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3 years ago

The Area of the Rectangle

Mathematics

1 answer:

pshichka [43]3 years ago

3 0

Answer:

450 I believe is correct

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Solve for y. Answer to the nearest tenth

AysviL [449]

Answer:

y = 11.5

Step-by-step explanation:

Given 2 secants from an external point to the circle.

Then the product of the external part and the whole of one secant is equal to the product of the external part and the whole of the other secant, that is

6(6 + y) = 5(5 + 16)

36 + 6y = 5 × 21 = 105 ( subtract 36 from both sides )

6y = 69 ( divide both parts by 6 )

y = 11.5

7 0

2 years ago

What is £300 divided into a ratio of 3:7

3241004551 [841]

A) sum 3 + 7 =10
b) divide £300 ÷ 10 = £30
c) multiply:
3 x 30 = £90
7 x 30 = £210

Values are: £90 and £210

5 0

3 years ago

Read 2 more answers

What is the domain of the relation shown on the graph?

rosijanka [135]

Answer:

where is the graph?........

7 0

2 years ago

Factor completely 2x^3y^4-8x^2y^3+6xy^2

Maru [420]

Answer:

2 x y^2 (x y - 1) (x y - 3)

Step-by-step explanation:

2x^3y^4-8x^2y^3+6xy^2

each term contains 2xy^2 so factor that out

2xy^2(x^2y^2 -4xy+3)

then lets factor the inside

what terms multiply together to give us +3 and add to -4

-3* -1 = 3 -3+-1 =-4

2 x y^2 (x y - 1) (x y - 3)

6 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