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3 years ago

How do you do this question?

Mathematics

1 answer:

Basile [38]3 years ago

6 0

Answer:

b=3

Step-by-step explanation:

When multiplying square roots, you multiply numbers inside the radical.

$\sqrt{3} \times \sqrt{6} = \sqrt{18}$

Next, simplify the answer, and you will get

$3 \sqrt{2}$

giving 3 as your b value.

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To get the echo of a positive integer, we write it twice in a row without a space. For example, the echo of 2022 is 20222022. Is

Lynna [10]

The echo number 20222022202220222022 is the perfect square of 4496890281.

<h3>What echo number is a perfect square</h3>

An echo number has a perfect square if its square root is also a natural number. After some iterations we found that echo number 20222022202220222022 is a perfect square:

$\sqrt{20222022202220222022} = 4496890281$

The echo number 20222022202220222022 is the perfect square of 4496890281. $\blacksquare$

To learn more on natural numbers, we kindly invite to check this verified question: brainly.com/question/17429689

8 0

3 years ago

Mr. Crawford’s 4th-Period math class I ready question

drek231 [11]

Answer:

yes

Step-by-step explanation:

yes

6 0

3 years ago

Find the Value of 2a2+5b2

Valentin [98]

Answer:

I get 92 but not really sire

3 0

3 years ago

Write down the explicit solution for each of the following: a) x’=t–sin(t); x(0)=1

Kay [80]

Answer:

a) x=(t^2)/2+cos(t), b) x=2+3e^(-2t), c) x=(1/2)sin(2t)

Step-by-step explanation:

Let's solve by separating variables:

$x'=\frac{dx}{dt}$

a) x’=t–sin(t), x(0)=1

$dx=(t-sint)dt$

Apply integral both sides:

$\int {} \, dx=\int {(t-sint)} \, dt\\\\x=\frac{t^2}{2}+cost +k$

where k is a constant due to integration. With x(0)=1, substitute:

$1=0+cos0+k\\\\1=1+k\\k=0$

Finally:

$x=\frac{t^2}{2} +cos(t)$

b) x’+2x=4; x(0)=5

$dx=(4-2x)dt\\\\\frac{dx}{4-2x}=dt \\\\\int {\frac{dx}{4-2x}}= \int {dt}\\$

Completing the integral:

$-\frac{1}{2} \int{\frac{(-2)dx}{4-2x}}= \int {dt}$

Solving the operator:

$-\frac{1}{2}ln(4-2x)=t+k$

Using algebra, it becomes explicit:

$x=2+ke^{-2t}$

With x(0)=5, substitute:

$5=2+ke^{-2(0)}=2+k(1)\\\\k=3$

Finally:

$x=2+3e^{-2t}$

c) x’’+4x=0; x(0)=0; x’(0)=1

Let $x=e^{mt}$ be the solution for the equation, then:

$x'=me^{mt}\\x''=m^{2}e^{mt}$

Substituting these equations in c)

$m^{2}e^{mt}+4(e^{mt})=0\\\\m^{2}+4=0\\\\m^{2}=-4\\\\m=2i$

This becomes the solution m=α±βi where α=0 and β=2

$x=e^{\alpha t}[Asin\beta t+Bcos\beta t]\\\\x=e^{0}[Asin((2)t)+Bcos((2)t)]\\\\x=Asin((2)t)+Bcos((2)t)$

Where A and B are constants. With x(0)=0; x’(0)=1:

$x=Asin(2t)+Bcos(2t)\\\\x'=2Acos(2t)-2Bsin(2t)\\\\0=Asin(2(0))+Bcos(2(0))\\\\0=0+B(1)\\\\B=0\\\\1=2Acos(2(0))\\\\1=2A\\\\A=\frac{1}{2}$

Finally:

$x=\frac{1}{2} sin(2t)$

7 0

3 years ago

16 4 12 36 9 27 44 11 complete the pattern

olasank [31]

16, 4, 12, 36, 9, 27, 44, 11, next is 33. 44-11=33

5 0

4 years ago