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Luden [163]
3 years ago
7

What are the solutions to the nonlinear equations below y=4x x^2 + y^2=17

Mathematics
1 answer:
wolverine [178]3 years ago
8 0

Answer:

x = 1 and x = -1

Step-by-step explanation:

Next time, please separate the equations with a comma or semicolon.  Thanks.

We can substitute 4x for y in the quadratic equation x^2 + y^2=17:

x² + (4x)²=17.

Then x² + 16x² = 17, or

17x² = 17, or x² = 1.  There are two solutions:  x = 1 and x = -1.

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Akeem is carrying water in a bucket, but is unsure how many gallons it holds. He knows that the bucket and water weigh 32.49 pou
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Answer:

Step-by-step explanation:

8.3g+2.61=32.49

8.3g=29.88

g=3.6

5 0
2 years ago
The rectangle has a length of 5x - 10 and a width of 2x. The perimeter is 120 units. Solve for x and use it to identify the actu
Ahat [919]

Answer: x= 10

Step-by-step explanation:

l= 40

w=20

P=l+w+l+w

120=5x-10+2x+5x-10+2x

120= 14x-20

120+20=14

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4 0
2 years ago
Read 2 more answers
A computer system uses passwords that are six characters, and each character is one of the 26 letters (a–z) or 10 integers (0–9)
Blababa [14]

First of all, since we have 36 characters available per spot (26 letters and 10 digits), and we have 6 spots, we have a total of

36^6

possible passwords.

Event A happens if the password starts with either a, e, i, o or u. If we fix the first character, we're left with 36 characters available for each of the remaining 5 spots, leading to a total of

5\cdot 36^5

possible passwords.

So, the probability of event A, computed as the ratio between "good" cases and all possible cases, is

\dfrac{5\cdot 36^5}{36^6}=\dfrac{5}{36}

Event B works exactly the same, since we're fixing the last spot, leaving 36 characters available for each of the first 5 spots. So, we have

P(A)=P(B)=\dfrac{5}{36}

As for the intersection, we want the first character to be a vowel, and the last character to be an even digits. There are 25 passwords satisfying this request:

axxxx0,\ axxxx2,\ axxxx4,\ axxxx6,\ axxxx8

exxxx0,\ exxxx2,\ exxxx4,\ exxxx6,\ exxxx8

ixxxx0,\ ixxxx2,\ ixxxx4,\ ixxxx6,\ ixxxx8

oaxxxx0,\ oxxxx2,\ oxxxx4,\ oxxxx6,\ oxxxx8

uxxxx0,\ uxxxx2,\ uxxxx4,\ uxxxx6,\ uxxxx8

Where x can be any of the 36 characters.

So, we have 25 cases with 4 vacant slots, leading to a probability of

P(A\cap B)=\dfrac{25\cdot 36^4}{36^6}=\dfrac{25}{1296}

Finally, you can compute the probability of the union using the formula

P(A\cup B)=P(A)+P(B)-P(A\cap B)

Since we already computed all these quantities.

7 0
2 years ago
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Naddika [18.5K]

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Answer:

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