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3 years ago

I need the full work shown and solution!

Mathematics

1 answer:

Alexus [3.1K]3 years ago

3 0

Answer:

the diameter of Britain flag is given = 28mm

so the radius = 14mm

perimeter of the flag occupied = 2πr = (2×22/7×14) mm

= 88mm

area of the flag= πr^2 = (22/7 × 196) mm^2 = 616 mm^2

Hope it helps

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In the right triangle shown, m∠A = 30 degrees, and BC = 6√2, how long is AC?

mart [117]

They're trying to trick you with √2. Remember in the 30/60/90 triangle the sides are in ratio 1:√3:2, with the "1" opposite the 30 degrees.

Here we have

1:√3:2 = 6√2:x:hypotenuse

x/(6√2) = √3/ 1

x = 6√2×√3 = 6√6

Answer: AC=6√6

3 0

3 years ago

If we sub (3x-5y) from (6x +8y)then the result is plz repy

konstantin123 [22]

Answer:

3x + 13y

Explanation:

6x + 8y − (3x − 5y)

Distribute the Negative Sign

6x + 8y + −1(3x − 5y)

6x + 8y + −1(3x) + −1(−5y)

6x + 8y + −3x + 5y

Combine Like Terms

6x + 8y + −3x + 5y

(6x + −3x) + (8y + 5y)

= 3x + 13y

8 0

3 years ago

Read 2 more answers

I'm having trouble with #2. I've got it down to the part where it would be the integral of 5cos^3(pheta)/sin(pheta). I'm not sur

Butoxors [25]

$\displaystyle\int\frac{\sqrt{25-x^2}}x\,\mathrm dx$

Setting $x=5\sin\theta$ , you have $\mathrm dx=5\cos\theta\,\mathrm d\theta$ . Then the integral becomes

$\displaystyle\int\frac{\sqrt{25-(5\sin\theta)^2}}{5\sin\theta}5\cos\theta\,\mathrm d\theta$
$\displaystyle\int\sqrt{25-25\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\sqrt{1-\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$

Now, $\sqrt{x^2}=|x|$ in general. But since we want our substitution $x=5\sin\theta$ to be invertible, we are tacitly assuming that we're working over a restricted domain. In particular, this means $\theta=\sin^{-1}\dfrac x5$ , which implies that $\left|\dfrac x5\right|\le1$ , or equivalently that $|\theta|\le\dfrac\pi2$ . Over this domain, $\cos\theta\ge0$ , so $\sqrt{\cos^2\theta}=|\cos\theta|=\cos\theta$ .

Long story short, this allows us to go from

$\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$

to

$\displaystyle5\int\cos\theta\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\dfrac{\cos^2\theta}{\sin\theta}\,\mathrm d\theta$

Computing the remaining integral isn't difficult. Expand the numerator with the Pythagorean identity to get

$\dfrac{\cos^2\theta}{\sin\theta}=\dfrac{1-\sin^2\theta}{\sin\theta}=\csc\theta-\sin\theta$

Then integrate term-by-term to get

$\displaystyle5\left(\int\csc\theta\,\mathrm d\theta-\int\sin\theta\,\mathrm d\theta\right)$
$=-5\ln|\csc\theta+\cot\theta|+\cos\theta+C$

Now undo the substitution to get the antiderivative back in terms of $x$ .

$=-5\ln\left|\csc\left(\sin^{-1}\dfrac x5\right)+\cot\left(\sin^{-1}\dfrac x5\right)\right|+\cos\left(\sin^{-1}\dfrac x5\right)+C$

and using basic trigonometric properties (e.g. Pythagorean theorem) this reduces to

$=-5\ln\left|\dfrac{5+\sqrt{25-x^2}}x\right|+\sqrt{25-x^2}+C$