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3 years ago

I need the answer fast!!!!!!!!

Mathematics

2 answers:

AlexFokin [52]3 years ago

4 0

Angle A should be equal to 47

umka2103 [35]3 years ago

3 0

Yessssssssssssdddddbbbbbbbb

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Find the area of each figure. A = _ in²

Sholpan [36]

Answer:

1824in^2

Step-by-step explanation:

area of rectangle:

L x W = A

32 x 48 = 1539

area of triangle:

(L x W) / 2 = A

(48 x 12) / 2 = 288

add:

1539 + 288 = 1824in ^2

5 0

3 years ago

1/2h(a+b)=1 solve for h

Cloud [144]

H equals eighti'll just keep on writing because it needs to have 20 characters to answer

6 0

4 years ago

Let U= U= Universal set ={0, 1, 2, 3, 4, 5, 6, 7, 8, 9}={0, 1, 2, 3, 4, 5, 6, 7, 8, 9} , A={1, 3, 4, 5, 7, 9} A={1, 3, 4, 5, 7,

blagie [28]

Answer:

a) $A' = [0,2,6,8]$

b) $(AUB)' = [0,2,6]$

c) $(AUB')' = [9]$

d) A∩B′ $= [3,5,9]$

Step-by-step explanation:

Assuming this problem: "Let U= U= Universal set ={0, 1, 2, 3, 4, 5, 6, 7, 8, 9} , A={1, 3, 4, 5, 7, 9} , and B={1, 4, 7, 8} . List the elemetns of the following sets in the increasing order: a) A′= b) (A∪B)′={ , , }} c) (A∪B′)′={ }} d) A∩B′={ , , }}"

Part a

For this case we just need to find the elements in the universal set that are not in A. And we see that:

$A' = [0,2,6,8]$

And that represent the complement for A

Part b

For this case we need to find first the Union AUB who are the elements on A or B without repetition and we got:

$AUB = [1,3,4,5,7,8,9]$

And now the complement for (AUB)' are the elements that are not in AUB but are on the universal set and we got:

$(AUB)' = [0,2,6]$

Part c

For this case we need to find B' who are the elements on the universal set that are not in B

$B' = [0,2,3,5,6,9]$

Then we can find the union between AUB' and we got:

$AUB' = [0,1,2,3,4,5,6,7,9]$

And then the complment is just:

$(AUB')' = [9]$

Part d

For this case we just need to see the elements in common between A and B' and we got:

A∩B′ $= [3,5,9]$

6 0

4 years ago

Give me the numbers that both add to negative 8 and multiply to -20

svp [43]

Answer:

- 10 and 2

Step-by-step explanation:

-10 x 2 = -20

-10 + 2 = -8

3 0

3 years ago

Find the product of ( 2a2– 3b2) and ( a +b2- 3z)

Bond [772]

Answer:

$2a^3+2a^2b^2-6a^2z-3ab^2-3b^4+9b^2z$

Step-by-step explanation:

$(2a^2-3b^2)(a+b^2-3z)$

Start by distributing the $2a^2$ into all of the terms in between the second parentheses.

$2a^3+2a^2b^2-6a^2z$

Next, distribute the $-3b^2$ into all of the terms in between the second parentheses.

$-3ab^2-3b^4+9b^2z$

For the sake of simpler interpretation, I separated the two above, but they are together:

$2a^3+2a^2b^2-6a^2z-3ab^2-3b^4+9b^2z$

Since the terms are already in descending order, we do not need to rearrange them.

4 0

3 years ago