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muminat
3 years ago
7

What 2v + 42.? What is the answer

Mathematics
1 answer:
seraphim [82]3 years ago
5 0
Before 1901 Australia was ruled by the
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Find the average value of the function f(t)=cos13(5t)sin(5t) f(t)=cos13⁡(5t)sin⁡(5t) on the interval [4,10].
Vilka [71]
The average value is given by

\displaystyle\frac1{10-4}\int_4^{10}\cos^{13}5t\sin5t\,\mathrm dt

Setting y=\cos5t, you get \mathrm dy=-5\sin5t\,\mathrm dt, and the integral becomes

\displaystyle-\frac1{30}\int_{\cos20}^{\cos50}y^{13}\,\mathrm dy
=-\dfrac1{30}\dfrac{y^{14}}{14}\bigg|_{y=\cos20}^{y=\cos50}
=-\dfrac{\cos^{14}50-\cos^{14}20}{420}\approx-0.00145
6 0
4 years ago
What are the next two terms in the pattern 3,6,5,10,9,18,17
Sunny_sXe [5.5K]
First you must find out the pattern The pattern is 3+3 = 6 minus one equals 5+5 = 10 minus one equals 9+9 = 18 minus one equals 17+17 The next two would be 17+17 = 34. 34 minus one equals 33 The next two answers are 34 and 33
6 0
3 years ago
Factor<br> x ^ 3 + 2x ^ 2 – 5x - 10 = 0
aleksklad [387]

Answer:

Step-by-step explanation:

When you have four terms and 3rd degree equation (the highest power is 3) you will want to try to "factor by grouping"

Group together two terms, watch out for the negative signs!

x^3 + 2x^2-5x-10=0

(x^3 + 2x^2) - (5x+10)=0

Find a common factor in each group and factor it out. You're hoping that what is left in the parenthesis is the same in both cases.

x^2(x+2)-5(x+2)=0

Now you can factor out that (x+2) because it is both terms.

x^2(x + 2) - 5(x + 2)=0

~~~~ ~~~~

Pull these out.

What will be left is the x^2 and the - 5 (dont lose that - in front of the 5)

(x + 2)(x^2 - 5) = 0

If all you have to do is factor, then you're done. It is factored. But if you have to "solve" also, then put x+2=0 and x^2-5=0 and solve.

x = -2 and x = +- sqrt5

7 0
2 years ago
Find the length of side x in simplest radical form with a rational denominator.
labwork [276]

Answer:

Step-by-step explanation:

given that both angles are the same, it is an equilateral right triangle.

we can use Pythagorean's theorem

a^2 + b^2 = c^2\\(\sqrt{6})^2+(\sqrt{6})^2= c^2\\  12 = c^2 \\c = \sqrt{12} = \sqrt{4*3} = 2\sqrt{3}

3 0
3 years ago
What’s the differential equation of
worty [1.4K]

Answer:

y=\frac{1}{x^{2}-6x+13 }

Step-by-step explanation:

We have given,

                        \frac{dy}{dx}=y^{2}(6-2x)

and initial condition x=3,\  y=\frac{1}{4}

Now,

\frac{dy}{dx}=y^{2}(6-2x)

Rearranging the variables, we get

\frac{dy}{y^{2} }=(6-2x).dx

Applying integration both sides, we get

\int\ {\frac{dy}{y^{2} } } \,=\int\ {(6-2x).} \, dx

⇒\frac{-1}{y} =6x-\frac{2x^{2} }{2}

⇒ \frac{-1}{y}=6x-x^{2}  +C                  

Putting the initial condition (i.e., x=3,\  y=\frac{1}{4}), we get

⇒ -4=6\times3-(3)^{2}+C

⇒ -4=18-9+C

∴ C=-13

We have,  \frac{-1}{y}=6x-x^{2}  +C    

now putting the value of C in above equation, we get

⇒ \frac{-1}{y}=6x-x^{2}  -13

⇒  \frac{1}{y}=-6x+x^{2}  +13

y=\frac{1}{x^{2}-6x+13 }

5 0
3 years ago
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