3 and b = 6

The co-vertices are

(h - a, k) and (h + a, k)

So, we have:

(0 - 3, 0) and (0 + 3, 0)

Evaluate

(-3,0) and (3, 0)

See attachment for the graph of the hyperbola