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vlada-n [284]
3 years ago
10

The perimeter of a semicircle is 15.42 kilometers. What is the semicircle's radius?

Mathematics
1 answer:
olya-2409 [2.1K]3 years ago
4 0

Answer:

r = 4.9 km

Step-by-step explanation:

Perimeter of circle = πr

Where P=15.42 km , π = 3.14

15.42 = (3.14)(r)

r = 15.42 / 3.14

r = 4.9 km

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Angle 1 measure 75 degree what is its complement
dalvyx [7]
I think it’s 76 because I added it that’s a way i solved it but you can also multiple it by 1 and it will give the the same answer
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A point is located in quadrant II of a
givi [52]

Answer:

Choose a point with a negative x coordinate and a positive y coordinate.

Step-by-step explanation:

The quadrants are labeled counter clockwise 1, 2, 3, and 4.

Quadrant I - has x and y coordinate both positive.

Quadrant 2 - has x coordinate negative and y coordinates positive.

Quadrant 3  - has x and y coordinates both negative.

Quadrant 4 - has x coordinates positive and y coordinates negative.

Since the point is in quadrant 2, choose a point where x is negative but y is positive like (-3, 2).

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3 years ago
The price of pair of shoes is $63.20. The sales tax rate is 4.5 percent. How much sales tax would you pay if you bought these sh
olga nikolaevna [1]
The answer to the problem is as follows:

We can convert 4.5% into a decimal form by moving the decimal to the left two places. This becomes: .045

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4 0
3 years ago
Read 2 more answers
1 + 2 + (-5) + 4 <br><br> A . -6<br> B . -5 <br> C . 0<br> D . 2<br> E . 4<br> F . 12
Dovator [93]

Answer:

Option D. 2

Step-by-step explanation:

1+2+(-5)+4

= 1+2-5+4

= 1+2+4-5

= 7-5

= 2

5 0
3 years ago
In the derivation of Newton’s method, to determine the formula for xi+1, the function f(x) is approximated using a first-order T
dimaraw [331]

Answer:

Part A.

Let f(x) = 0;

suppose x= a+h

such that f(x) =f(a+h) = 0

By second order Taylor approximation, we get

f(a) + hf'±(a) + \frac{h^{2} }{2!}f''(a) = 0

h = \frac{-f'(a) }{f''(a)} ± \frac{\sqrt[]{(f'(a))^{2}-2f(a)f''(a) } }{f''(a)}

So, we get the succeeding equation for Newton's method as

x_{i+1} = x_{i} + \frac{1}{f''x_{i}}  [-f'(x_{i}) ± \sqrt{f(x_{i})^{2}-2fx_{i}f''x_{i} } ]

Part B.

It is evident that Newton's method fails in two cases, as:

1.  if f''(x) = 0

2. if f'(x)² is less than 2f(x)f''(x)    

Part C.

In case  x_{i+1} is close to x_{i}, the choice that shouldbe made instead of ± in part A is:

f'(x) = \sqrt{f'(x)^{2} - 2f(x)f''(x)}  ⇔ x_{i+1} = x_{i}

Part D.

As given x_{i+1} = x_{i} = h

or                 h = x_{i+1} - x_{i}

We get,

f(a) + hf'(a) +(h²/2)f''(a) = 0

or h² = -hf(a)/f'(a)

Also,             (x_{i+1}-x_{i})² = -(x_{i+1}-x_{i})(f(x_{i})/f'(x_{i}))

So,                f(a) + hf'(a) - (f''(a)/2)(hf(a)/f'(a)) = 0

It becomes   h = -f(a)/f'(a) + (h/2)[f''(a)f(a)/(f(a))²]

Also,             x_{i+1} = x_{i} -f(x_{i})/f'(x_{i}) + [(x_{i+1} - x_{i})f''(x_{i})f(x_{i})]/[2(f'(x_{i}))²]

6 0
3 years ago
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