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Aneli [31]
3 years ago
8

Solve for h: p= 62.4h

Mathematics
2 answers:
krok68 [10]3 years ago
8 0
Because there are 2 variables in just one equation, we cannot get an integer for h
p=62.4h
there is only one step to this problem, and it is to divide both sides by 62.4:
h= \frac{p}{62.4}
There is your answer.
If you have any other questions about my answer, just ask me in the comments
nata0808 [166]3 years ago
3 0
P = 62.4h

Flip equation

62.h = p

Divide both sides by 62.4

62.4h/62.4 = p / 62.4

h = 0.016026<span>p</span>
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Anita plans to cover a solid cone with construction paper for a science project. The cone has a diameter of 11 inches and a slan
galben [10]

Answer:

587.18 in²

Step-by-step explanation:

In the above question, we are given the following values

Diameter = 11 inches

Radius = Diameter/2 = 11 inches/2 = 5.5 inches

Slant height = 28.5 inches.

We were asked to find how many square inches of paper will she need to cover the ENTIRE cone.

To solve for this, we would use formula for Total Surface Area of a Cone

Total Surface Area of a Cone = πrl + πr²

= πr(r + l)

Using 3.14 for π

Total Surface Area of a Cone

= 3.14 × 5.5( 5.5 + 28.5)

= 3.14 × 5.5 × (34)

= 587.18 in²

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7 0
3 years ago
A) Use the definition of Laplace transform to find L{f }. (Do the integrals.) For what values of s is L{f } defined?f(t) = (2t+1
kiruha [24]

For the given function f(t) = (2t + 1) using definition of Laplace transform the required solution is L(f(t))s = [ ( 2/s²) + ( 1/s) ].

As given in the question,

Given function is equal to :

f(t) = 2t + 1

Simplify the given function using definition of Laplace transform we have,

L(f(t))s = \int\limits^\infty_0 {f(t)e^{-st} } \, dt

          =  \int\limits^\infty_0[2t +1] e^{-st} dt

          = 2\int\limits^\infty_0 te^{-st} + \int\limits^\infty_0e^{-st} dt

         = 2 L(t) + L(1)

L(1) = \int\limits^\infty_0e^{-st} dt

     = (-1/s) ( 0 -1 )

     = 1/s , ( s >  0)

2L ( t ) = 2\int\limits^\infty_0 te^{-st}

        =  2[t\int\limits^\infty_0 e^{-st} - \int\limits^\infty_0 ({(d/dt)(t) \int\limits^\infty_0e^{-st} \, dt )dt]

        =  2/ s²

Now ,

L(f(t))s = 2 L(t) + L(1)

          = 2/ s² + 1/s

Therefore, the solution of the given function using Laplace transform the required solution is L(f(t))s = [ ( 2/s²) + ( 1/s) ].

Learn more about Laplace transform here

brainly.com/question/14487937

#SPJ4

8 0
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hodyreva [135]
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