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olya-2409 [2.1K]
2 years ago
14

y varies directly with x, and y = 9 when x = 2. What is the value of y when x = 3? Enter your answer in the box.

Mathematics
1 answer:
Reptile [31]2 years ago
7 0
Let y=kx
9=2k
k=4.5
when x=3
y=kx becomes
y=4.5 x 3
=13.5
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I am so lost on this question. I’ve never heard of a matrices so I have no clue where to even start to solve this problem.
balu736 [363]

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Step-by-step explanation:

5 0
2 years ago
Use Newton's Method to approximate the zero(s) of the function. Continue the iterations until two successive approximations diff
liubo4ka [24]

Answer:

There is only one real zero and it is located at x = 1.359

Step-by-step explanation:

After the 4th iteration the solution was repeating the first 3 decimal places.  The formula for Newton's Method is

x_{n}-\frac{f(x_{n}) }{f'(x_{n}) }

If our function is

f(x)=x^5+x-6

then the first derivative is

f'(x)=5x^4+1

I graphed this on my calculator to see where the zero(s) looked like they might be, and saw there was only one real one, somewhere between 1 and 2.  I started with my first guess being x = 1.

When I plugged in a 1 for x, I got a zero of 5/3.  

Plugging in 5/3 and completing the process again gave me 997/687

Plugging in 997/687 and completing the process again gave me 1.36976

Plugging in 1.36976 and completing the process again gave me 1.359454

Plugging in 1.359454 and completing the process again gave me 1.359304

Since we are looking for accuracy to 3 decimal places, there was no need to go further.

Checking the zeros on the calculator graphing program gave me a zero of 1.3593041 which is exactly the same as my 5th iteration!

Newton's Method is absolutely amazing!!!

5 0
3 years ago
5.Last Sunday, 4 of the houses had some cars parked outside. If there
LUCKY_DIMON [66]
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3 years ago
In a class of 50 students, everyone has either a pierced nose or a pierced ear. The professor asks everyone with a pierced nose
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Answer:  3

Step-by-step explanation:

Let E be the event of that student pierces ear and N be the event of that student pierces nose.

Given: n(E\cup N=50)

n(E)=46\\\\n(N)=7

For any two event A and B, we have

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

Similarly , n(E\cup N)=n(E)+n(N)-n(E\cap N)

50=46+7-n(E\cap N)\\\\\Rightarrow\ n(E\cap N)=53-50=3

Hence, 3 students have piercings both on their ears and their noses.

7 0
3 years ago
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