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worty [1.4K]
3 years ago
5

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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n less than 7, is represented as this:

n>7

Step-by-step explanation:

You just need to think about what the question is, if it's less than, you use the > symbol to prove one variable is smaller than the other.

i hope this helps!

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(b) The perimeter of a rectangular window is 298 cm.
Ronch [10]

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68

Step-by-step explanation:

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3 0
3 years ago
How can you find a unit rate when given a rate?
ivanzaharov [21]
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7 0
3 years ago
HELP PLEASE!!! I need help with 94 if you could show the steps that would be very helpful!
aksik [14]
A combination is an unordered arrangement of r distinct objects in a set of n objects. To find the number of permutations, we use the following equation:

n!/((n-r)!r!)

In this case, there could be 0, 1, 2, 3, 4, or all 5 cards discarded. There is only one possible combination each for 0 or 5 cards being discarded (either none of them or all of them). We will be the above equation to find the number of combination s for 1, 2, 3, and 4 discarded cards.

5!/((5-1)!1!) = 5!/(4!*1!) = (5*4*3*2*1)/(4*3*2*1*1) = 5

5!/((5-2)!2!) = 5!/(3!2!) = (5*4*3*2*1)/(3*2*1*2*1) = 10

5!/((5-3)!3!) = 5!/(2!3!) = (5*4*3*2*1)/(2*1*3*2*1) = 10

5!/((5-4)!4!) = 5!/(1!4!) = (5*4*3*2*1)/(1*4*3*2*1) = 5

Notice that discarding 1 or discarding 4 have the same number of combinations, as do discarding 2 or 3. This is being they are inverses of each other. That is, if we discard 2 cards there will be 3 left, or if we discard 3 there will be 2 left.

Now we add together the combinations

1 + 5 + 10 + 10 + 5 + 1 = 32 choices combinations to discard.

The answer is 32.

-------------------------------

Note: There is also an equation for permutations which is:

n!/(n-r)!

Notice it is very similar to combinations. The only difference is that a permutation is an ORDERED arrangement while a combination is UNORDERED.

We used combinations rather than permutations because the order of the cards does not matter in this case. For example, we could discard the ace of spades followed by the jack of diamonds, or we could discard the jack or diamonds followed by the ace of spades. These two instances are the same combination of cards but a different permutation. We do not care about the order.

I hope this helps! If you have any questions, let me know :)








7 0
3 years ago
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