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Karo-lina-s [1.5K]
3 years ago
8

Help. I got it wrong and have no clue how to get to the right answer!

Mathematics
1 answer:
natima [27]3 years ago
7 0
\bf 2x+\sqrt{4x^2}+\sqrt[3]{8x^3}\qquad 
\begin{cases}
4\to 2^2\\
8\to 2^3
\end{cases}\qquad thus
\\\\\\
2x+\sqrt{2^2x^2}+\sqrt[3]{2^3x^3}\implies 2x+\sqrt{(2x)^2}+\sqrt[3]{(2x)^3}
\\\\\\
2x+2x+2x=\boxed{6x}
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6+7m<6m or 3m-7<5+6m
gogolik [260]

Step-by-step explanation:

6+7m<6m or 3m-7<5+6m

6<7m+6m or 3m-6m<5+7

6<13m or -3m<13

m<13/6 or m<13/-3

m<2.16 or m<-4.333

m<2.2 or m<-4.34

5 0
3 years ago
Name two triangles that are congruent by the ASA Postulate.
lubasha [3.4K]
Is there a picture to this question?
8 0
3 years ago
Complete the square to find the minimum value of the expression 4x2 + 8x + 23.
Tcecarenko [31]
So you need to come up with a perfect square that works for the x coefficients.
like.. (2x + 2)^2
(2x+2)(2x+2) = 4x^2 + 8x + 4
Compare this to the equation given. Our perfect square has +4 instead of +23. The difference is: 23 - 4 = 19

I'm going to assume the given equation equals zero..

So, If we add subtract 19 from both sides of the equation we get the perfect square.

4x^2 + 8x + 23 - 19 = 0 - 19
4x^2 + 8x + 4 = - 19
complete the square and move 19 over..
(2x+2)^2 + 19 = 0
factor the 2 out becomes 2^2 = 4
ANSWER: 4(x+1)^2 + 19 = 0

for a short cut, the standard equation
ax^2 + bx + c = 0 becomes a(x - h)^2 + k = 0
Where "a, b, c" are the same and ..
h = -b/(2a)
k = c - b^2/(4a)

Vertex = (h, k)
this will be a minimum point when "a" is positive upward facing parabola and a maximum point when "a" is negative downward facing parabola.


3 0
3 years ago
Solve for x in the equation x squared + 2 x + 1 = 17.
Papessa [141]

Answer:

x = - 1 + \sqrt{17}\\and\\x = - 1 - \sqrt{17}\\

Step-by-step explanation:

given equation

x^2 +2x +1 = 17

subtracting 17 from both sides

x^2 +2x +1 = 17\\x^2 +2x +1 -17= 17-17\\x^2 +2x - 16 = 0\\

the solution for quadratic equation

ax^2 + bx + c = 0 is given by

x = x = -b + \sqrt{b^2 - 4ac} /2a \\\\and \ \\-b - \sqrt{b^2 - 4ac} /2a

________________________________

in our problem

a = 1

b = 2

c = -16

x =( -2 + \sqrt{2^2 - 4*1*-16}) /2*1 \\x =( -2 + \sqrt{4  + 64}) /2\\x =( -2 + \sqrt{68} )/2\\x = ( -2 + \sqrt{4*17} )/2\\x =  ( -2 + 2\sqrt{17} )/2\\x = - 1 + \sqrt{17}\\and\\\\x = - 1 - \sqrt{17}\\

thus value of x is

x = - 1 + \sqrt{17}\\and\\x = - 1 - \sqrt{17}\\

3 0
2 years ago
True or false? The quadratic equation 10p2 - 5p = -8 has one solution.
pav-90 [236]

Answer: False

=============================================

Explanation:

I'll use x in place of p.

The original equation 10x^2-5x = -8 becomes 10x^2-5x+8 = 0 after moving everything to one side.

Compare this to ax^2+bx+c = 0

We have

  • a = 10
  • b = -5
  • c = 8

Plug those three values into the discriminant formula below

d = b^2 - 4ac

d = (-5)^2 - 4(10)(8)

d = 25 - 40*8

d = 25 - 320

d = -295

The discriminant is negative, which means we have no real solutions. If your teacher has covered complex or imaginary numbers, then you would say that the quadratic has 2 complex roots. If your teacher hasn't covered this topic yet, then you'd simply say "no real solutions".

Either way, this quadratic doesn't have exactly one solution. That only occurs when d = 0. Therefore, the original statement is false.

6 0
2 years ago
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