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Westkost [7]
1 year ago
5

Which expression is equivalent to x² + 2x + 2?

Mathematics
1 answer:
melisa1 [442]1 year ago
4 0

Answer:

(x+1-i)(x+1+i)

Step-by-step explanation:

(x+1-i)(x+1+i)\\\\=(x+1)^2 -i^2~~~~~~~~~~~~~;[a^2 -b^2 = (a+b)(a-b)]\\\\=x^2 +2x +1 -(-1)\\\\=x^2 +2x +1+1\\\\=x^2 +2x +2

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Which equation represents the amount of money that Mr. Cooney earns in x hours? math
padilas [110]

Nobody can really answer this unless you can just give a random equation like 5*x=25 something like that

4 0
3 years ago
Read 2 more answers
Question is in picture.
Norma-Jean [14]

Answer:

<h2>B. 2x + y = 4</h2>

Step-by-step explanation:

Having the system of equations in its simplest form

\left\{\begin{array}{ccc}ax+by=c\\dx+ey=f\end{array}\right

If

a=d,\ b=e,\ c=f

then the system of equations has infinitely many solutions.

If

a=d,\ b=e,\ c\neq f

then the system of equations has no solution.

If

a\neq d\ or\ b\neq e

then the system of equations has one solution.

We have the equation

y=-2x+4

Convert to the standard form Ax + By = C<em>:</em>

<em />y=-2x+4<em>              add 2x to both sides</em>

y+2x=-2x+2x+4\\\\2x+y=4

3 0
3 years ago
Mathematical induction, prove the following two statements are true
adelina 88 [10]
Prove:
1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+n\left(\frac12\right)^{n-1}=4-\dfrac{n+2}{2^{n-1}}
____________________________________________

Base Step: For n=1:
n\left(\frac12\right)^{n-1}=1\left(\frac12\right)^{0}=1
and
4-\dfrac{n+2}{2^{n-1}}=4-3=1
--------------------------------------------------------------------------

Induction Hypothesis: Assume true for n=k. Meaning:
1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+k\left(\frac12\right)^{k-1}=4-\dfrac{k+2}{2^{k-1}}
assumed to be true.

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

Induction Step: For n=k+1:
1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+k\left(\frac12\right)^{k-1}+(k+1)\left(\frac12\right)^{k}

by our Induction Hypothesis, we can replace every term in this summation (except the last term) with the right hand side of our assumption.
=4-\dfrac{k+2}{2^{k-1}}+(k+1)\left(\frac12\right)^{k}

From here, think about what you are trying to end up with.
For n=k+1, we WANT the formula to look like this:
1+2\left(\frac12\right)+...+k\left(\frac12\right)^{k-1}+(k+1)\left(\frac12\right)^{k}=4-\dfrac{(k+1)+2}{2^{(k+1)-1}}

That thing on the right hand side is what we're trying to end up with. So we need to do some clever Algebra.

Combine the (k+1) and 1/2, put the 2 in the bottom,
=4-\dfrac{k+2}{2^{k-1}}+\dfrac{(k+1)}{2^{k}}

We want to end up with a 2^k as our final denominator, so our middle term is missing a power of 2. Let's multiply top and bottom by 2,
=4+\dfrac{-2(k+2)}{2^{k}}+\dfrac{(k+1)}{2^{k}}

Distribute the -2 and combine the fractions together,
=4+\dfrac{-2k-4+(k+1)}{2^{k}}

Combine like-terms,
=4+\dfrac{-k-3}{2^{k}}

pull the negative back out,
=4-\dfrac{k+3}{2^{k}}

And ta-da! We've done it!
We can break apart the +3 into +1 and +2,
and the +0 in the bottom can be written as -1 and +1,
=4-\dfrac{(k+1)+2}{2^{(k-1)+1}}
3 0
3 years ago
PLSSS ANSWER!!!<br>tnk u!!!​
8090 [49]

Answer:

1.true

2.false

3.true

4.true

5.true

6.fales

7.true

8.true

9.false

10.false

Step-by-step explanation:

sana makatulog

5 0
2 years ago
HA=2-bA , solve the equation for A
yaroslaw [1]
HA + bA = 2
A(h+b) = 2
A = 2/h+b
7 0
2 years ago
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