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3 years ago

12−(−2) over a line 12−6÷2

Mathematics

1 answer:

joja [24]3 years ago

7 0

Given 12−(−2) over a line 12−6÷2

We can use PEMDAS rule and solve the expression in the order as given below:-

1. Parentheses

2. Exponents

3. Multiplication

4. Division

5. Addition

6. Subtraction

Now solving for given expression:-

$\frac{12-(-2)}{12-(\frac{6}{2})}\\\\=\frac{12+2}{12-3}\\\\=\frac{14}{9}$

So final answer is 14/9 i.e. 14 over 9.

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There are 50 chairs in the auditorium. 15 of the chairs are red and the rest are blue. What is the ratio of the number of blue c

navik [9.2K]

Answer: 7:3

Step-by-step explanation:

15 chairs are red

35 are blue

35:15

Simplifies to

7:3

7 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

How do you add mixed fractions?

ycow [4]

If u have 4\3 and 4\5 u add 4+4=8 and then do 5+3=8 and now you have 8\8 and that is how to add ixed fractions

3 0

3 years ago

Yahoo creates a test to classify emails as spam or not spam based on the contained words. This test accurately identifies spam (

Amiraneli [1.4K]

Answer and Step-by-step explanation:

The computation is shown below:

Let us assume that

Spam Email be S

And, test spam positive be T

Given that

P(S) = 0.3

$P(\frac{T}{S}) = 0.95$

$P(\frac{T}{S^c}) = 0.05$

Now based on the above information, the probabilities are as follows

i. P(Spam Email) is

= P(S)

= 0.3

$P(S^c) = 1 - P(S)$

= 1 - 0.3

= 0.7

ii. $P(\frac{S}{T}) = \frac{P(S\cap\ T}{P(T)}$

$= \frac{P(\frac{T}{S}) . P(S) }{P(\frac{T}{S}) . P(S) + P(\frac{T}{S^c}) . P(S^c) }$

$= \frac{0.95 \times 0.3}{0.95 \times 0.3 + 0.05 \times 0.7}$

= 0.8906

iii. $P(\frac{S}{T^c}) = \frac{P(S\cap\ T^c}{P(T^c)}$

$= \frac{P(\frac{T^c}{S}) . P(S) }{P(\frac{T^c}{S}) . P(S) + P(\frac{T^c}{S^c}) . P(S^c) }$

$= \frac{(1 - 0.95)\times 0.3}{ (1 -0.95)0.95 \times 0.3 + (1 - 0.05) \times 0.7}$

= 0.0221

We simply applied the above formulas so that the each part could come

8 0

3 years ago

7 tenths times 10 is another problem that i need help on

Elza [17]

Hi there!

7 tenths times 10 is pretty easy and can be done on a calculator.
7 tenths is 0.7 in standard form.

0.7 × 10 = 7

Because if you think about it...
Tenths is like being dropped down by ten... then you multiply by ten which brings it back up to whole number seven. It may not make sense but that's what it is.

Hope this helps!

8 0

3 years ago