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babymother [125]
3 years ago
6

If the sum of three consecutive numbers is 852, what is the middle one?

Mathematics
1 answer:
denis-greek [22]3 years ago
5 0

Divide the total by the quantity of numbers:

852/3 = 284

Because there are an odd quantity of numbers (3) the answer is the middle number.

The answer is 284

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Simplify the ratio 9/12
Alex777 [14]
3:4 This is the answer
5 0
2 years ago
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The vertices of a polygon are P(0,4), Q(5,4), R(5,−3) and S(0,−3). What is
andrezito [222]

Answer:

 Perimeter of the polygon p = 24

Step-by-step explanation:

Step(i):-

Given the vertices of a polygon are

P(0,4) ,Q( 5,4) ,R( 5,-3) and S(0,-3)

The distance of PQ

 a = PQ = \sqrt{(4-4)^{2} +(5-0)^{2} }  = \sqrt{25} =5

The distance of QR

b = Q R= \sqrt{(-3-4)^{2} +(5-5)^{2} }  = \sqrt{49} =7

The distance of RS

c = RS = \sqrt{(0-5)^{2} +(-3+3)^{2} }  = \sqrt{25} =5

The distance of PS

d = PS = \sqrt{(0-0)^{2} +(-3-4)^{2} }  = \sqrt{49} =7

<u><em>Step(ii):-</em></u>

Perimeter of the polygon

       = sum of all sides of polygon

p = a+ b+ c+ d

p = 5+7+5+7

p = 24

<u><em>Final answer:-</em></u>

 Perimeter of the polygon p = 24

5 0
3 years ago
Find the value of x so that f(x)=7
3241004551 [841]

Answer:

-3

Step-by-step explanation:

7 0
3 years ago
This one is extremely confusing I’ve worked it out 6 times and still no answer can somebody plz help
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It should be A
Angle b=angle g
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College calculus. I've tried it a couple times but i can't seem to get the right answer
boyakko [2]
The area is the sum of 'n' rectangle areas.
The width of the rectangle is size of interval, (domain size)/n
The height of each rectangle is f(interval) as interval moves along domain.

For this example, domain size = 3-1 = 2
size of interval = 2/n
height varies from f(1) to f(3), increasing by 2/n each time.
f(1+(2/n)i)

Putting this together, the area is the sum of:
\frac{2}{n}*f(1+\frac{2i}{n})

Since you are given the function f(x). Sub input into f(x) to get area in terms of n and i.

f(1+\frac{2i}{n}) = \frac{3(1+\frac{2i}{n})}{(1+\frac{2i}{n})^2 +8} \\  \\ =\frac{3n+6i}{n}*\frac{n^2}{(n+2i)^2 +8n^2} \\  \\ =\frac{3n^2 +6ni}{9n^2+4ni+4i^2}

Finally, the summation is:
\frac{2}{n}*\frac{3n^2 +6ni}{9n^2+4ni+4i^2} = \frac{6n +12i}{9n^2+4ni+4i^2} \\  \\  A =\lim_{n \to \infty} \sum_{i=1}^n \frac{6n +12i}{9n^2+4ni+4i^2}
7 0
3 years ago
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