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Mathematics

2 answers:

My name is Ann [436]3 years ago

7 0

(-2,8),(-1,5),(0,2),(1,-1),(2,-4) are points of the line.

Natasha_Volkova [10]3 years ago

6 0

−7y-21x=<span>−6-8
</span>−7y-21x =<span>−14</span>

You might be interested in

(a) Use the reduction formula to show that integral from 0 to pi/2 of sin(x)^ndx is (n-1)/n * integral from 0 to pi/2 of sin(x)^

Sedbober [7]

Hello,

a)
$I= \int\limits^{ \frac{\pi}{2} }_0 {sin^n(x)} \, dx = \int\limits^{ \frac{\pi}{2} }_0 {sin(x)*sin^{n-1}(x)} \, dx \\

= [-cos(x)*sin^{n-1}(x)]_0^ \frac{\pi}{2}+(n-1)*\int\limits^{ \frac{\pi}{2} }_0 {cos(x)*sin^{n-2}(x)*cos(x)} \, dx \\

=0 + (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {cos^2(x)*sin^{n-2}(x)} \, dx \\

= (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {(1-sin^2(x))*sin^{n-2}(x)} \, dx \\
= (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {sin^{n-2}(x)} \, dx - (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {sin^n(x) \, dx\\

$
$I(1+n-1)= (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {sin^{n-2}(x)} \, dx \\
I= \dfrac{n-1}{n} *\int\limits^{ \frac{\pi}{2} }_0 {sin^{n-2}(x)} \, dx \\
$

b)
$\int\limits^{ \frac{\pi}{2} }_0 {sin^{3}(x)} \, dx \\
= \frac{2}{3} \int\limits^{ \frac{\pi}{2} }_0 {sin(x)} \, dx \\
= \dfrac{2}{3}\ [-cos(x)]_0^{\frac{\pi}{2}}=\dfrac{2}{3} \\




$

$\int\limits^{ \frac{\pi}{2} }_0 {sin^{5}(x)} \, dx \\
= \dfrac{4}{5}*\dfrac{2}{3} \int\limits^{ \frac{\pi}{2} }_0 {sin(x)} \, dx = \dfrac{8}{15}\\





$

c)

$I_n= \dfrac{n-1}{n} * I_{n-2} \\

I_{2n+1}= \dfrac{2n+1-1}{2n+1} * I_{2n+1-2} \\
= \dfrac{2n}{2n+1} * I_{2n-1} \\
= \dfrac{(2n)*(2n-2)}{(2n+1)(2n-1)} * I_{2n-3} \\
= \dfrac{(2n)*(2n-2)*...*2}{(2n+1)(2n-1)*...*3} * I_{1} \\\\

I_1=1\\

$

3 0

3 years ago

The maximum afternoon temperature in Granderson is modeled by t=60-30 cos (x<img src="https://tex.z-dn.net/?f=%20%5Cpi%20" id="T

kicyunya [14]

The equation is $t=60-30 cos (x \frac{\pi}{6})$

the -30 expression, is subtracting, if negative, from the 60
if the cosine returned is negative, you'd end up with a +30,
negative * negative = positive

and then the 30 expression will ADD to the 60 amount
so the temperature "t" is highest, when the 30 expression, is
positive and and it's highest

when does that happen, when cosine is negative and at its highest,
well, cosine range is $-1\le cos(\theta ) \le 1$
so the lowest value cosine can provide is -1, when is cosine -1?
well, at $\pi$

so... let's find a value that makes that expression to $cos(\pi)$

$\bf t=60-30 cos (x \frac{\pi}{6}) \qquad x=6
\\\\
thus
\\\\
t=60-30 cos (6 \frac{\pi}{6})\implies t=60-30 cos (\pi )
\\\\
t=60-30[-1]\implies t=60+30\implies t=90$

----------------------------------------------------------------------------------------
so...for August, that'll mean

$\bf t=60-30 cos (x \frac{\pi}{6}) \qquad x=7
\\\\
t=60-30 cos (\frac{7\pi}{6})\implies t=60-30\left( -\cfrac{\sqrt{3}}{2} \right)
\\\\
t=60+(15\cdot \sqrt{3})\implies t\approx85.98^o$

4 0

3 years ago

4. Liam can work at most 20 hours a week but he needs to earn at least $125. His dog-walking job pays

Oxana [17]

Two possible solutions are: (4.5, 8.5) and (8, 6.3)

<h3>How to graph the inequality?</h3>

Let x represents hours dog walking and y represents hours washing cars.

So, we have:

Earnings = Rate of dog walk * x + Rate of car wash * y

This gives

Earnings = 7x + 11y

He wants to earn at least $125.

This means that:

7x + 11y ≥ 125

See attachment for the graph of the inequality

From the attached graph, two possible solutions are: (4.5, 8.5) and (8, 6.3)