All categories

3 years ago

the point A(7,1) is reflected over the point (4,0) and its image is point B. What are the coordinates of point B?

Mathematics

1 answer:

umka2103 [35]3 years ago

6 0

Answer: The coordinates of point be are (1,-1)

Step-by-step explanation:

The reflected point will be equidistant from the point over which it is reflected as the original point and in the same line as the original point and the reflected point.

Find the difference between the x and y values of the given points and add that value (distance) to the reflection point.

x-values: 4-7 = -3 y-values: 0-1 = -1

Add to reflection point x: 4-3 = 1 y: 0 -1 = -1

You might be interested in

Find the slope of the line that passes through<br> (10, 1) and (6,2)

Daniel [21]

Answer:

m=-1/4

Step-by-step explanation:

2-1/6-10=1/-4

-1/4 is the slope

7 0

2 years ago

Read 2 more answers

Solve x plssssss :)))))

blondinia [14]

Answer:

x = 43

Step-by-step explanation:

180 - 65 = 115

115 = x + 72

43 = x

x = 43

7 0

2 years ago

What is the solution of

kobusy [5.1K]

Answer:

Third option: x=0 and x=16

Step-by-step explanation:

$\sqrt{2x+4}-\sqrt{x}=2$

Isolating √(2x+4): Addind √x both sides of the equation:

$\sqrt{2x+4}-\sqrt{x}+\sqrt{x}=2+\sqrt{x}\\ \sqrt{2x+4}=2+\sqrt{x}$

Squaring both sides of the equation:

$(\sqrt{2x+4})^{2}=(2+\sqrt{x})^{2}$

Simplifying on the left side, and applying on the right side the formula:

$(a+b)^{2}=a^{2}+2ab+b^{2}; a=2, b=\sqrt{x}$

$2x+4=(2)^{2}+2(2)(\sqrt{x})+(\sqrt{x})^{2}\\ 2x+4=4+4\sqrt{x}+x$

Isolating the term with √x on the right side of the equation: Subtracting 4 and x from both sides of the equation:

$2x+4-4-x=4+4\sqrt{x}+x-4-x\\ x=4\sqrt{x}$

Squaring both sides of the equation:

$(x)^{2}=(4\sqrt{x})^{2}\\ x^{2}=(4)^{2}(\sqrt{x})^{2}\\ x^{2}=16 x$

This is a quadratic equation. Equaling to zero: Subtract 16x from both sides of the equation:

$x^{2}-16x=16x-16x\\ x^{2}-16x=0$

Factoring: Common factor x:

x (x-16)=0

Two solutions:

1) x=0

2) x-16=0

Solving for x: Adding 16 both sides of the equation:

x-16+16=0+16

x=16

Let's prove the solutions in the orignal equation:

1) x=0:

$\sqrt{2x+4}-\sqrt{x}=2\\ \sqrt{2(0)+4}-\sqrt{0}=2\\ \sqrt{0+4}-0=2\\ \sqrt{4}=2\\ 2=2$

x=0 is a solution

2) x=16

$\sqrt{2x+4}-\sqrt{x}=2\\ \sqrt{2(16)+4}-\sqrt{16}=2\\ \sqrt{32+4}-4=2\\ \sqrt{36}-4=2\\ 6-4=2\\ 2=2$

x=16 is a solution

Then the solutions are x=0 and x=16

5 0

3 years ago

Which description below the equation <br> X/4=55

Mariana [72]

x/4 = 55

=> x = 55 × 4

=> x = 220

4 0

2 years ago

Read 2 more answers

There are 2.54 centimeters in 1 inch. There are 100 centimeters in 1 meter .to the nearest meter ,how many meters are in 241 inc

matrenka [14]

Answer:

6.12 meters

Step-by-step explanation:

5 0

3 years ago