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OverLord2011 [107]
3 years ago
13

Hey if anyone could explain this to me that would be awesome

Mathematics
2 answers:
Ne4ueva [31]3 years ago
6 0
The answer to this question is (1,-4). An ordered pair is of the form (x,y), where x is the input into the equation and y is the output. It is asking you what is the value of y, which you can find using the given equation.
krok68 [10]3 years ago
3 0
3 times 1 is 3. Take 3 minus 7 equals -4.

The answer is (1 , -4) 
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If the order of three loci is A B C, and the map distance between A and B is 15 m.u., and the map distance between B and C is 20
zysi [14]

Answer:

Therefore, we conclute that the map distance between A and C is 35 m.u..

Step-by-step explanation:

We know that the order of three loci is A B C, and the map distance between A and B is 15 m.u., and the map distance between B and C is 20 m.u. We calculate the map distance between A and C.

Therefore, we get

AC=AB+BC\\\\AC=15+20\\\\AC=35\\

Therefore, we conclute that the map distance between A and C is 35 m.u..

4 0
3 years ago
Find the distance between the points.<br><br> (5,-2),(-6,-2)
damaskus [11]
As they are both on the same y axis, (-2), that means you just need to find the difference between 5 and - 6 which is 11. Therefore the answer is 11 units. Hope this helps! :)
8 0
3 years ago
If an object is launched at an angle of 28° with an initial velocity of 133 ft./s from a height of 6 feet how many seconds will
Tema [17]

Answer:

Hence after  3.98 sec  i.e  4 sec Object will hit the ground  .

Step-by-step explanation:

Given:

Height= 6 feet

Angle =28 degrees.

V=133 ft/sec

To Find:

Time in seconds after which it will hit the ground?

Solution:

<em>This problem is related to projectile motion for objec</em>t

First calculate the Range for object  and it is given by ,

R=v^2Sin(2Ф)/g

Here R= range  g= acceleration due to gravity =9.8 m/sec^2

1m =3.2 feet

So 9.8 m, equals to 9.8 *3.2=31.36 ft

So g=31.36 ft/sec^2. and 2Ф=2(28)=56

R=133^2*Sin(56)/31.36

R=14664.84/31.36

R=467.62  fts

Now using Formula for time and range as

R=VxT

Vx is horizontal velocity

Vx=V*cosФ

Vx=133*cos(28)

Vx=117.43  ft/sec

So above equation becomes as ,

467.62=117.43*T

T=467.62/117.43

T=3.98 sec

T is approximately equals to 4 sec.

4 0
3 years ago
Suppose that the length of a side of a cube X is uniformly distributed in the interval 9
Nastasia [14]

Answer:

f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.

Step-by-step explanation:

Given

9 < x < 10 --- interval

Required

The probability density of the volume of the cube

The volume of a cube is:

v = x^3

For a uniform distribution, we have:

x \to U(a,b)

and

f(x) = \left \{ {{\frac{1}{b-a}\ a \le x \le b} \atop {0\ elsewhere}} \right.

9 < x < 10 implies that:

(a,b) = (9,10)

So, we have:

f(x) = \left \{ {{\frac{1}{10-9}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.

Solve

f(x) = \left \{ {{\frac{1}{1}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.

f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.

Recall that:

v = x^3

Make x the subject

x = v^\frac{1}{3}

So, the cumulative density is:

F(x) = P(x < v^\frac{1}{3})

f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right. becomes

f(x) = \left \{ {{1\ 9 \le x \le v^\frac{1}{3} - 9} \atop {0\ elsewhere}} \right.

The CDF is:

F(x) = \int\limits^{v^\frac{1}{3}}_9 1\  dx

Integrate

F(x) = [v]\limits^{v^\frac{1}{3}}_9

Expand

F(x) = v^\frac{1}{3} - 9

The density function of the volume F(v) is:

F(v) = F'(x)

Differentiate F(x) to give:

F(x) = v^\frac{1}{3} - 9

F'(x) = \frac{1}{3}v^{\frac{1}{3}-1}

F'(x) = \frac{1}{3}v^{-\frac{2}{3}}

F(v) = \frac{1}{3}v^{-\frac{2}{3}}

So:

f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.

8 0
2 years ago
7) Will you ever get 100% red M&amp;M's in a packet? Explain your answer.
andreev551 [17]
No. A machine dispenses all the colors on a conveyor belt that mixes all the available colors.
8 0
2 years ago
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