All categories

3 years ago

Polygon ABCDE is reflection to produce polygon A’B’C’D E’. What is the equation for the line of reflection

Mathematics

1 answer:

lina2011 [118]3 years ago

3 0

Answer:

D Hb=HB

Step-by-step explanation:

i'm pretty sure thats the awnser

You might be interested in

23. The nth term of a sequence is n2 + 3. Find the first 3 terms of this sequence.

Setler79 [48]

23.
assuming you mean $n^2+3$

first term is n=1, so 1²+3=1+3=4
2nd is n=2 so 2²+3=4+3=7
3rd is n=3 so 3²+3=9+3=12

the first 3 terms are 4,7,12

24.
5th term is 5²+3=25+3=28
10th term is 10²+3=100+3=103

difference is 103-28=75

7 0

3 years ago

Simpifly (-5x2 - 3x - 7) + (-2x3 + 6x2 - 8)

klasskru [66]

Answer:

-2x³ + x² - 3x - 15

Step-by-step explanation:

Simply combine like terms together:

-5x² - 3x - 7 - 2x³ + 6x² - 8

-2x³ + (-5x² + 6x²) - 3x + (-7 - 8)

-2x³ + x² - 3x + (-7 - 8)

-2x³ + x² - 3x - 15

6 0

4 years ago

Read 2 more answers

Maci and I are making a small kite. Two sides are 10". Two sides are 5". The shorter diagonal is 6". Round all your answers to t

Art [367]

Answer:

A. 4".

B. Approximately 9.54".

C. Approximately 13.54".

Step-by-step explanation:

Please find the attachment.

Let x be the distance from the peak of the kite to the intersection of the diagonals and y be the distance from the peak of the kite to the intersection of the diagonals.

We have been given that two sides of a kite are 10 inches and two sides are 5 inches. The shorter diagonal is 6 inches.

A. Since we know that the diagonals of a kite are perpendicular and one diagonal (the main diagonal) is the perpendicular bisector of the shorter diagonal.

We can see from our attachment that point O is the intersection of both diagonals. In triangle AOD the side length AD will be hypotenuse and side length DO will be one leg.

We can find the value of x using Pythagorean theorem as:

$(AO)^2=(AD)^2-(DO)^2$

$x^{2}=5^2-3^2$

$x^{2}=25-9$

$x^{2}=16$

Upon taking square root of both sides of our equation we will get,

$x=\sqrt{16}$

$x=\pm 4$

Since distance can not be negative, therefore, the distance from the peak of the kite to the intersection of the diagonals is 4 inches.

B. We can see from our attachment that point O is the intersection of both diagonals. In triangle DOC the side length DC will be hypotenuse and side length DO will be one leg.

We can find the value of y using Pythagorean theorem as:

$(OC)^2=(DC)^2-(DO)^2$

Upon substituting our given values we will get,

$y^2=10^2-3^2$

$y^2=100-9$

$y^2=91$

Upon taking square root of both sides of our equation we will get,

$y=\sqrt{91}$

$y\pm 9.539392$

$y\pm\approx 9.54$

Since distance can not be negative, therefore, the distance from intersection of the diagonals to the top of the tail is approximately 9.54 inches.

C. We can see from our diagram that the length of longer diagram will be the sum of x and y.

$\text{The length of the longer diagonal}=x+y$