The diagonals of rectangle nopq intersect at point r. if qr=3x-4 and np=5x+20, solve for x.

Mathematics

2 answers:

FrozenT [24]3 years ago

7 0

2(3x - 4) = 5x + 20
6x - 8 = 5x + 20
subtract 5x from both sides
x - 8 = 20
add 8 to both sides
x = 28

Yuki888 [10]3 years ago

7 0

Answer:

The value of "x" is 28.

Step-by-step explanation:

As can see in the attach the length of both diagonals is the same. From the problem, you have the length of segment qr = 3x -4, which represents half of the diagonal. Then, length of segment np = 5x + 20, is the completa diagonal. So, you have to multiply the equation for qr by 2 and then equal it to equation of segment np.

2·(3x - 4) = 5x + 20 ⇒ 6x - 8 = 5x + 20 ⇒ 6x - 5x = 20 + 8 ⇒ x = 28.

Summarizing the value of "x" is 28.

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Which value, when placed in the box, would result in a system of equations with infinitely many solutions?

Fantom [35]

Answer:

<h2>If we placed the number 10 in the box, we obtain a system of equations with infinitely many solutions.</h2>

Step-by-step explanation:

The given system is

$y=2x-5\\2y-4x=a$

<em>It's important to know that a system with infinitely many solutions, it's a system that has the same equation</em>, that is, both equation represent the same line, or as some textbooks say, one line is on the other one, so they have inifinitely common solutions.

Having said that, the first thing we should do here is reorder the system

$y=2x-5\\2y=4x+a$

This way, you can compare better both equations. If you look closer, observe that the second equation is double, that is, it can be obtained by multiplying a factor of 2 to the first one, that is

$y=2x-5\\2y=4x-10$

So, by multiplying such factor, we obtaine the second equation. Observe that $a$ must be equal to 10, that way the system would have infinitely solutions.