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Marina CMI [18]
3 years ago
5

Helppppppppppppppppppppppp

Mathematics
2 answers:
PolarNik [594]3 years ago
7 0
Answer:
The second choice is the correct one

Explanation:
(2x+3)^2 + 8(2x+3) + 11 = 0
To use the u substitution, we will assume that:
2x + 3 = u

Substitute with this in the given expression, we will get:
u^2 + 8u + 11 = 0 

The general form of the second degree equation is:
ax^2 + bx + c = 0

Comparing the expression we reached with the general one, we will find that:
a = 1
b = 8
c = 11

The roots can be found using the rule found in the attached picture.
This means that, for the given expression:
u = -4 ± √5 

Now, we have:
u = 2x+3
This means that:
at u = -4 + √5 
2x + 3 = -4 + √5 
2x = -7 + √5
x = (-7 + √5) / 2

at u = -4 - √5 
2x + 3 = -4 - √5 
2x = -7 - √5
x = (-7 - √5) / 2

This means that, for the given expression:
x = (-7 ± √5 ) / 2

Hope this helps :)

tatuchka [14]3 years ago
5 0
(2x + 3) ^ 2 + 8 (2x + 3) + 11 = 0
 Using substitution we have:
 u = 2x + 3
 We rewrite:
 (u) ^ 2 + 8 (u) + 11 = 0
 We use resolvent:
 u = (- b +/- root (b ^ 2 - 4ac)) / (2a)
 We replace:
 u = (- (8) +/- root ((8) ^ 2 - 4 (1) (11))) / (2 (1))
 u = (- (8) +/- root (64 - 44)) / (2)
 u = (- (8) +/- root (20)) / (2)
 u = (- (4) +/- root (5))
 We return the change:
 u = (- (4) +/- root (5)) = 2x + 3
 We clear x:
 x = (- (7) +/- root (5)) / 2
 Answer:
 x = (- (7) +/- root (5)) / 2
 (option2)
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4. DE is the perpendicular bisector of JL. Which statement must be true?
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Since DE is the perpendicular bisector of JL.

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Read 2 more answers
which expressions are equivalent to the first one? I don't understand how to determine that so please explain. Thanks!​
coldgirl [10]

9514 1404 393

Answer:

  (a) -(x+7)/y

  (b) (x+7)/-y

Step-by-step explanation:

There are several ways you can show expressions are equivalent. Perhaps the easiest and best is to put them in the same form. For an expression such as this, I prefer the form of answer (a), where the minus sign is factored out and the numerator and denominator have positive coefficients.

The given expression with -1 factored out is ...

  \dfrac{-x-7}{y}=\dfrac{1(x+7)}{y}=\boxed{-\dfrac{x+7}{y}} \quad\text{matches A}

Likewise, the expression of (b) with the minus sign factored out is ...

  \dfrac{x+7}{-y}=\boxed{-\dfrac{x+7}{y}}

On the other hand, simplifying expression (c) gives something different.

  \dfrac{-x-7}{-y}=\dfrac{-(x+7)}{-(y)}=\dfrac{x+7}{y} \qquad\text{opposite the given expression}

__

Another way you can write the expression is term-by-term with the terms in alpha-numeric sequence (so they're more easily compared).

  Given: (-x-7)/y = (-x/y) +(-7/y)

  (a) -(x+7)/y = (-x/y) +(-7/y)

  (b) (x+7)/(-y) = (-x/y) +(-7/y)

  (c) (-x-7)/(-y) = (x/y) +(7/y) . . . . not the same.

__

Of course, you need to know the use of the distributive property and the rules of signs.

  a(b+c) = ab +ac

  -a/b = a/(-b) = -(a/b)

  -a/(-b) = a/b

__

<u>Summary</u>: The given expression matches (a) and (b).

_____

<em>Additional comments</em>

Sometimes, when I'm really stuck trying to see if two expressions are equal, I subtract one from the other. If the difference is zero, then I know they are the same. Looking at (b), we could compute ...

  \left(\dfrac{-x-7}{y}\right)-\left(\dfrac{x+7}{-y}\right)=\dfrac{-y(-x-7)-y(x+7)}{-y^2}\\\\=\dfrac{xy+7y-xy-7y}{-y^2}=\dfrac{0}{-y^2}=0

Yet another way to check is to substitute numbers for the variables. It is a good idea to use (at least) one more set of numbers than there are variables, just to make sure you didn't accidentally find a solution where the expressions happen to be equal. We can use (x, y) = (1, 2), (2, 3), and (3, 5) for example.

The given expression evaluates to (-1-7)/2 = -4, (-2-7)/3 = -3, and (-3-7)/5 = -2.

(a) evaluates to -(1+7)/2 = -4, -(2+7)/3 = -3, -(3+7)/5 = -2, same as given

(b) evaluates to (1+7)/-2 = -4, (2+7)/-3 = -3, (3+7)/-5 = -2, same as given

(c) evaluates to (-1-7)/-2 = 4, different from given

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