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3 years ago

Given A={x ∈ Z : x is even}, B={x ∈ Z : x is prime number}, C={x ∈ Z : x is odd}, and D={5, 7, 8, 12, 13, 15}.

Mathematics

1 answer:

Delvig [45]3 years ago

6 0

Answer:

a) {15}

b) { }

c) D or {5, 7, 8, 12, 13, 15}

d) no

e) yes

Step-by-step explanation:

a) AUB: even or prime

D - (AUB) = {15}

b) AUC: even or odd; basically all integers

D - (AUC) = { }

c) A∩B : even and prime, that's only 2

D - (A∩B) = {5, 7, 8, 12, 13, 15} = D

d) A and B, both contain 2

Not disjoint

e) A and C are disjoint because no number can be both, even and odd

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4 years ago

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3 years ago

Find the surface area and show work

Law Incorporation [45]

Answer: $1,270in^2$

Step-by-step explanation:

The surface area of a rectangle can be calculated with:

$Area_{rectangle}=lenght*width$

The surface area of a trapezoid can be calculated with:

$Area_{trapezoid}=\frac{height}{2}(Larger\ base+Smaller\ base)$

Calculate the area of the base, wich is a rectangle:

$A_{1}=(34in)(15in)=510in^2$

There are two equal rectangles, therefore, multiply the area of any of them by 2:

$A_{2}=2(15in)(13in)=390in^2$

Calculate the area of the other rectangle:

$A_{3}=(15in)(10in)=150in^2$

There are two equal trapezoids, therefore, multiply the area of any of them by 2:

$A_4=\frac{2(5in)}{2}(34in+10in)=220in^2$

Therefore, the surface area will be the sum of all the areas calculated:

$SA=A_1+A_2+A_3+A_4\\\\SA=510in^2+390in^2+150in^2+220in^2\\\\SA=1,270in^2$

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3 years ago

How do you solve 3x^6- 39x^4+ 108x^2= 0

ArbitrLikvidat [17]

You can first take out an x^2 from each number , 3x^6 - 39x^4 + 108x^2 = x^2 ( 3x^4 - 39x^2 + 108) Then just factor and solve for the zero's of the equation

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