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2 years ago

For the set D = (4,6,8,10), determine n(D)

Mathematics

1 answer:

pentagon [3]2 years ago

8 0

n(D) = 4

Step-by-step explanation:

Given Data:

Set D = {4, 6, 8, 10}

The number of elements in a set is called the cardinal elements (or) cardinality of the sets. This is denoted by n(A).

Here it is denoted by n(D).

The number of elements in the set D is 4.

The cardinality of the set is 4.

Hence, n(D) = 4.

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Evaluate the following pic

GREYUIT [131]

Answer:

1) $\sqrt{1225}+\sqrt{1024}=67$

2) $\sqrt[3]{-1331}=-11$

3) Evaluating $2:p :: p:8$ we get $p=\pm 4$

4) $x^3+y^2+z \ when \ x=3, y=-2, x=-6 \ we \ get \ \mathbf{25}$

5) $\frac{(-6)^4\times(-2)^3\times(3)^3}{(-6)^6}=-6$

Step-by-step explanation:

1) $\sqrt{1225}+\sqrt{1024}$

Prime factors of 1225 : 5x5x7x7

Prime factors of 1024: 2x2x2x2x2x2x2x2x2x2

$\sqrt{1225}+\sqrt{1024}\\=\sqrt{5\times5\times7\times7}+\sqrt{2\times2\times2\times2\times2\times2\times2\times2\times2\times2}\\=\sqrt{5^2\times7^2}+\sqrt{2^2\times2^2\times2^2\times2^2\times2^2}\\=5\times7+(2\times2\times2\times2\times2)\\=35+32\\=67$

$\sqrt{1225}+\sqrt{1024}=67$

2) $\sqrt[3]{-1331}$

We know that $\sqrt[n]{-x}=-\sqrt[n]{x} \ ( \ if \ n \ is \ odd)$

Applying radical rule:

$\sqrt[3]{-1331}\\=-\sqrt[3]{1331} \\=-\sqrt[3]{11\times\11\times11}\\=-\sqrt[3]{11^3} \\Using \ \sqrt[n]{x^n}=x \\=-11$

So, $\sqrt[3]{-1331}=-11$

3) $2:p :: p:8$

It can be written as:

$p*p=2*8\\p^2=16\\Taking \ square \ root \ on \ both \ sides\\\sqrt{p^2}=\sqrt{16}\\p=\pm 4$

Evaluating $2:p :: p:8$ we get $p=\pm 4$

4) $x^3+y^2+z \ when \ x=3, y=-2, x=-6$

Put value of x, y and z in equation and solve:

$x^3+y^2+z \\=(3)^3+(-2)^2+(-6)\\=27+4-6\\=25$

So, $x^3+y^2+z \ when \ x=3, y=-2, x=-6 \ we \ get \ \mathbf{25}$

5) $\frac{(-6)^4\times(-2)^3\times(3)^3}{(-6)^6}$

We know (-a)^n = (a)^n when n is even and (-a)^n = (-a)^n when n is odd

$\frac{(-6)^4\times(-2)^3\times(3)^3}{(-6)^6}\\\\=\frac{1296\times-8\times 27}{46656}\\\\=\frac{-279936}{46656} \\\\=-6$

So, $\frac{(-6)^4\times(-2)^3\times(3)^3}{(-6)^6}=-6$

8 0

2 years ago

Write the slope-intercept form of the equation of each line

Vera_Pavlovna [14]

First let's talk about the blue line.

You can see its rising so its slope is certainly positive. But by how much is it rising? You can observe that each unit it rises it goes 1 forward and 1 up so its slope is the ratio of 1 up and 1 forward which is just 1.

We have thusly,

$y=x+n$

Now look at where blue line intercepts y-axis, -1. That is our n.

So the blue line has the equation of,

$y=x-1$

Next the black lines. The black lines are axes so their equations are a bit different.

First let's deal with x-axis, does it have slope? Yes but it is 0. The x-axis is still, not rising nor falling. Where does x-axis intercept y-axis? At 0. So the equation would be,

$y=0x+0=0$

Now we have y-axis. Does y axis have a slope? Yes but it is $\infty$ . The y-axis rises infinitely in no run. Where does it intercept y-axis? Everywhere! So what should the equation be? What if we ask where does y-axis intercept x-axis and write its equation in terms of x. Y-axis intercepts x-axis at 0 which means its equation is,

$x=0$

That is, every point of a form $(0,a)$ lies on y-axis.

Hope this helps :)

8 0

2 years ago

I need help can you please explain where does the 8 come from

BaLLatris [955]

Since you want to cancel something out, the book makes the 24 3 times eight so that both of the threes can be canceled out

5 0

3 years ago

Write the equation of the line that passes through (2, 3) and is parallel to the line 12x – 5y = 2.

vladimir1956 [14]

$k:12x-5y=2\to-5y=-12x+2\ \ \ /:(-5)\to y=\frac{12}{5}y-\frac{2}{5}\\\\l:y=mx+b\\\\k\ ||\ l\iff m=\frac{12}{5}\\\\ l:y=\frac{12}{5}x+b;\ (2;\ 3)\\\\\frac{12}{5}\cdot2+b=3\\\\\frac{24}{5}+b=3\\\\b=\frac{15}{5}-\frac{24}{5}\\\\b=-\frac{9}{5}\\\\Answer:y=\frac{12}{5}x-\frac{9}{5}\to\frac{12}{5}x-y=\frac{9}{5}\ \ \ \ /\cdot5\to12x-5y=9$