Answer:
Given
Step-by-step explanation:
Given that: △RST ~ △VWX, TU is the altitude of △RST, and XY is the altitude of △VWX.
Comparing △RST and △VWX;
TU ~ XY (given altitudes of the triangles)
<TUS = <XYW (all right angles are congruent)
<UTS ≅ <YXW (angle property of similar triangles)
Thus;
ΔTUS ≅ ΔXYW (congruent property of similar triangles)
<UTS + <TUS + < UST = <YXW + <XTW + <XWY =
(sum of angles in a triangle)
Therefore by Angle-Angle-Side (AAS), △RST ~ △VWX
So that:
=
(corresponding side length proportion)
let the other number be " n "
now, according to above condition :
therefore, the other number would be 23.
You need to know the area if you're going to
1st selection: cover the surface of the dartboard.
(Простите, пожалуйста, мой английский. Русский не мой родной язык. Надеюсь, у вас есть способ перевести это решение. Если нет, возможно, прилагаемое изображение объяснит достаточно.)
Use the shell method. Each shell has a height of 3 - 3/4 <em>y</em> ², radius <em>y</em>, and thickness ∆<em>y</em>, thus contributing an area of 2<em>π</em> <em>y</em> (3 - 3/4 <em>y</em> ²). The total volume of the solid is going to be the sum of infinitely many such shells with 0 ≤ <em>y</em> ≤ 2, thus given by the integral

Or use the disk method. (In the attachment, assume the height is very small.) Each disk has a radius of √(4/3 <em>x</em>), thus contributing an area of <em>π</em> (√(4/3 <em>x</em>))² = 4<em>π</em>/3 <em>x</em>. The total volume of the solid is the sum of infinitely many such disks with 0 ≤ <em>x</em> ≤ 3, or by the integral

Using either method, the volume is 6<em>π</em> ≈ 18,85. I do not know why your textbook gives a solution of 90,43. Perhaps I've misunderstood what it is you're supposed to calculate? On the other hand, textbooks are known to have typographical errors from time to time...