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PIT_PIT [208]
3 years ago
9

Drag the tiles to the boxes to form correct pairs. Not all tiles will be used. Match the systems of linear equations with their

solutions. Tiles x + y = -1 -6x + 2y = 14 infinite solutions x − 2y = -5 5x + 3y = 27 (-1, -6) -4x + y = -9 5x + 2y = 3 (-2, 1) 6x + 3y = -6 2x + y = -2 (3, 4) -x + 2y = 4 -3x + 6y = 11 no solution -7x + y = 1 14x − 7y = 28

Mathematics
1 answer:
puteri [66]3 years ago
3 1
Hello,
Please, see the detailed solution in the attached files.
Thanks.

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A computer program contains one error. In order to find the error, we split the program into 6 blocks and test two of them, sele
Aleksandr-060686 [28]
There are \dbinom62 ways of selecting two of the six blocks at random. The probability that one of them contains an error is

\dfrac{\dbinom11\dbinom51}{\dbinom62}=\dfrac5{15}=\dfrac13

So X has probability mass function

f_X(x)=\begin{cases}\dfrac13&\text{for }x=1\\\\\dfrac23&\text{for }x=0\end{cases}

These are the only two cases since there is only one error known to exist in the code; any two blocks of code chosen at random must either contain the error or not.

The expected value of finding an error is then

\displaystyle\sum_{x=0}^1xf_X(x)=0\times\dfrac23+1\times\dfrac13=\dfrac13
7 0
3 years ago
Solve.
baherus [9]
The correct answer is:  [C]:  " 5 " .
__________________________________________________________
                                →  " a = 5 " .
__________________________________________________________
Explanation:
__________________________________________________________ 

Given:  " a + 1 <span>− 2 = 4 " ; Solve for "a" ; 

4 + 2 = 6 ; 

6 </span>− 1 = 5 ;   →  a = 5 ;

To check our work:

5 + 1 − 2 = ? 4 ?? ;  

5 + 1 = 6 ; 

6 − 2 = 4.  Yes!

So the answer is:  [C]:  " 5 ". 
_________________________________________________________
                          →   " a = 5 " . 
_________________________________________________________
4 0
3 years ago
What operation is being done to the variable in the equation -5m = -40?
Jlenok [28]

Answer:

it's being divided by -5

Step-by-step explanation:

-5m= -40

m=8

8 0
3 years ago
Does anyone know the answer?
GarryVolchara [31]
You multiply the constants 3 * 2 = 6
Add you add the exponents 6 + 1/2 = 13/2

So the answer is 6x^(13/2)
(D)
3 0
3 years ago
Read 2 more answers
Distance between two ships At noon, ship A was 12 nautical miles due north of ship B. Ship A was sailing south at 12 knots (naut
frozen [14]

Answer:

a)\sqrt{144-288t+208t^2} b.) -12knots, 8 knots c) No e)4\sqrt{13}

Step-by-step explanation:

We know that the initial distance between ships A and B was 12 nautical miles. Ship A moves at 12 knots(nautical miles per hour) south. Ship B moves at 8 knots east.

a)

We know that at time t , the ship A has moved 12\dot t (n.m) and ship B has moved 8\dot t (n.m). We also know that the ship A moves closer to the line of the movement of B and that ship B moves further on its line.

Using Pythagorean theorem, we can write the distance s as:

\sqrt{(12-12\dot t)^2 + (8\dot t)^2}\\s=\sqrt{144-288t+144t^2+64t^2}\\s=\sqrt{144-288t+208t^2}

b)

We want to find \frac{ds}{dt} for t=0 and t=1

\sqrt{144-288t+208t^2}|\frac{d}{dt}\\\\\frac{ds}{dt}=\frac{1}{2\sqrt{144-288t+208t^2}}\dot (-288+416t)\\\\\frac{ds}{dt}=\frac{208t-144}{\sqrt{144-288t+208t^2}}\\\\\frac{ds}{dt}(0)=\frac{208\dot 0-144}{\sqrt{144-288\dot 0 + 209\dot 0^2}}=-12knots\\\\\frac{ds}{dt}(1)=\frac{208\dot 1-144}{\sqrt{144-288\dot 1 + 209\dot 1^2}}=8knots

c)

We know that the visibility was 5n.m. We want to see whether the distance s was under 5 miles at any point.

Ships have seen each other = s\leq 5\\\\\sqrt{144-288t+208t^2}\leq 5\\\\144-288t+208t^2\leq 25\\\\199-288t+208t^2\leq 0

Since function f(x)=199-288x+208x^2 is quadratic, concave up and has no real roots, we know that 199-288x+208x^2>0 for every t. So, the ships haven't seen each other.

d)

Attachedis the graph of s(red) and ds/dt(blue). We can see that our results from parts b and c were correct.

e)

Function ds/dt has a horizontal asympote in the first quadrant if

                                                \lim_{t \to \infty} \frac{ds}{dt}

So, lets check this limit:

\lim_{t \to \infty} \frac{ds}{dt}=\lim_{t \to \infty} \frac{208t-144}{\sqrt{144-288t+208t^2}}\\\\=\lim_{t \to \infty} \frac{208-\frac{144}{t}}{\sqrt{\frac{144}{t^2}-\frac{288}{t}+208}}\\\\=\frac{208-0}{\sqrt{0-0+208}}\\\\=\frac{208}{\sqrt{208}}\\\\=4\sqrt{13}

Notice that:

4\sqrt{13}=\sqrt{12^2+5^2}=√(speed of ship A² + speed of ship B²)

5 0
3 years ago
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