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4 years ago

How many dekameters are in 225 millimeters? Use the metric table to help answer the question.

Mathematics

1 answer:

Marina CMI [18]4 years ago

5 0

Answer:

0.0225

Step-by-step explanation:

Have a great rest of your day or night hope this helps

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Match each differential equation to a function which is a solution.

inessss [21]

Answer: 1 - C

2 - E

3 - no answer

4 - B

Step-by-step explanation:

A. $y = 3x+x^2$

$y' = 3 + 2x\\y'' = 2$

Replace in 1:

$y'' + y = 0$

$2 + 3x + x^2 \neq 0$

So, A is not an answer for 1

Replace in 2:

$y' = 3y$

$3+2x = 3(3x + x^2)$

So, A is not an answer for 2

Replace in 3

$2x^2y'' + 3xy'= y$

$2x^2(2) + 3x(3 + 2x) = 4x^2 + 9x + 6x^2 \neq 3x+x^2$

So, A is not an answer for 3

Replace in 4

$y'' + 6y' + 8y = 0$

$2 + 6(3+2x)+8(3x + x^2) = 2+18+12x+24x+8x^2 \neq 0$

So, A is not an answer for 4

B. $y = e^{-4x}$

$y' = -4e^{-4x}$

$y'' = 16e^{-4x}$

Replace in 1

$y'' + y = 0$

$16e^{-4x} -4e^{-4x} = 12e^{-4x} \neq 0$

So, B is not an answer for 1

Replace in 2

$y' = 3y$

$-4e^{-4x} \neq 3e^{-4x}$

So, B is not an answer for 2

Replace in 3

$2x^2y'' + 3xy' = y$

$2x^2(16e^{-4x}) +3x(-4e^{-4x}) \neq e^{-4x}$

So, B is not an answer for 3

Replace in 4

$y'' + 6y' +8y = 0$

$16e^{-4x} + 6(-4e^{-4x}) + 8e^{-4x} = e^{-4x}(16-24+8) = 0$

So, B is an answer for 4

C. $y = sin(x)$

$y' = cos(x)$

$y'' = -sin(x)$

Replace in 1

$y'' + y = 0$

$-sin(x) + sin(x) = 0$

So, C is an answer for 1

We jump to

D. $y = x^{12}$

$y' = 12x^{11}$

$y'' = 132 x^{10}$

Replace in 2

$y' = 3y$

$12x^{11} \neq 3x^{12}$

So, D is not an answer for 2

Replace in 3

$2x^2y'' + 3xy' = y$

$2x^2(132x^{10}) + 3x(12x^{11}) = 300x^{12} \neq x^{12}$

So, D is not an answer for 3

E. $y = 6e^{3x}$

$y' = 18e^{3x}$

$y'' = 54e^{3x}$

Replace in 2

$y' = 3y$

$18e^{3x} = 3(6e^{3x}) = 18e^{3x}$

So, E is an answer for 2

3 0

3 years ago

Sin(x)^4+ cos(x)^4=1/2

posledela

I'll assume that what was meant was $\sin ^4 x + \cos ^4 x = \dfrac{1}{2}$ .

The exponent in the funny place is just an abbreviation: $\sin ^4 x = (\sin x)^4$ .

I hope that's what you meant. Let me know if I'm wrong.

Let's start from the old saw

$\cos^2 x + \sin ^2x = 1$

Squaring both sides,

$(\cos^2 x + \sin ^2x)^2 = 1^2$

$\cos^4 x + 2 \cos ^2 x \sin ^2x +\sin ^4x = 1$

$\cos^4 x + \sin ^4x = 1 - 2 \cos ^2 x \sin ^2x$

So now the original question

$\sin ^4 x + \cos ^4 x = \dfrac{1}{2}$

becomes
$1 - 2 \cos ^2 x \sin ^2x = \dfrac{1}{2}$

$4 \cos ^2 x \sin ^2x = 1$

Now we use the sine double angle formula

$\sin 2x = 2 \sin x \cos x$

We square it to see

$\sin^2 2x = 4\sin^2 x \cos^2 x = 1$

Taking the square root,

$\sin 2x = \pm 1$

Not sure how you want it; we'll do it in degrees.

When we know the sine of an angle, there's usually two angles on the unit circle that have that sine. They're supplementary angles which add to $180^\circ$ . But when the sine is 1 or -1 like it is here, we're looking at $90^\circ$ and $-90^\circ$ , which are essentially their own supplements, slightly less messy.

That means we have two equations:

$\sin 2x = 1 = \sin 90^\circ$

$2x = 90^\circ + 360^\circ k \quad$ integer $k$

$x = 45^\circ + 180^\circ k$

or

$\sin 2x = -1 = \sin -90^\circ$

$2x = -90^\circ+ 360^\circ k$

$x = - 45^\circ + 180^\circ k$

We can combine those for a final answer,

$x = \pm 45^\circ + 180^\circ k \quad$ integer $k$

Check. Let's just check one, how about

$x=-45^\circ + 180^\circ = 135^\circ$

$\sin(135)= 1/\sqrt{2}$

$\sin ^4(135)=(1/\sqrt{2})^4 = 1/4$

$\cos ^4(135)=(-1/\sqrt{2})^4 = 1/4$

$\sin ^4(135^\circ) +\cos ^4(135^\circ) = 1/2 \quad\checkmark$

6 0

3 years ago

(02.01 LC)

horsena [70]

Answer:

B. Number the students on the school roster. Use a table of random numbers to choose 160 students from this roster for the survey.

Step-by-step explanation:

4 0

4 years ago

Given ABCE is a rectangle, D is the midpoint of CE<br> Prove AD =~ BD

Bas_tet [7]

Since the congruent operator is â‰… and since AD is congruent to BD, I'm going to assume that you want to prove that AD is congruent to BD.
1. DE is equal to CD by definition since D is the midpoint of CE.
2. AE is equal to BC since opposite sides of a rectangle are equal to each other.
3. Angle AEC is equal to Angle BCE since all angles in a rectangle are right angles and all right angles are equal to each other.
4. Triangles ADE and BDC are congruent to each other because we have SAS congruence for both triangles.
5. AD is congruent to BC since they're corresponding sides of congruent triangles.

5 0

3 years ago

Please! Help! The table below shows preferences of dancing or playing sports for male and female students:

Kryger [21]

Answer:

D) He calculated the joint relative frequency of female students who prefer playing sports. The conditional relative frequency for female students who prefer playing sports is 34%.

Step-by-step explanation:

The table is given as:

Playing sports Dancing Row totals

Male students 18 16 34

Female students 18 35 53

Column totals 36 51 87

We know that the joint relative frequency of an outcome is calculated as dividing the frequency of the outcome by the grand total.

Hence, when we divide the frequency of the female students who prefer playing sports i.e. 18 by the grand total i.e. 87 ; we obtain:

18/87=0.20689

which is approximately equal to 21%.

Hence, in order to calculate the conditional relative frequencies she should have divided the required frequency by the row total;.

Hence, here we divide the frequency of the female students who prefer playing sports i.e. 18 by the row total i.e. 53 ; we obtain:

18/53-0.3396

which is approximately equal to 34%.

Hence, option: D is correct.

5 0

3 years ago

Read 2 more answers