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4 years ago

What is the value of x is?

Mathematics

1 answer:

marin [14]4 years ago

3 0

The applicable theorem is the one that tells you the exterior angle is equal to the sum of the opposite interior angles.
9x = (5x) +(9 +x)
9x = 6x +9 . . . . . . . . collect terms
3x = 9 . . . . . . . . . . . . subtract 6x
x = 3 . . . . . . . . . . . . . divide by 3

The value of x is 3.

_____
The angles (clockwise from left) are then
(9*3)° = 27°
(5*3)° = 15°
(9 +3)° = 12°
This is an obtuse triangle, obviously not drawn to scale.

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How to find the perimeter of the triangle

Sliva [168]

Answer:

A. 26.2

Step-by-step explanation:

To find the perimeter of the triangle, you have to find the distances of all three lines and add them up.

Line AB

Let's start off by finding the distance of line AB.

We will use the formula, $d = \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$ .

Point A is (-2,5) and Point B is (4,-3).

To substitute the values, it will get to $d = \sqrt{(4--2)^{2}+(-3-5)^{2}}$ which in other words is $d = \sqrt{(4+2)^{2}+(-3-5)^{2}}$ .

Now we have to solve the parenthesis to get $d = \sqrt{(6)^{2}+(-8)^{2}}$ .

Now we have to solve the exponents which would get to $d = \sqrt{36+64}$ .

Now we have to simplify the square root to $d = \sqrt{100}$ . In other words, that is $d = 10$ .

Line AB = 10

Line BC

Now let's find the distance of line BC.

We will use the same formula, $d = \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$ .

Point B is (4,-3) and Point C is (0,-6).

To substitute the values, it will get to $d = \sqrt{(0-4)^{2}+(-6--3)^{2}}$ which in other words is $d = \sqrt{(0-4)^{2}+(-6+3)^{2}}$ .

Now we have to solve the parentheses to get $d = \sqrt{(-4)^{2}+(-3)^{2}}$ .

Now we have to solve the exponents which would get to $d = \sqrt{16+9}$ .

Now we have to simplify the square root to $d = \sqrt{25}$ . In other words, that is $d = 5$ .

Line BC = 5.

Line AC

Now let's fine the distance of line AC.

We will use the same formula, $d = \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$ .

Point A is (-2,5) and Point C is (0,-6).

To substitute the values, it will get to $d = \sqrt{(0--2)^{2}+(-6-5)^{2}}$ which in other words is $d = \sqrt{(0+2)^{2}+(-6-5)^{2}}$ .

Now we have to solve the parentheses to get $d = \sqrt{(2)^{2}+(-11)^{2}}$ .

Now we have to solve the exponents which would get to $d = \sqrt{4+121}$ .

Now we have to simplify the square root to $d = \sqrt{125}$ . In other words, that is $d = 5\sqrt{5}$ . To round that, $d = 11.2$ .

Line AC = 11.2.

Perimeter of Triangle ABC

Perimeter of Triangle ABC = Line AB + Line BC + Line AC.

Perimeter of Triangle ABC = 10 + 5 + 11.2

Perimeter of Triangle ABC = 26.2

Hope this helped! If not, please let me know <3

3 0

3 years ago

Which function is a horizontal compression of its parent function f(x)=x^7 by a factor of 4?

nasty-shy [4]

Note that what we want is to compress any point (a, b) in the parent function, to (a, b/4).

(a, b) is a point of the graph of the function $f(x)=x^7$ , thus we can rewrite the point as:
$(a, a^7)$ .

This point is compressed to

$\displaystyle{ (a, \frac{a^7}{4} )$ .

This means that the function we are looking for is $\displaystyle{ t(x)= \frac{x^7}{4}$ .

Answer: t(x)=1/4x^7

6 0

3 years ago

Read 2 more answers

How do you find the area of a trapezoid?

Anika [276]