Sets Let A = {1, 2, 3, 8} and {x | x is an even whole number less than 9}, find An B

Mathematics

2 answers:

spin [16.1K]2 years ago

6 0

A={1,2,3,8}

B={x|x is an even whole no less than 9}

B={2,4,6,8}

Now

$\\ \sf\longmapsto A\cap B=\{2,8\}$

Tomtit [17]2 years ago

4 0

Answer:

A = {1, 2, 3, 8}

• Let set B include the values of x

B = {0, 2, 4, 6, 8}

• Therefore;

${ \underline{ \rm{ \: \:A \cap B \: \dashrightarrow \: \: \{2, \: 8 \} \: \: }}}$

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a regular rectangular pyramid has a base and lateral faces that are congruent equilateral triangles. it has a lateral surface ar

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A pyramid is regular if its base is a regular polygon, that is a polygon with equal sides and angle measures.
(and the lateral edges of the pyramid are also equal to each other)

Thus a regular rectangular pyramid is a regular pyramid with a square base, of side length say x.

The lateral faces are equilateral triangles of side length x.

The lateral surface area is 72 cm^2, thus the area of one face is 72/4=36/2=18 cm^2.

now we need to find x. Consider the picture attached, showing one lateral face of the pyramid.

by the Pythagorean theorem:

$h= \sqrt{ x^{2} - (x/2)^{2}}= \sqrt{ x^{2}- x^{2}/4}= \sqrt{3x^2/4}= \frac{ \sqrt{3} }{2}x$

thus,

$Area_{triangle}= \frac{1}{2}\cdot base \cdot height\\\\18= \frac{1}{2}\cdot x \cdot \frac{ \sqrt{3} }{2}x\\\\ \frac{18 \cdot 4}{ \sqrt{3}}=x^2$

thus:

$x^2 =\frac{18 \cdot 4}{ \sqrt{3}}= \frac{18 \cdot 4 \cdot\ \sqrt{3} }{3}=24 \sqrt{3}$ (cm^2)

but $x^{2}$ is exactly the base area, since the base is a square of sidelength = x cm.

So, the total surface area = base area + lateral area = $24 \sqrt{3}+72$ cm^2

Answer: $24 \sqrt{3}+72$ cm^2