Elinas friend Luke said that he could make mor rectangules with 24 tiles than with Elinas 10 tiles do you agree with Luke

Leno4ka [110]

My guess would be A ! Hope this helps

5 0

3 years ago

Find, correct to four decimal places, the length of the curve of intersection of the cylinder 4x2 1 y2 − 4 and the plane x 1 y 1

charle [14.2K]

Answer- Length of the curve of intersection is 13.5191 sq.units

Solution-

As the equation of the cylinder is in rectangular for, so we have to convert it into parametric form with

x = cos t, y = 2 sin t (∵ 4x² + y² = 4 ⇒ 4cos²t + 4sin²t = 4, then it will satisfy the equation)

Then, substituting these values in the plane equation to get the z parameter,

cos t + 2sin t + z = 2

⇒ z = 2 - cos t - 2sin t

∴ $\frac{dx}{dt} = -\sin t$

$\frac{dy}{dt} = 2 \cos t$

$\frac{dz}{dt} = \sin t-2cos t$

As it is a full revolution around the original cylinder is from 0 to 2π, so we have to integrate from 0 to 2π

∴ Arc length

$= \int_{0}^{2\pi}\sqrt{(\frac{dx}{dt})^{2}+(\frac{dy}{dt})^{2}+(\frac{dz}{dt})^{2}$

$=\int_{0}^{2\pi}\sqrt{(-\sin t)^{2}+(2\cos t)^{2}+(\sin t-2\cos t)^{2}$

$=\int_{0}^{2\pi}\sqrt{(2\sin t)^{2}+(8\cos t)^{2}-(4\sin t\cos t)$

Now evaluating the integral using calculator,

$=\int_{0}^{2\pi}\sqrt{(2\sin t)^{2}+(8\cos t)^{2}-(4\sin t\cos t)$ $= 13.5191$

8 0

3 years ago

The Blue devils had won 37.5 games. they had played 24 games altogether. how may games had they won?

lubasha [3.4K]

The Blue Devils are trying to sell us a bridge and a bill of goods.
But we are sharp, and we are not falling for their story.

We are smart enough to spot two glaring weaknesses in their story.

1). There is no way to win .5 of a game.

2). There is no way to win 37 games if you've
only played 24 so far.

4 0

3 years ago

Question 1

vivado [14]

Answer:

For Lin's answer

Step-by-step explanation:

When you have a triangle, you can flip it along a side and join that side with the original triangle, so in this case the triangle has been flipped along the longest side and that longest side is now common in both triangles. Now since these are the same triangle the area remains the same.

Now the two triangles form a quadrilateral, which we can prove is a parallelogram by finding out that the opposite sides of the parallelogram are equal since the two triangles are the same(congruent), and they are also parallel as the alternate interior angles of quadrilateral are the same. So the quadrilaral is a paralllelogram, therefore the area of a parallelogram is bh which id 7 * 4 = 7*2=28 sq units.

Since we already established that the triangles in the parallelogram are the same, therefore their areas are also the same, and that the area of the parallelogram is 28 sq units, we can say that A(Q)+A(Q)=28 sq units, therefore 2A(Q)=28 sq units, therefore A(Q)=14 sq units, where A(Q), is the area of triangle Q.

7 0

3 years ago

Read 2 more answers

What is the answer to the problem <img src="https://tex.z-dn.net/?f=%28x%20%7B3%7D%29%5E%7B3%7D%20%3D%20x%5En" id="TexFo

11Alexandr11 [23.1K]

Solve for x:
x^9 = n x

Subtract n x from both sides:
x^9 - n x = 0

Factor x and constant terms from the left hand side:
-x (n - x^8) = 0

Multiply both sides by -1:
x (n - x^8) = 0

Split into two equations:
x = 0 or n - x^8 = 0

Subtract n from both sides:
x = 0 or -x^8 = -n

Multiply both sides by -1:
x = 0 or x^8 = n

Taking 8^th roots gives n^(1/8) times the 8^th roots of unity:
Answer: x = 0 or x = -n^(1/8) or x = -i n^(1/8) or x = i n^(1/8) or x = n^(1/8) or x = -(-1)^(1/4) n^(1/8) or x = (-1)^(1/4) n^(1/8) or x = -(-1)^(3/4) n^(1/8) or x = (-1)^(3/4) n^(1/8)

3 0

3 years ago