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Eddi Din [679]
4 years ago
9

|x + 1| + |x + 2| = 3

Mathematics
2 answers:
lina2011 [118]4 years ago
7 0

Answer: x=0 or -3

Step-by-step explanation: those two lines are absolute value, which is the distance a number is away from zero so it is always positive and since 1+2=3 x is zero

It can also be -3 because  l -3+1 l = l-2 l = 2 and l-3+2 l = l-1 l = 1 and 2+1+3

Olin [163]4 years ago
4 0

Answer:

0

Step-by-step explanation:

Since there is no negative sign with x, you could just take out the absolute signs.

So after you take them out you have:

x+1+x+2=3

Now, combine like terms:

x+x=2x

1+2=3

Combine into an equation:

2x+3=3

Subtract 3 on both sides:

2x+3-3=3-3

2x=0

Divide 2 on both sides:

x=0

(It would be undefined if it was \frac{x}{0})

But because it is \frac{0}{2}, it would just be 0.

Hope this helped!

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Answer:

well yes if the back sticks out or if its turned like a x / it can go inside

Step-by-step explanation:

4 0
3 years ago
23 puppies are for sale at the pet store 14 puppies are black and 6 puppies are brown the rest of the puppies have spots how man
lord [1]

Answer:

3 puppies have spots

Step-by-step explanation:

23-14=9

9-6=3

6 0
3 years ago
Read 2 more answers
Suppose we play the following game based on tosses of a fair coin. You pay me $10, and I agree to pay you $n 2 if heads comes up
Artyom0805 [142]

Answer:

In the long run, ou expect to  lose $4 per game

Step-by-step explanation:

Suppose we play the following game based on tosses of a fair coin. You pay me $10, and I agree to pay you $n^2 if heads comes up first on the nth toss.

Assuming X be the toss on which the first head appears.

then the geometric distribution of X is:

X \sim geom(p = 1/2)

the probability function P can be computed as:

P (X = n) = p(1-p)^{n-1}

where

n = 1,2,3 ...

If I agree to pay you $n^2 if heads comes up first on the nth toss.

this implies that , you need to be paid \sum \limits ^{n}_{i=1} n^2 P(X=n)

\sum \limits ^{n}_{i=1} n^2 P(X=n) = E(X^2)

\sum \limits ^{n}_{i=1} n^2 P(X=n) =Var (X) + [E(X)]^2

\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{1-p}{p^2}+(\dfrac{1}{p})^2        ∵  X \sim geom(p = 1/2)

\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{1-p}{p^2}+\dfrac{1}{p^2}

\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{1-p+1}{p^2}

\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{2-p}{p^2}

\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{2-\dfrac{1}{2}}{(\dfrac{1}{2})^2}

\sum \limits ^{n}_{i=1} n^2 P(X=n) =\dfrac{ \dfrac{4-1}{2} }{{\dfrac{1}{4}}}

\sum \limits ^{n}_{i=1} n^2 P(X=n) =\dfrac{ \dfrac{3}{2} }{{\dfrac{1}{4}}}

\sum \limits ^{n}_{i=1} n^2 P(X=n) =\dfrac{ 1.5}{{0.25}}

\sum \limits ^{n}_{i=1} n^2 P(X=n) =6

Given that during the game play, You pay me $10 , the calculated expected loss = $10 - $6

= $4

∴

In the long run, you expect to  lose $4 per game

3 0
4 years ago
Gabriel is at the grocery store, and wants to figure out his total cost before he gets to the register. He bought 2.5 pounds of
Readme [11.4K]

Answer:

32 - 2.5x - 2y

Step-by-step explanation:

He bought 2.5 pounds weight of apples for x dollars a pound. Thus, cost is 2.5x.

He bought 2 bags of lettuce for y dollars each. Thus, cost is 2y

Total bill cost = 2.5x + 2y

We are told he has a coupon for $2 off his bill.

This means the amount he will pay will now be;

2.5x + 2y - 2

Gabriel has $30 to pay for the groceries.

Thus, his change will be;

30 - (2.5x + 2y - 2)

This gives;

32 - 2.5x - 2y

6 0
3 years ago
Read 2 more answers
A.
Basile [38]

Answer:

d

Step-by-step explanation:

by substitution what is before x² is considered as a

what is before x is b and the number alone(without x) with it's sign included is c

8 0
3 years ago
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