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dlinn [17]
3 years ago
8

0.1-3=5 find mia's mistake

Mathematics
1 answer:
EleoNora [17]3 years ago
7 0

Answer:

1-3=5 does not equal what it is supposed to.

1-3=-2. So mia is completely wrong.

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One method of indirect measurement involves setting up a right triangle and measuring one of the
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Obligue I think it should be
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3 years ago
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Given: △RST ~ △VWX
tatiyna

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 = 180^{0} (sum of angles in a triangle)

Therefore by Angle-Angle-Side (AAS), △RST ~ △VWX

So that:

\frac{TU}{XY} = \frac{US}{YW} (corresponding side length proportion)

4 0
3 years ago
The product of two numbers is 161 , one of number is 7,find the other number. ​
Diano4ka-milaya [45]

let the other number be " n "

now, according to above condition :

  • n × 7 = 161

  • n = 161 ÷ 7

  • n = 23

therefore, the other number would be 23.

4 0
3 years ago
In which case would it be necessary to find the area of this circle
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You need to know the area if you're going to
  1st selection: cover the surface of the dartboard.
7 0
3 years ago
ПОЖАЛУЙСТА ПОМОГИТЕ РЕШИТЬ ДАЮ 23 ОЧКА!!!
posledela

(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

\displaystyle 2\pi \int_0^2 y \left(3-\frac34 y^2\right) \,\mathrm dy = 2\pi \left(\frac32 y^2 - \frac3{16} y^4\right)\bigg|_0^2 = 6\pi

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

\displaystyle \pi \int_0^3 \left(\sqrt{\frac43x}\right)^2 \,\mathrm dx = \frac{2\pi}3 x^2\bigg|_0^3 = 6\pi

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...

8 0
3 years ago
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