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2 years ago

Draw the image of quadrilateral ABCD under the translation (x,y) (x-5,y)

Mathematics

1 answer:

cupoosta [38]2 years ago

5 0

Answer:

move the shape 5 left in the x axis

Step-by-step explanation:

take the coordinates and -5 from each x value (X,,y)

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Please HELP

eduard

Answer:

Let B = (x, y)

Given A = (8, 9)

$distance = \sqrt{(x_1 - x_2)^2 + (y_1-y_2)^2} \\\\10 = \sqrt{(x_1 - 8)^2 + (y_1-9)^2}$

option 3

7 0

2 years ago

Use the Product Property for Exponents to explain why x · x = x2.

Lubov Fominskaja [6]

Whenever there is no exponent on a variable,

you can give it an exponent of 1.

So we can rewrite the x's in this problem as x¹.

When we multiply two terms together

with like bases, we add their exponents.

So now just add their exponents to get x².

3 0

2 years ago

What is 13.8÷10 and a half

KengaRu [80]

The answer is 1.314

hope it helps

8 0

2 years ago

Read 2 more answers

Which would most likely be graphed using a continuous graph? Check all that apply. (Medal for correct answer.

zlopas [31]

We need an image to be able to answer this question

7 0

2 years ago

Prove that :( 1 + 1/<img src="https://tex.z-dn.net/?f=tan%5E%7B2%7DA" id="TexFormula1" title="tan^{2}A" alt="tan^{2}A" align="ab

Aliun [14]

Answer:

See explanation

Step-by-step explanation:

Simplify left and right parts separately.

$\left(1+\dfrac{1}{\tan^2A}\right)\left(1+\dfrac{1}{\cot ^2A}\right)\\ \\=\left(1+\dfrac{1}{\frac{\sin^2A}{\cos^2A}}\right)\left(1+\dfrac{1}{\frac{\cos^2A}{\sin^2A}}\right)\\ \\=\left(1+\dfrac{\cos^2A}{\sin^2A}\right)\left(1+\dfrac{\sin^2A}{\cos^2A}\right)\\ \\=\dfrac{\sin^2A+\cos^2A}{\sin^2A}\cdot \dfrac{\cos^2A+\sin^A}{\cos^2A}\\ \\=\dfrac{1}{\sin^2A}\cdot \dfrac{1}{\cos^2A}\\ \\=\dfrac{1}{\sin^2A\cos^2A}$

<u>Right part:</u>

$\dfrac{1}{\sin^2A-\sin^4A}\\ \\=\dfrac{1}{\sin^2A(1-\sin^2A)}\\ \\=\dfrac{1}{\sin^2A\cos^2A}$

Since simplified left and right parts are the same, then the equality is true.

3 0

2 years ago