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3 years ago

H - 20/j = k solve for j

Mathematics

1 answer:

mina [271]3 years ago

4 0

Answer: j = 20/k-h

Step-by-step explanation: Multiply to remove your variable from your denominator.

I hope this helps you out.

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3(x-4)=12x distributing or diving

frutty [35]

Distribution; 3x-12=12x

7 0

3 years ago

Given the function f(x)= e^3x, write an expression that represents the derivative of f using the limit shown below.

Nimfa-mama [501]

Answer:

f'(x)=6x+8

Step-by-step explanation:

8 0

2 years ago

The rate at which people learn about a secret is directly proportional to the number of people who know the secret. If two peopl

fenix001 [56]

Answer:

between 222 and 223 days

Step-by-step explanation:

This relation includes the point (0, 2).

A proportional relation must include the point (0, 0).

This is not proportional, but it may be linear.

Points: (0, 2), (2, 20)

y = mx + b

20 = (20 - 2)/(2 - 0) × 2 + b

20 = 18/2 × 2 + b

b = 2

y = 18x + 2

When 4000 people know, y = 4000.

y = 18x + 2

4000 = 18x + 2

3998 = 18x

x = 3998/18

x = 222.111...

Answer: between 222 and 223 days

8 0

2 years ago

17. The radius of a circular path is 7m. Find the cost of fencing the path if the cost per meter is Rs.35

Elis [28]

The cost of fencing the path is Rs 1539.38.

It is required to find the cost of fencing the path.

<h3>What is circle?</h3>

A circle is a closed two-dimensional figure in which the set of all the points in the plane is equidistant from a given point called “centre”. The perimeter around the circle is known as the circumference.

Given that:

radius of a circular path is 7m.

Rate = Rs .35 per meter

We know that

Circumference of circle = $2\pi r$

$2*\pi *7\\\\=43.98 m$

Circumference of circle=43.98 m

Now,

Cost = Circumference × Rate

Cost =43.98× 35

Cost = Rs 1539.38

Therefore, the cost of fencing the path is Rs 1539.38.

Learn more about circle here:

brainly.com/question/11833983

#SPJ1

6 0

1 year ago

5. Find the general solution to y'''-y''+4y'-4y = 0

CaHeK987 [17]

For any equation,

$a_ny^(n)+\dots+a_1y'+a_0y=0$

assume solution of a form, $e^{yt}$

Which leads to,

$(e^{yt})'''-(e^{yt})''+4(e^{yt})'-4e^{yt}=0$

Simplify to,

$e^{yt}(y^3-y^2+4y-4)=0$

Then find solutions,

$\underline{y_1=1}, \underline{y_2=2i}, \underline{y_3=-2i}$

For non repeated real root y, we have a form of,

$y_1=c_1e^t$

Following up,

For two non repeated complex roots $y_2\neq y_3$ where,

$y_2=a+bi$

and,

$y_3=a-bi$

the general solution has a form of,

$y=e^{at}(c_2\cos(bt)+c_3\sin(bt))$

Or in this case,

$y=e^0(c_2\cos(2t)+c_3\sin(2t))$

Now we just refine and get,

$\boxed{y=c_1e^t+c_2\cos(2t)+c_3\sin(2t)}$

Hope this helps.

r3t40

5 0

3 years ago