All categories

3 years ago

Given that Segment BE is 5 units, find the indicated measures for each

Mathematics

1 answer:

zalisa [80]3 years ago

6 0

Answer:

The indicated measure of the length of AE = 5 units

Step-by-step explanation:

To determine:

What is the indicated measure of length?

Information Fetching and Solution Steps:

From this single diagram, it seems the indicated measure of length which we have to determine is the length of AE.

Given that Segment BE is 5 units. i.e. BE = 5 units

As the square has two equal diagonals AC and DB which meet at point E.

So, point E being the common mid-point of both diagonals.

As both of the diagonals of square are equal in length.

And the length of BE is 5 units, which is half the length of BD. It means the diagonal BD has a length of 10 units.

And since both diagonals AC and DB are equal in length, it means the length of AC must be 10 units.

Since, AC has the length of 10 units, it means the length of AE must be half of the length of AC. Thus, the indicated measure of the length of AE is 5 units.

Therefore, the indicated measure of the length of AE = 5 units

Keywords: square, diagonals, length, line segment

Learn more about square diagonals from brainly.com/question/628263

#learnwithBrainly

You might be interested in

The square of 9 less than a number is 3 less than the number. What is the number?

dmitriy555 [2]

I think it’s maybe C

8 0

3 years ago

Which value, when placed in the box, would result in a system of equations with infinitly many solutions? Y=2x - 5 2y-4x=

Nezavi [6.7K]

Given:

The system of equations is

$y=2x-5$

$2y-4x=?$

To find:

The missing value for which the given system of equations have infinitely many solutions.

Solution:

Let the missing value be k.

We have,

$y=2x-5$

$2y-4x=k$

Taking all the terms on the left side, the given equations can be rewritten as

$-2x+y+5=0$

$-4x+2y-k=0$

The system of equations $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$ have infinitely many solutions if

$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$

We have,

$a_1=-2,b_1=1,c_1=5$

$a_2=-4,b_2=2,c_2=-k$

Now,

$\dfrac{-2}{-4}=\dfrac{1}{2}=\dfrac{5}{-k}$

$\dfrac{1}{2}=\dfrac{1}{2}=\dfrac{5}{-k}$

$\dfrac{1}{2}=\dfrac{5}{-k}$

On cross multiplication, we get

$-k=10$

$k=-10$

Therefore, the missing value is -10.

6 0

3 years ago

Please help me solve this!!!!!!

BabaBlast [244]

Answer:

C. 9

Step-by-step explanation:

y = -4x^2 + 2kx

subtitute 3:

y = -4(9)^2 + 2k(9)

y = -4(81) + 18k

let y = 0

=> 0 = -324 + 18k

=> k = 324/18 = 9

this is my calculations hope it helps

4 0

2 years ago

Read 2 more answers

Simplify the following expression.

tankabanditka [31]

The answer is 2/3

or in decimal form: 0.6 repeated

7 0

3 years ago

Find the value of X answer in simplest form

Arlecino [84]

Answer:

14.14

Step-by-step explanation:

You can find the value of x using the trig ratio Sine = opposite/hypotenuse.

The sin 45 = 10/x. Solve for x.

x= 10/ sin45

x = 10/0.707

x = 14.14

5 0

3 years ago