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AlladinOne [14]

3 years ago

10

During track practice, Jonathan ran 5 kilometers. Rob 4,250 meters. Who ran further?

2 answers:

vagabundo [1.1K]3 years ago

7 0

Answer:

Johnathan

Step-by-step explanation:

motikmotik3 years ago

4 0

Answer: The answer is: Jonathan ran further.

Step-by-step explanation: First, 1 kilometer (km )= 1,000 meters (m)

Then,multiply 5 times 1,000 because Jonathan ran 5 kilometers and were using meters. (We’re converting. )The answer for that is 5,000 meters. Remember to change kilometers to meters! This is greater than of that of Rob did. ( 4,250 meters. ) So,we know that Jonathan ran further then Rob.

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About how many students do not prefer non- fiction?

slamgirl [31]

Answer:

B. 75%

Step-by-step explanation:

If about a quarter ( 25% ) do like non-fiction, then we know that the rest do not prefer it ( 75% )

75 + 25 = 100

8 0

2 years ago

In a class there are 15 boys and 15 girls 5/6 of the children have brown hair if someone is chosen at random what is the probabi

Cloud [144]

Probability someone chosen at random will not have brown hair = 1/6

<h3>Probability of two variables</h3>

The number of girls = 15

The number of boys = 15

The total number of children = 15 + 15 = 30

Probability that someone chosen at random will have a brown hair, P(Brown hair) = 5/6

P(Brown hair) + P(Not Brown hair) = 1

5/6 + P(Not brown hair) = 1

P(not brown hair) = 1 - 5/6

P(not brown hair) = 1/6

Therefore, if someone is chosen at random, probability that they will not have brown hair = 1/6

Learn more on probability here: brainly.com/question/27899440

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2 years ago

Which is the value of this expression when a = 5 and k = negative 2?

Brums [2.3K]

Answer:

what is the expression? please provide an expression.

4 0

3 years ago

Write the given differential equation in the form L(y) = g(x), where L is a linear differential operator with constant coefficie

melamori03 [73]

Answer:

The complete solution is

$\therefore y= Ae^{3x}+Be^{-\frac13 x}-\frac43$

Step-by-step explanation:

Given differential equation is

3y"- 8y' - 3y =4

The trial solution is

$y = e^{mx}$

Differentiating with respect to x

$y'= me^{mx}$

Again differentiating with respect to x

$y''= m ^2 e^{mx}$

Putting the value of y, y' and y'' in left side of the differential equation

$3m^2e^{mx}-8m e^{mx}- 3e^{mx}=0$

$\Rightarrow 3m^2-8m-3=0$

The auxiliary equation is

$3m^2-8m-3=0$

$\Rightarrow 3m^2 -9m+m-3m=0$

$\Rightarrow 3m(m-3)+1(m-3)=0$

$\Rightarrow (3m+1)(m-3)=0$

$\Rightarrow m = 3, -\frac13$

The complementary function is

$y= Ae^{3x}+Be^{-\frac13 x}$

y''= D², y' = D

The given differential equation is

(3D²-8D-3D)y =4

⇒(3D+1)(D-3)y =4

Since the linear operation is

L(D) ≡ (3D+1)(D-3)

For particular integral

$y_p=\frac 1{(3D+1)(D-3)} .4$

$=4.\frac 1{(3D+1)(D-3)} .e^{0.x}$ [since $e^{0.x}=1$ ]

$=4\frac{1}{(3.0+1)(0-3)}$ [ replace D by 0 , since L(0)≠0]

$=-\frac43$

The complete solution is

y= C.F+P.I

$\therefore y= Ae^{3x}+Be^{-\frac13 x}-\frac43$

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3 years ago

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The Answer to your question is A

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3 years ago