Is 26 a perfect square? Why or why not?

2 answers:

7 0

No it is not because it can't be expressed as the product of 2 integers but instead would be expressed by the product of 2 numbers between 5 and 6, closer to 5.

Hope this helps :)

VladimirAG [237]3 years ago

6 0

No, because square root of 26 is 5.099

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I want the answer of question 27 with explanation

m_a_m_a [10]

Answer:

6√2

Step-by-step explanation:

√32+√18-1√4

√16×2+√9×2-1√2

4√2+3√2-1√2

6√2

7 0

3 years ago

Find area and circumference of the circle (12.) pls NO links

xxTIMURxx [149]

Answer:

The area is 452.39

The Circumference is 75.36

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Please answer this correctly without making mistakes

IrinaVladis [17]

Answer:

44 5/6 "the second one"

5 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 y ², radius y, and thickness ∆y, thus contributing an area of 2π y (3 - 3/4 y ²). The total volume of the solid is going to be the sum of infinitely many such shells with 0 ≤ y ≤ 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 x), thus contributing an area of π (√(4/3 x))² = 4π/3 x. The total volume of the solid is the sum of infinitely many such disks with 0 ≤ x ≤ 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 6π ≈ 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...