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prohojiy [21]
3 years ago
7

Y^2 − x^2 = 100 is that a function or no?

Mathematics
2 answers:
Ugo [173]3 years ago
7 0

Answer:

No, it is not an equation

Step-by-step explanation:

See attached graph:

bearhunter [10]3 years ago
5 0

No. If you solve it for y, you have

y=\pm\sqrt{x^2+100}

So, given a certain value for x, say x=0, you have

y=\pm\sqrt{100}=\pm10

So, you're associating two y-values for each x value. This implies that this is not a function.

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Ose has 50 coins in a jar. 20% are nickels, 40% are quarters, and the rest are pennies. (6.RP.3c-Benchmark #3)
cupoosta [38]

Answer:

Part A: 40% are pennies

Part B: NIckels 10, Quarters 20, Pennis 20

5 0
3 years ago
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What is the product of all constants $k$ such that the quadratic $x^2 + kx +15$ can be factored in the form $(x+a)(x+b)$, where
Setler79 [48]

Answer:

k=-16,k=-8,k=8,k=16

Step-by-step explanation:

We are given quadratic equations as

x^2+kx+15

and it can be factored as

=(x+a)(x+b)

now, we can multiply factor term

(x+a)(x+b)=x^2+(a+b)x+ab

now, we can compare

x^2+(a+b)x+ab=x^2+kx+15

so, we get

k=a+b

ab=15

we are given that

'a' and 'b' are integers

so, we can find all possible factors

15=(-1\times -15),(1\times 15)

15=(-3\times -5),(3\times 5)

so, we can find k

At (-1\times -15):

k=a+b

we can plug values

k=-1-15

k=-16

At (1\times 15):

k=a+b

we can plug values

k=1+15

k=16

At (-3\times -5):

k=a+b

we can plug values

k=-3-5

k=-8

At (3\times 5):

k=a+b

we can plug values

k=3+5

k=8

So, values of k are

k=-16,k=-8,k=8,k=16

6 0
3 years ago
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Degger [83]

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Step-by-step explanation:

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7 0
2 years ago
12/2 as a whole number or mixed number
bogdanovich [222]
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