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1 year ago

A. P is equal to the set containing t,v,c and d

Mathematics

1 answer:

Angelina_Jolie [31]1 year ago

3 0

The answers to the questions are:

a. p = { t, v, c, q}

b. {2, 6} ⊂ {2,6,8,12}

c. {0, 3} ⊄ {3,2,4,6}

<h3>How to write the mathematical expression s using mathematical symbols.</h3>

To represent sets we use

p = { }

Hence

a. p = { t, v, c, q}

b. If 2 and 6 is a proper subset of [ 2,6,8,12},

This can be represented as

{2, 6} ⊂ {2,6,8,12}

c. the set consisting of the elements 0 and 3 is not a subset of {3,2,4,6} is written as

{0, 3} ⊄ {3,2,4,6}

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Find the midpoint of the segment with the given endpoints (5,-9) and (-2,-2)

QveST [7]

Another way to solve this is to use the Midpoint Formula. The midpoint of a segment joining points $(x_1,y_1)$ and $(x_2,y_2)$ is

$\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2} \right)$

So the midpoint of your segment is

$\left(\frac{5+(-2)}{2},\frac{-9+(-2)}{2}\right) = \left(\frac{3}{2},-\frac{11}{2} \right)$

Perhaps it helps to see that the x-coordinate of the midpoint is just the average of the x-coordinates of the points. Ditto for the y-coordinate of the midpoint; just average the y's.

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2 years ago

Use the algebraic procedure explained in section 8.9 in your book to find the derivative of f(x)=1/x. Use h for the small number

Triss [41]

Answer:

By definition, the derivative of f(x) is

$lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

Let's use the definition for $f(x)=\frac{1}{x}$

$lim_{h\rightarrow 0} \frac{\frac{1}{x+h}-\frac{1}{x}}{h}=\\lim_{h\rightarrow 0} \frac{\frac{x-(x+h)}{x(x+h)}}{h}=\\lim_{h\rightarrow 0} \frac{\frac{(-1)h}{x^2+xh}}{h}=\\lim_{h\rightarrow 0} \frac{(-1)h}{h(x^2+xh)}=\\lim_{h\rightarrow 0} \frac{-1}{x^2+xh)}=\frac{-1}{x^2+x*0}=\frac{-1}{x^2}$

Then, $f'(x)=\frac{-1}{x^2}$

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3 years ago

How to do this question plz answer me step by step plzz plz please

Alina [70]

Answer:

See the attached

Step-by-step explanation:

The solution and filled in Venn diagram is attached

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3 years ago

What are all values of x for which the series n 3^n/x^n converges

Eva8 [605]

The interval of the convergence is x < -3 or x > 3 if the series n 3^n/x^n goes infinitely.

<h3>What is convergent of a series?</h3>

A series is convergent if the series of its partial sums approaches a limit; that really is, when the values are added one after the other in the order defined by the numbers, the partial sums getting closer and closer to a certain number.

We can find the interval for the convergent by root test.

Like the Ratio Test, the root Test is used to determine absolute convergence (or not) with factorials, the ratio test is useful.

For the given series:

$\sum^{\infty}_{n=0}\dfrac{3^n}{x^n}$

As the series goes infinitely, we can use root test.

By the root test, the convergence interval will be;

The interval of convergence is:

x < -3 or x > 3 we can write this as:

|x| < 3

Thus, the interval of the convergence is x < -3 or x > 3 if the series n 3^n/x^n goes infinitely.

Learn more about the convergent of a series here:

brainly.com/question/15415793

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1 year ago

Line h contains the points (3,0) and (0,4). which point lies on a line that is perpendicular to line h and passes through point

Iteru [2.4K]

(3,0)(0,4)
slope = (4 - 0) / (0 - 3) = -4/3
A perpendicular line will have a negative reciprocal slope. So our perpendicular line has a slope of 3/4

y = mx + b
slope(m) = 3/4
(-6,-5)....x = -6 and y = -5
now sub into the formula and find b, the y int
-5 = 3/4(-6) + b
-5 = -18/4 + b
-5 + 18/4 = b
-20/4 + 18/4 = b
-2/4 = b
so ur perpendicular line is : y = 3/4x - 2/4....or 3x - 4y = 2

and ur point (6,4) lies on the perpendicular line <===

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2 years ago