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3 years ago

On a coordinate plane, a triangle has points J (negative 3, 9), L (negative 3, 2), and K (5, 2).

Mathematics

2 answers:

crimeas [40]3 years ago

7 0

Answer: Y-axis.

Step-by-step explanation:

Given : On a coordinate plane, a triangle has points J (negative 3, 9), L (negative 3, 2), and K (5, 2).

After reflection the coordinates of point K' = (-5 , 2)

Here the sign of x-coordinate changed from K to K' but the y-coordinated remained exactly same.

We know that when we reflect a figure across y-axis the sign of x-coordinate changes but y-coordinate remains same.

Rule for reflection across y-axis = (x,y) → (-x , y)

∴ The line of reflection maps point K to point K' at (–5, 2) = Y-axis.

sergeinik [125]3 years ago

7 0

Answer:

What line of reflection maps point K to point K' at (–5, 2)?

✔ y-axis

What line of reflection maps point L to point L' at (–2, 3)?

✔ y = -x

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Read 2 more answers

Suppose ABC is a right triangle with sides a, b, and c and right angle at C. Find the unknown side length using the Pythagorean

Leno4ka [110]

Answer:

a = $\sqrt{7}$

Step-by-step explanation:

I think your question is missed of key information, allow me to add in and hope it will fit the original one.:

Suppose ABC is a right triangle with sides a, b, and c and right angle at C. Find the unknown side length using the Pythagorean theorem and then find the values of the six trigonometric functions for angle B. when b=3 and c=4.

My answer:

We will Pythagoras theorem, which states that the sum of squares of two legs of a right triangle is equal to the square of the hypotenuse of right triangle. Because the question says that ABC is a right triangle.

$a^2+b^2=c^2$

Given that: b=3 and c=4

$a^2+3^2=4^2a^2+9=16a^2+9-9=16-9a^2=7$

so a = $\sqrt{7}$

We know that tangent relates opposite side of a right triangle with adjacent side.

$\text{tan}(B)=\frac{a}{b}\\\text{tan}(B)=\frac{\sqrt{7}}{3}$

Please have a look at the attached photos.

3 0

3 years ago

in a right triangle the length of a hypotenuse is c and the length of one leg is a, and the length of the other leg is b what is

puteri [66]

Answer:

Step-by-step explanation:

a=12 (height)

c=13 (hypotenous)

b=? (base)

By using Pythagoras theorm

We have, hypotenous sq=base sq+height sq

Let b =base SO, base sq=hypotenous Sq-height Sq

=>base Sq =13sq-12sq

Base Sq =169-144

Base Sq =25

Base =5 (shifting the sq)....

Hope it's helpful

5 0

2 years ago

PLEASE HELP MEEEE HURRRY!!! :)

sammy [17]

Answer:

Option D

Step-by-step explanation:

We are given the following equations -

$\begin{bmatrix}-5x-12y-43z=-136\\ -4x-14y-52z=-146\\ 21x+72y+267z=756\end{bmatrix}$

It would be best to solve this equation in matrix form. Write down the coefficients of each terms, and reduce to " row echelon form " -

$\begin{bmatrix}-5&-12&-43&-136\\ -4&-14&-52&-146\\ 21&72&267&756\end{bmatrix}$ First, I swapped the first and third rows.

$\begin{bmatrix}21&72&267&756\\ -4&-14&-52&-146\\ -5&-12&-43&-136\end{bmatrix}$ Leading coefficient of row 2 canceled.

$\begin{bmatrix}21&72&267&756\\ 0&-\frac{2}{7}&-\frac{8}{7}&-2\\ -5&-12&-43&-136\end{bmatrix}$ The start value of row 3 was canceled.

$\begin{bmatrix}21&72&267&756\\ 0&-\frac{2}{7}&-\frac{8}{7}&-2\\ 0&\frac{36}{7}&\frac{144}{7}&44\end{bmatrix}$ Matrix rows 2 and 3 were swapped.

$\begin{bmatrix}21&72&267&756\\ 0&\frac{36}{7}&\frac{144}{7}&44\\ 0&-\frac{2}{7}&-\frac{8}{7}&-2\end{bmatrix}$ Leading coefficient in row 3 was canceled.

$\begin{bmatrix}21&72&267&756\\ 0&\frac{36}{7}&\frac{144}{7}&44\\ 0&0&0&\frac{4}{9}\end{bmatrix}$

And at this point, I came to the conclusion that this system of equations had no solutions, considering it reduced to this -

$\begin{bmatrix}1&0&-1&0\\ 0&1&4&0\\ 0&0&0&1\end{bmatrix}$

The positioning of the zeros indicated that there was no solution!

Hope that helps!

6 0

3 years ago