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2 years ago

Simplify this problem!!!

Mathematics

2 answers:

DedPeter [7]2 years ago

8 0

The answer is c hope that helps you

Nonamiya [84]2 years ago

4 0

Answer:

Step-by-step explanation:

HOPE THAT HELPSBDNQBDB

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Chords AC and DB intersect at point E. Find the value of x.

Elina [12.6K]

Answer:

the answer would be 9 so option B

4 0

3 years ago

1011+111 in binary form

Ann [662]

Answer:

10010

Step-by-step explanation:

$1011=1(2)^3+0(2)^2+1(2)^1+1(2)^0$

$111=1(2)^2+1(2)^1+1(2)^0$

So $1011+111$ gives us:

$1(2)^3+0(2)^2+1(2)^1+1(2)^0$

$+$

$1(2)^2+1(2)^1+1(2)^0$

-----------------------------------------------------

Combine like terms:

$1(2)^3+(0+1)(2)^2+(1+1)(2)^1+(1+1)(2)^0$

$1(2)^3+1(2)^2+(2)(2)^1+(2)(2)^0$

We aren't allowed to have a coefficient bigger than 1.

I'm going to replace $2^0$ with 1 and $2$ with $(2)^1$ :

$1(2)^3+1(2)^2+(2)^2+(2)^1(1)$

I want a $2^0$ number:

$1(2)^3+1(2)^2+1(2)^2+1(2)^1+0(2)^0$

Combine like terms:

$1(2)^3+2(2)^2+1(2)^1+0(2)^0$

$2(2)^2=2^3$ :

$1(2)^3+2^3+1(2)^1+0(2)^0$

Combine like terms:

$2(2)^3+1(2)^1+0(2)^0$

We can rewrite the first term by law of exponents:

$2^4+1(2)^1+0(2)^0$

$1(2)^4+1(2)^1+0(2)^0$

So the binary form is:

$10010$

Maybe you like this way more:

Keep in mind 1+1=10 and that 1+1+1=11:

Setup:

1 0 1 1

+ 1 1 1

------------------------------

(1) (1) (1)

1 0 1 1

+ 1 1 1

------------------------------

1 0 0 1 0

I had to do some carry over with my 1+1=10 and 1+1+1=11.

8 0

3 years ago

40 is a rational number.

Sauron [17]

Answer: if its a true or false question then yes 40 is a rational number

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Im really bad with graphs, pls help

Gnoma [55]

Answer:

Step-by-step explanation:

8 0

2 years ago

There are 11 candidates for three postions at a restaraunt. One postion is for a cook. The second position is for a food server

LenaWriter [7]

Answer:

165 combinations possible

Step-by-step explanation:

This is a combination problem as opposed to a permutation, because the order in which we fill these positions is not important. We are merely looking for how many ways each of these 11 people can be rearranged and matched up with different candidates, each in a different position each time. The formula can be filled in as follows:

₁₁C₃ = $\frac{11!}{3!(11-3)!}$

which simplifies to

₁₁C₃ = $\frac{11*10*9*8!}{3*2*1(8!)}$

The factorial of 8 will cancel out in the numerator and the denominator, leaving you with

₁₁C₃ = $\frac{990}{6}$

which is 165

6 0

3 years ago