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kaheart [24]
3 years ago
11

|p|<1.3? how do I find this?

Mathematics
2 answers:
wlad13 [49]3 years ago
7 0
|p|
natulia [17]3 years ago
7 0
|p|
You might be interested in
G(n)=-n^3+3n^2, find g(-4)
Sauron [17]

Answer:

g(-4) = 112

Step-by-step explanation:

Hi there!

g(n)=-n^3+3n^2

To find g(-4), replace all n's with -4:

g(-4)=-(-4)^3+3(-4)^2\\g(-4)=-(-64)+3(16)\\g(-4)=64+48\\g(-4)=112

I hope this helps!

5 0
3 years ago
Exercise is mixed —some require integration by parts, while others can be integrated by using techniques discussed in the chapte
wlad13 [49]

Answer:

\frac{2x^{\frac{7}{2}}}{7}+4x+c

Step-by-step explanation:

We are asked to find the value of the indefinite integral \int \:x^2\sqrt{x}+4\:dx.

Let us solve our given problem applying sum rule as:

\int \:x^2\sqrt{x}\:dx+\int \:4\:dx

\int \:x^2*x^{\frac{1}{2}}\:dx+\int \:4\:dx

\int \:x^{\frac{4}{2}}*x^{\frac{1}{2}}\:dx+\int \:4\:dx

\int \:x^{\frac{5}{2}}\:dx+\int \:4\:dx

Now, we will use power rule.

=\frac{x^{\frac{5}{2}+1}}{\frac{5}{2}+1}}+4x+c

=\frac{x^{\frac{7}{2}}}{\frac{7}{2}}+4x+c

=\frac{2x^{\frac{7}{2}}}{7}+4x+c

Therefore, our required integral would be \frac{2x^{\frac{7}{2}}}{7}+4x+c.  

7 0
3 years ago
Sandra calculated her taxable income as $39,250. She paid $6,000 in federal withholding tax. And you are single Your tax is If l
True [87]

Answer: 56

Step-by-step explanation:

6,000-5,944=

because 6,000 is how much she paid, but she only owed 5,944 so the the difference is what her refund is.

4 0
3 years ago
Find the area of this sector
slavikrds [6]

Answer:

1492.885

Step-by-step explanation:

the whole circle has 360°  and an area of pi*r² =pi*24²=pi* 576

for 360°  and A=pi*576

for 360-63= 297°  and A= (297*576*pi) / 360 = 1,492.885 m²

6 0
3 years ago
As the domain values approach infinity, the range values approach infinity. As the domain values approach negative infinity, the
Firdavs [7]

Limits at infinity truly are not so difficult once you've become familiarized with then, but at first, they may seem somewhat obscure. The basic premise of limits at infinity is that many functions approach a specific y-value as their independent variable becomes increasingly large or small. We're going to look at a few different functions as their independent variable approaches infinity, so start a new worksheet called 04-Limits at Infinity, then recreate the following graph.

plot(1/(x-3), x, -100, 100, randomize=False, plot_points=10001) \ .show(xmin=-10, xmax=10, ymin=-10, ymax=10) Toggle Explanation Toggle Line Numbers

In this graph, it is fairly easy to see that as x becomes increasingly large or increasingly small, the y-value of f(x) becomes very close to zero, though it never truly does equal zero. When a function's curve suggests an invisible line at a certain y-value (such as at y=0 in this graph), it is said to have a horizontal asymptote at that y-value. We can use limits to describe the behavior of the horizontal asymptote in this graph, as:

 and 

Try setting xmin as -100 and xmax as 100, and you will see that f(x) becomes very close to zero indeed when x is very large or very small. Which is what you should expect, since one divided by a large number will naturally produce a small result.

The concept of one-sided limits can be applied to the vertical asymptote in this example, since one can see that as x approaches 3 from the left, the function approaches negative infinity, and that as x approaches 3 from the right, the function approaches positive infinity, or:

 and 

Unfortunately, the behavior of functions as x approaches positive or negative infinity is not always so easy to describe. If ever you run into a case where you can't discern a function's behavior at infinity--whether a graph isn't available or isn't very clear--imagining what sort of values would be produced when ten-thousand or one-hundred thousand is substituted for x will normally give you a good indication of what the function does as x approaches infinity.

6 0
4 years ago
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