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3 years ago

6

What are the missing sides and angles of this triangle? What is the area?

1 answer:

Butoxors [25]3 years ago

4 0

This does not appear to be a right triangle. However, we know 2 sides and the included angle, so can find the unknown side length. Let x represent this length. Then:

x^2 = (9 m)^2 + (12 m)^2 - 2(9m)(12 m)*cos 30 degrees, or

x^2 = 81 + 144 - 216(sqrt(3) / 2). Please solve for x^2 and then solve the result for x, making sure to choose the positive value. The result will be the length of the side opposite the 30 degree angle.

With 1 of 3 angles known, and 3 of 3 sides known, you can use the Law of Sines to find the other two angles. As a reminder, the Law of Sines looks like this:

a b c
-------- = --------- = ----------
sin A sin B sin C.

You can give the 30-deg angle any name you want; then a, the length of the side opposite the 30-deg angle, which you have just found. And so on.

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Help with math please!

My name is Ann [436]

Answer: 8^10

Add up the exponents 6 and 4 to get 6+4 = 10. This is using the rule a^b*a^c = a^(b+c). In this case, a = 8, b = 6, c = 4.

7 0

3 years ago

Read 2 more answers

Tom says that he needs 6 rolls to obtain each possible outcome on a 6-sided die. On the fourth roll, he rolls his second "3”. To

laila [671]

Tom is incorrect

He keeps getting the same number of 1:6 on each roll.

8 0

3 years ago

A kite on a 100-foot string is fastened to the ground, making a 25° angle with the ground. To the nearest tenth of a foot, how h

lozanna [386]

Answer: 42.3 ft

<u>Step-by-step explanation:</u>

Draw a picture (see attachment)

Given: angle = 25°, hypotenuse = 100, opposite = ???

$\sin\theta=\dfrac{opposite}{hypotenuse}\quad \rightarrow \quad \sin25^o=\dfrac{y}{100}\quad \rightarrow \quad 100\ \sin 25^o=y\\$

$\large\boxed{42.3\ ft}=y$

5 0

3 years ago

The parabola with equation $y=ax^2+bx+c$ is graphed below:

Mashutka [201]

Answer:

$m-n=2$

Step-by-step explanation:

Instead of using the standard form, we can use the vertex form of a quadratic equation:

$f(x)=a(x-h)^2+k$

Where a is the leading coefficient, and (h, k) is our vertex.

Our vertex point is at (2, -4). So, let’s substitute 2 for h and -4 for k:

$f(x)=a(x-2)^2-4$

Now, we need to determine a.

We know that it passes through the point (4, 12). So, when x is 4, y must be 12. In other words:

$12=a((4)-2)^2-4$

Solve for a. Subtract within the parentheses:

$12=a(2)^2-4$

Add 4 to both sides:

$16=a(2)^2$

Square:

$16=4a$

Solve:

$a=4$

Thererfore, the value of a is 4.

So, our function is:

$f(x)=4(x-2)^2-4$

Now, let’s find our roots. Set the equation to 0 and solve for x:

$0=4(x-2)^2-4$

$4=4(x-2)^2\\1=(x-2)^2\\x-2=\pm1 \\ x=2\pm1 \\ x=3\text{ or } 1$

So, our roots are 1 and 3.

The greater root is 3 and the lesser root is 1.

Therefore, m-n, where m>n, is 3-1 or 2.

Our final answer is 2.

4 0

3 years ago

Read 2 more answers

What are the factors of x^2 + 3x -4 ?

algol13

Answer:

1 and -4

Step-by-step explanation:

cause when you factor the equation its (x-1)(x+4)

4 0

3 years ago

Read 2 more answers