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3 years ago

The sum of two consecutive odd numbers is 76.Find the numbers.

Mathematics

1 answer:

devlian [24]3 years ago

5 0

Answer:

The 2 numbers are 37 and 39

Step-by-step explanation:

Let the first number be x

The next number that is odd is 2 away from it ( 1 is odd, 3 is the next number that is odd )

x, x+2 are the two numbers

Add them together to get 76

x + x+2 = 76

Combine like terms

2x+2 = 76

Subtract 2 from each side

2x+2-2 = 76-2

2x = 74

Divide by 2

2x/2 = 74/2

x= 37

x+2 = 39

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18 points help me fast please

Phantasy [73]

The answer is d positive

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3 years ago

Find the radius of convergence, r, of the series. ? n2xn 7 · 14 · 21 · ? · (7n) n = 1

defon

I'm guessing the series is supposed to be

$\displaystyle\sum_{n=1}^\infty\frac{n^2x^n}{7\cdot14\cdot21\cdot\cdots\cdot(7n)}$

By the ratio test, the series converges if the following limit is less than 1.

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(n+1)^2x^{n+1}}{7\cdot14\cdot21\cdot\cdots\cdot(7n)\cdot(7(n+1))}}{\frac{n^2x^n}{7\cdot14\cdot21\cdot\cdots\cdot(7n)}}\right|$

The first $n$ terms in the numerator's denominator cancel with the denominator's denominator:

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(n+1)^2x^{n+1}}{7(n+1)}}{n^2x^n}\right|$

$|x^n|$ also cancels out and the remaining factor of $|x|$ can be pulled out of the limit (as it doesn't depend on $n$ ).

$\displaystyle|x|\lim_{n\to\infty}\left|\frac{\frac{(n+1)^2}{7(n+1)}}{n^2}\right|=|x|\lim_{n\to\infty}\frac{|n+1|}{7n^2}=0$

which means the series converges everywhere (independently of $x$ ), and so the radius of convergence is infinite.

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4 years ago

Help me i need to submit to my teacher by 8pm

Vladimir [108]

Please see the figure. We'll first work out half the area of the rounded triangle, half the unshaded part, then double it, then subtract it from the big square.

Half the area is the circular sector PTQ (with center P, arc TQ) minus the right triangle PUT.

A/2 = area(sector PTQ) - area(triangle PUT)

The triangle is half of equilateral triangle PQT, so a 30/60/90 right triangle so we know the sides are in ratio 1:√3:2 so

TU = (7/2)√3

area(PUT) = (1/2) (7/2)(7/2)√3 = (49/8)√3

area(sector PTQ) = (angle TQP / 360°) πr^2

We know angle TQP is 60° because TQP is equilateral. r=7.

area(sector PTQ) = (60°/360°) π (7²) = 49π/6

Putting it together,

A/2 = area(sector PTQ) - area(triangle PUT)

A = 2(49π/6 - (49/8)√3)

A = 49(π/3 - √3/4) square cm

I hate ruining a nice exact answer with an approximation, but they seem to be asking.

A ≈ 30.095057615914535

Check:

I'm not sure how to check it. I'd estimate it's about 25% bigger than equilateral triangle PQT with area (√3/4)7² ≈ 21.2, so around 27. 30 seems reasonable.

Now the real area we seek is the big square PQRS minus A, so

area = 7² - 30.095057615914535 = 18.904942384086 sq cm

They want square meters for some reason; we scale by (1/100)²

Answer: 0.00189 square meters