8 more than 4 times a number is 36

AB=6CM AC= 12cm <br> calculate the length of CD

Brrunno [24]

AB=6cm AC=12cm

AC² = AB + BC²

12² = 6² + CB²

CB² = 12² - 6²

CB² = 144 - 36

CB² = 108

CB = √ 108 cm.

sin 55° = CB / CD

0.81915 = √108 / CD

CD = 10.392 / 0.81915

CD = 12.686 cm

7 0

3 years ago

Read 2 more answers

Gavin goes to the market and buys one rectangle shaped board. The length of the board is 16 cm and width is 10 cm. if he wants t

Ira Lisetskai [31]

Check the picture below.

doesn't that make it just a 20 x 14? well, surely you know what that area is.

8 0

3 years ago

URGENT! PLEASE HELP.

Harrizon [31]

The distance BC from Tower 2 to the plane will be 14,065.5 ft and the height of the plane from the ground will be 5,720.9 ft.

<h3>What is a right-angle triangle?</h3>

It is a type of triangle in which one angle is 90 degrees and it follows the Pythagoras theorem and we can use the trigonometry function.

A plane is located at C on the diagram.

There are two towers located at A and B.

The distance between the towers is 7600 feet and angles of elevation are given.

Then in the right-angle triangle ΔADC, we have

$\rm \tan 16 = \dfrac{H}{7600 + X}\\\\\\H = 0.2867(7600 +X)\\\\\\H = 0.2756\ X + 2179.2649$ ...1

Then in the right-angle triangle ΔBDC, we have

$\rm \tan24 = \dfrac{H}{ X}\\\\\\H = 0.4452\ X$ ...2

From equations 1 and 2, we have

0.4452X = 0.2756 X + 2179.2649

0.1696X = 2179.2649

X = 12849.439 ≅ 12,849.4 ft

Then the distance BC from Tower 2 to the plane will be

$\rm BC = \dfrac{12849.4}{\cos 24}\\\\\\BC = 14065.5 \ ft$

Then the height of the plane from the ground will be

$\rm H = 12849.4 \times \tan 24 \\\\\\H = 5720.9 \ ft$

The figure is shown below.

More about the right-angle triangle link is given below.

brainly.com/question/3770177

#SPJ1

3 0

2 years ago

What is the solution of x=2+ square root x-2?

alex41 [277]

So I'm going to assume that this question is asking for <u>non extraneous solutions</u>, or solutions that are found in the equation <em>and</em> are valid solutions when plugged back into the equation. So firstly, subtract 2 on both sides of the equation:

$x-2=\sqrt{x-2}$

Next, square both sides:

$(x-2)^2=x-2\\(x-2)(x-2)=x-2\\x^2-4x+4=x-2$

Next, subtract x and add 2 to both sides of the equation:

$x^2-5x+6=0$

Now we are going to be factoring by grouping to find the solution(s). Firstly, what two terms have a product of 6x^2 and a sum of -5x? That would be -3x and -2x. Replace -5x with -2x - 3x:

$x^2-2x-3x+6=0$

Next, factor x^2 - 2x and -3x + 6 separately. Make sure that they have the same quantity on the inside of the parentheses:

$x(x-2)-3(x-2)=0$

Now you can rewrite the equation as $(x-3)(x-2)=0$

Now, apply the Zero Product Property and solve for x as such:

$x-3=0\\x=3\\\\x-2=0\\x=2$

Now, it may appear that the answer is C, however we need to plug the numbers back into the original equation to see if they are true as such:

$2=2+\sqrt{2-2}\\2=2+\sqrt{0}\\2=2+0\\2=2\ \textsf{true}\\\\3=2+\sqrt{3-2}\\3=2+\sqrt{1}\\3=2+1\\3=3\ \textsf{true}$

Since both solutions hold true when x = 2 and x = 3, <u>your answer is C. x = 2 or x = 3.</u>

8 0

3 years ago

What is 4/3 times 3/7=?

Sidana [21]

The answer is 4/7 in simplest form

3 0

3 years ago