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3 years ago

If a= square a over square b=x then what stamens is true

Mathematics

1 answer:

Dahasolnce [82]3 years ago

6 0

Answer:

C I thinks

Step-by-step explanation:

If you square both sides that is what you get.

(Also can I please have Brainliest? I need it to level up!)

You might be interested in

Which of the following represents the most accurate estimation of 38 + 23?

Arte-miy333 [17]

The Correct Answer is

Ex. 38+ 23=30 + 20 +8 +3=50+8+3=58+3=61.

3 0

4 years ago

Use the method of variation of parameters to find a particular solution of the given differential equation. Then check your answ

olga_2 [115]

Answer:

Therefore the complete primitive is

$y=c_1 e^{2y}+c_2e^{3t}+e^{t}$

Therefore the general solution is

$y=c_1e^{2t}+c_2e^{3t}+e^t$

Step-by-step explanation:

Given Differential equation is

$y''-5y'+6y=2e^t$

<h3>Method of variation of parameters:</h3>

Let $y=e^{mt}$ be a trial solution.

$y'= me^{mt}$

and $y''= m^2e^{mt}$

Then the auxiliary equation is

$m^2e^{mt}-5me^{mt}+6e^{mt}=0$

$\Rightarrow m^2-5m+6=0$

$\Rightarrow m^2 -3m -2m +6=0$

$\Rightarrow m(m -3) -2(m -3)=0$

$\Rightarrow (m-3)(m-2)=0$

$\Rightarrow m=2,3$

∴The complementary function is $C_1e^{2t}+C_2e^{3t}$

To find P.I

First we show that $e^{2t}$ and $e^{3t}$ are linearly independent solution.

Let $y_1=e^{2t}$ and $y_2= e^{3t}$

The Wronskian of $y_1$ and $y_2$ is $\left|\begin{array}{cc}y_1&y_2\\y'_1&y'_2\end{array}\right|$

$=\left|\begin{array}{cc}e^{2t}&e^{3t}\\2e^{2t}&3e^{3t}\end{array}\right|$

$=e^{2t}.3e^{3t}-e^{2t}.2e^{3t}$

$=e^{5t}$ ≠ 0

∴ $y_1$ and $y_2$ are linearly independent.

Let the particular solution is

$y_p=v_1(t)e^{2t}+v_2(t)e^{3t}$

Then,

$Dy_p= 2v_1(t)e^{2t}+v'_1(t)e^{2t}+3v_2(t)e^{3t}+v'_2(t)e^{3t}$

Choose $v_1(t)$ and $v_2(t)$ such that

$v'_1(t)e^{2t}+v'_2(t)e^{3t}=0$ .......(1)

So that

$Dy_p= 2v_1(t)e^{2t}+3v_2(t)e^{3t}$

$D^2y_p= 4v_1(t)e^{2t}+9v_2(t)e^{3t}+ 2v'_1(t)e^{2t}+3v'_2(t)e^{3t}$

Now

$4v_1(t)e^{2t}+9v_2(t)e^{3t}+ 2v'_1(t)e^{2t}+3v'_2(t)e^{3t}-5[2v_1(t)e^{2t}+3v_2(t)e^{3t}] +6[v_1e^{2t}+v_2e^{3t}]=2e^t$

$\Rightarrow 2v'_1(t)e^{2t}+3v'_2(t)e^{3t}=2e^t$ .......(2)

Solving (1) and (2) we get

$v'_2=2 e^{-2t}$ and $v'_1(t)=-2e^{-t}$

Hence

$v_1(t)=\int (-2e^{-t}) dt=2e^{-t}$

and $v_2=\int 2e^{-2t}dt =-e^{-2t}$

Therefore $y_p=(2e^{-t}) e^{2t}-e^{-2t}.e^{3t}$

$=2e^t-e^t$

$=e^t$

Therefore the complete primitive is

$y=c_1 e^{2y}+c_2e^{3t}+ e^{t}$

<h3>Undermined coefficients:</h3>

∴The complementary function is $C_1e^{2t}+C_2e^{3t}$

The particular solution is $y_p=Ae^t$

Then,

$Dy_p= Ae^t$ and $D^2y_p=Ae^t$

$\therefore Ae^t-5Ae^t+6Ae^t=2e^t$

$\Rightarrow 2Ae^t=2e^t$

$\Rightarrow A=1$

$\therefore y_p=e^t$

Therefore the general solution is

$y=c_1e^{2t}+c_2e^{3t}+e^t$

4 0

3 years ago

Find the product of −35∙−1.5.

LekaFEV [45]

Answer:

If I read the question right, then the answer is -36.5.

Step-by-step explanation:

Like I said, only if I read the question right. I read it as

(-35) - 1.5.

If I'm wrong, then I'm sorry.

5 0

3 years ago

Please help click on the image below

Elena L [17]

To solve is relatively easy. All you need to do is divide both sides by 3 to get the x variable alone.

When you do that what you are left with is x = 4 - 4/3y

5 0

4 years ago

One of the authors with too much time on his hands weighed each chocolate candy in a bag of 463 plain chocolate candies. Complet

wel

Answer:

a. One of the chocolate candies weighed 0.776 gram and it was heavier than 25 of the other chocolate candies. What is the percentile of this particular value?

It is the <u>5</u>th percentile.

b. One of the chocolate candies weighed 0.876 gram and it was heavier than 329 of the other chocolate candies. What is the percentile of this particular value?

It is the <u>71</u>th percentile.

c. One of the chocolate candies weighed 0.856 gram and it was heavier than 229 of the other chocolate candies. What is the percentile of this particular value?

It is the <u>49</u>th percentile.

Step-by-step explanation:

So when we are talking about percentile, what it means is the point below a certain score or percentage lies, or the rank.

The formula is simple:

Let's call percentile pth, pth= $\frac{p}{100}*(n+1)$

where p is the position in the data set you want to evaluate (in statistical analysis, these are called quartiles, and they are useful for evaluating groups, so 3 quartiles would divide sample into 25th, 50th and 75th, four quartiles would be 20th, 40th, 60th, 80th, and so on).

and n is the number of elements in the set

For this problem there are no quartiles specified, but we can then work with the definition of percentile, meaning, if you are the 5th percentile, then 95% of the group are above you.

a. So 0.776 was heavier than 25 others, meaning it is in the 26th position out of 463. But we need that in relation to 100, so $\frac{26}{463}$ is 0.0562 or about 5/100, (rounded down). That would mean that the percentile is:

pth= $\frac{26}{463}$ *100=5.62 or 5, rounded down

b. 0.876 was heavier than 329 others, meaning it is in the 330th position out of 463. In relation to 100, so $\frac{330}{463}$ is 0.7127 or about 71/100, (rounded down). That would mean that the percentile is:

pth= $\frac{330}{463}$ *100=71.27 or 71, rounded down

c. 0.856 was heavier than 229 others, meaning it is in the 230th position out of 463. In relation to 100, so $\frac{230}{463}$ is 0.4967 or about 49/100, (rounded down). That would mean that the percentile is:

pth= $\frac{230}{463}$ *100=49.67 or 49, rounded down

3 0

4 years ago