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3 years ago

12

Help needed asap please

2 answers:

Cerrena [4.2K]3 years ago

7 0

No because thier ain’t enough

Alenkasestr [34]3 years ago

6 0

Answer:

I don't understand the language you used

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Pls help me solve this problem!!

ddd [48]

Answer:

49.85

Step-by-step explanation:

41 is the mean, so half of the daily requests are above that number, half below, the question is about numbers above the mean, so the lower numbers won't be included in the percentage.

The difference between 59 and 41 is 18. The standard deviation given is 6.

18 ÷ 6 = 3 So that is 3 standard deviations. The empirical rule states that 99.7 of all values are within 3 standard deviations from the mean. But we are looking at only the upper half of those values so 99.7 ÷ 2 = 49.85 %

If the answer asks for approximate, you could round to 49.9 or 50.

And tell the math people to learn the correct spelling of fluorescent!

4 0

3 years ago

A particle moving in a planar force field has a position vector x that satisfies x'=Ax. The 2×2 matrix A has eigenvalues 4 and 2

andrey2020 [161]

Answer:

The required position of the particle at time t is: $x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}$

Step-by-step explanation:

Consider the provided matrix.

$v_1=\begin{bmatrix}-3\\1 \end{bmatrix}$

$v_2=\begin{bmatrix}-1\\1 \end{bmatrix}$

$\lambda_1=4, \lambda_2=2$

The general solution of the equation $x'=Ax$

$x(t)=c_1v_1e^{\lambda_1t}+c_2v_2e^{\lambda_2t}$

Substitute the respective values we get:

$x(t)=c_1\begin{bmatrix}-3\\1 \end{bmatrix}e^{4t}+c_2\begin{bmatrix}-1\\1 \end{bmatrix}e^{2t}$

$x(t)=\begin{bmatrix}-3c_1e^{4t}-c_2e^{2t}\\c_1e^{4t}+c_2e^{2t} \end{bmatrix}$

Substitute initial condition $x(0)=\begin{bmatrix}-6\\1 \end{bmatrix}$

$\begin{bmatrix}-3c_1-c_2\\c_1+c_2 \end{bmatrix}=\begin{bmatrix}-6\\1 \end{bmatrix}$

Reduce matrix to reduced row echelon form.

$\begin{bmatrix} 1& 0 & \frac{5}{2}\\ 0& 1 & \frac{-3}{2}\end{bmatrix}$

Therefore, $c_1=2.5,c_2=1.5$

Thus, the general solution of the equation $x'=Ax$

$x(t)=2.5\begin{bmatrix}-3\\1\end{bmatrix}e^{4t}-1.5\begin{bmatrix}-1\\1 \end{bmatrix}e^{2t}$

$x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}$

The required position of the particle at time t is: $x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}$

6 0

3 years ago

Determine the common ratio r of a geometric sequence with first term 5 and<br> fourth term -40.

viva [34]

Answer:

Step-by-step explanation:

8 0

3 years ago

Solve. -7x=x^2+12 help

Nat2105 [25]

Hey There!

The answer to your problem is $x=-3$ or $x=-4$ both answers are correct!

<u>Step 1: Subtract x²+12 from both sides.</u>

−7x−(x²+12)=x²+12−(x²+12)

−x²−7x−12=0

<u>Step 2: Factor left side of equation.</u>

$(-x-3)(x+4)=0$

<u>Step 3: Set factors equal to 0.</u>

$-x-3=0$ or $x+4=0$

<u>Get your answer:</u>

$x=-3$ or $x=-4$

3 0

3 years ago

Please take a look at the picture

Debora [2.8K]

$\tiny \text{yes \: looked}$

7 0

3 years ago

Read 2 more answers