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alexgriva [62]
3 years ago
10

I need help on these three PLEASE!!

Mathematics
2 answers:
Vitek1552 [10]3 years ago
4 0

Answer:

Hi... You should do like this

Sedaia [141]3 years ago
3 0
idek it looks difficult
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Given: ΔLMN ≅ ΔPQR Find: m
JulsSmile [24]

Answer:

what's the question .......

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3 years ago
How would I find the integral of <img src="https://tex.z-dn.net/?f=%5Cint%5Cfrac%7Btdt%7D%7Bt%5E4%2B2%7D" id="TexFormula1" title
kotegsom [21]
Let t=\sqrt y, so that t^2=y, t^4=y^2, and \mathrm dt=\dfrac{\mathrm dy}{2\sqrt y}. Then

\displaystyle\int\frac t{t^4+2}\,\mathrm dt=\int\frac{\sqrt y}{2\sqrt y(y^2+2)}\,\mathrm dy=\frac12\int\frac{\mathrm dy}{y^2+2}

Now let y=\sqrt2\tan z, so that \mathrm dy=\sqrt2\sec^2z\,\mathrm dz. Then

\displaystyle\frac12\int\frac{\mathrm dy}{y^2+2}=\frac12\int\frac{\sqrt2\sec^2z}{(\sqrt2\tan z)+2}\,\mathrm dz=\frac{\sqrt2}4\int\frac{\sec^2z}{\tan^2z+1}\,\mathrm dz=\frac1{2\sqrt2}\int\mathrm dz=\dfrac1{2\sqrt2}z+C

Transform back to y to get

\dfrac1{2\sqrt2}\arctan\left(\dfrac y{\sqrt2}\right)+C

and again to get back a result in terms of t.

\dfrac1{2\sqrt2}\arctan\left(\dfrac{t^2}{\sqrt2}\right)+C
3 0
4 years ago
Find the range of the function for the given domain, f(x)=3x+3; (-2,-1,0,1,2)
stich3 [128]

Answer:

Range: (-3,0,3,6,9)

Step-by-step explanation:

We need to evaluate each value in the domain of the function, so

f(-2)=3(-2)+3=-6+3=-3

f(-1)=3(-1)+3=-3+3=0

f(0)=3(0)+3=0+3=3

f(1)=3(1)+3=3+3=6

f(2)=3(2)+3=6+3=9

8 0
3 years ago
A circle and a square can intersect at how many points
SSSSS [86.1K]

If the circle has the same center as the diagonals of a square and the radius of the circle is smaller than 1/2 the diagonal of the square but larger than 1/2 the length of the side of a square, then there are 8 points of intersection -- 2 at each corner of the square.

If the radius of the circle is smaller than 1/2 the side length of the square and the center is as described above, there are no points of intersection.

If the circle is located outside the square it can have 1 tangent point or 2 intersection points depending on the location conditions of the circle in relation to the square.

3 0
3 years ago
ASAP NO ROCKY I NEED THIS NOOOWWWWW PLEASE AND THANK YOU
hram777 [196]

Answer:

Henry's balloon was farther from the town at the beginning and Henry's balloon traveled more quickly.

Step-by-step explanation:

The distance of Tasha's balloon from the town is represented by the function y = 8x+ 20

Where y is the distance in miles from the town and x represents the time of fly in hours.

So, at the start of the journey i.e. at x = 0, y = 20 miles {From equation (1)} from the town.

Again, Tasha's balloon is traveling at a rate of 8 miles per hour.

Now, Henry's balloon begins 30 miles from the town and is 48 miles from the town after 2 hours.

So, Henry's balloon traveling with the speed of  miles per hour.

Therefore, Henry's balloon was farther from the town at the beginning i.e. 30 miles from the town. And Henry's balloon traveled more quickly i.e at the rate of 9 miles per hour. (Answer)

happy to help:)

4 0
3 years ago
Read 2 more answers
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