Choose the graph which represents the solution to the inequality: 1 2 x + 8 ≤ 3 A) B) C) D)

2.50+ 1.50(m-1)=17.50

neonofarm [45]

Answer: m = 11

Step-by-step explanation: Isolate the variable by dividing each side by facotrs that don't contain the variable.

Hope this helps you out! ☺

4 0

3 years ago

Read 2 more answers

A metronome uses an inverted pendulum rod to control tempo. A weight can be slid up the pendulum rod to decrease tempo, or down

Dominik [7]

Answer:

Part 1

Multiply both sides by 2π

2πf = √(g/L)

Square both sides

4π²f²= g/L

Invert both sides

1/(4π²f²) = L/g

Multiply both side by g

g/(4π²f²) = L

We usually write an equation with the subject (L) n the left

L = g/(4π²f²)

________________________

Part 2

Using the above equation with the given values:

L = g/(4π²f²)

. .= 9.8 / (4π² x 1.6²)

. .= 0.097m (= 9.7cm)

________________________

By the way, where it says

“f=1.6 then there are 2 beats a second, or 192 beats per minute (bpm).”

this should say

“f=1.6 then there are 3.2 beats a second, or 192 beats per minute (bpm)”

Step-by-step explanation:

7 0

3 years ago

Help Me Number 6 or if you can help me with number 7 if you want too ☺️

Makovka662 [10]

Simply use the formula they give you. For question 6, 4 is the radius. So the area of the circle is 4pi. Now for seven you do the same thing, but they gave you the diameter instead and it’s a half circle. So try that one on your own. Let me know if you get stuck

7 0

3 years ago

Use theorem 7. 4. 2 to evaluate the given laplace transform. do not evaluate the convolution integral before transforming. (writ

irga5000 [103]

With convolution theorem the equation is proved.

According to the statement

we have given that the equation and we have to evaluate with the convolution theorem.

Then for this purpose, we know that the

A convolution integral is an integral that expresses the amount of overlap of one function as it is shifted over another function.

And the given equation is solved with this given integral.

So, According to this theorem the equation becomes the

$\mathscr{L} \left( \int_{0}^{t} e^{-\tau} \cos \tau d \tau \right) = \frac{ \mathscr{L} (e^{-\tau} \cos \tau ) }{s} \\\mathscr{L} \left( \int_{0}^{t} e^{-\tau} \cos \tau d \tau \right) = \frac{\frac{s+1}{(s+1)^2+1}}{s} \\\mathscr{L} \left( \int_{0}^{t} e^{-\tau} \cos \tau d \tau \right) = \frac{1}{s}\left (\frac{s+1}{(s+1)^2+1} \right).$