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2 years ago

Nipa exercises for 225 minutes a week. How many hours does she exercise in a year?

Mathematics

1 answer:

Lena [83]2 years ago

7 0

About 12,000 minutes a year.

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The height of a baseball hit from a bat is modeled by a quadratic function of time, h(t). What does the second coordinate of the

Brut [27]

The maximum height of the baseball. Hope this helped :)

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2 years ago

What is the value of the expression 5ab-2c, where a =2, b=3 and c=0?

ira [324]

Answer:

Step-by-step explanation:

5*2*3 - 2*0 = 30 - 0 = 30

5 0

2 years ago

Review the graph of function f(x).

s344n2d4d5 [400]

The two limits when x tends to zero are:

$\lim_{x \to \ 0^-} f(x) = 1\\\\ \lim_{x \to \ 0^+} f(x) = 0$

<h3 /><h3>How to get the limits when x tends to zero?</h3>

Notice that we have a jump at x = 0.

Then we can take two limits, one going from the negative side (where we will go along the blue line)

And other from the positive side (where we go along the orange line).

We will get:

$\lim_{x \to \ 0^-} f(x) = 1\\\\ \lim_{x \to \ 0^+} f(x) = 0$

Notice that the two limits are different, that means that the function is not a continuous function.

If you want to learn more about limits:

brainly.com/question/5313449

#SPJ1

6 0

2 years ago

What gives the range of the function y=|x-1|+7

riadik2000 [5.3K]

The rage of the function is x=-6

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3 years ago

The Laplace Transform of a function f(t), which is defined for all t > 0, is denoted by L{f(t)} and is defined by the imprope

lesya692 [45]

(1) D

$L_s\left\{t\right\} = \displaystyle\int_0^\infty te^{-st}\,\mathrm dt$

Integrate by parts, taking

$u = t \implies \mathrm du=\mathrm dt$

$\mathrm dv = e^{-st}\,\mathrm dt \implies v=-\dfrac1se^{-st}$

Then

$L_s\left\{t\right\} = \displaystyle \left[-\frac1ste^{-st}\right]\bigg|_{t=0}^{t\to\infty}+\frac1s\int_0^\infty e^{-st}\,\mathrm dt$

$L_s\left\{t\right\} = \displaystyle \frac1s\int_0^\infty e^{-st}\,\mathrm dt$

$L_s\left\{t\right\} = \displaystyle -\frac1{s^2}e^{-st}\bigg|_{t=0}^{t\to\infty}$

$L_s\left\{t\right\} = \displaystyle \boxed{\frac1{s^2}}$

(2) A

$L_s\left\{1\right\} = \displaystyle\int_0^\infty e^{-st}\,\mathrm dt$

$L_s\left\{1\right\} = \displaystyle\left[-\frac1se^{-st}\right]\bigg|_{t=0}^{t\to\infty}$

$L_s\left\{1\right\} = \displaystyle\boxed{\frac1s}$

7 0

3 years ago