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Katena32 [7]
1 year ago
3

Choose the inequality shown by this diagram

Mathematics
1 answer:
stiks02 [169]1 year ago
3 0

Answer: C

Step-by-step explanation:

Starting off we need to understand what the circles are in inequalities.

A closed circle means that the number on the inequality is included in the solution. An open circle means that the number in the inequality is not included in the solution. A closed circle would use \leq or \geq. An open circle would use < or >.

So, using these processes, -2<x\leq7 would be the answer.

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Hi, please help me out with this math problem.
GalinKa [24]

9514 1404 393

Answer:

  x - 4

Step-by-step explanation:

The expression simplifies to ...

  \sqrt{x^2-8x+16}=\sqrt{(x-4)^2}=|x-4|

For x ≥ 4, the argument of the absolute value function is non-negative, so it remains unchanged. The simplified expression is ...

  x - 4 . . . . for x ≥ 4

6 0
3 years ago
Dy/dxif y = 4x3 + 3x2 + 2
DIA [1.3K]
The function is 

                                \displaystyle{ y=4x^3+3x^2+2.

We recall that the derivative of each monomial ax^b is a\cdot b\cdot x^{b-1}, and the derivative of a constant (function) is 0.

According to this:

\frac{d}{dx}(4x^3+3x^2+2)=4\cdot 3\cdot x^2 + 3\cdot 2\cdot x+0 =12x^2+6x.


Answer: 12x^2+6x
7 0
4 years ago
Find the highest common factor of 18 and 12
I am Lyosha [343]

Answer:

Step-by-step explanation:

18 -> 1,2,3,6,9,18

12 -> 1,2,3,4,6,12

heighest common factor is 6.

It will be nice if you give me brainliest. Good luck!

4 0
4 years ago
What are the coordinates of the focus of the parabola?
Dima020 [189]
This is a tough one.  the general form of a parabola is (x-h) ^{2} =4p(y-k), where h and k are the coordinates of the vertex and p is the distance from the vertex to the focus.  In order to get our parabola into this form and solve for p (which will give us our focal point), we have to complete the square.  Set the parabola equal to 0, then move over the constant to get this equation: - \frac{1}{16}  x^{2} -x=-2.  In order to complete the square, the leading coefficient on the squared term has to be a +1.  Ours is a - \frac{1}{16}, so we have to factor that out of the x terms.  When you do that you end up with - \frac{1}{16} ( x^{2} +16x)=-2.  Now we can complete the square by taking half the linear term, squaring it, and adding it to both sides.  Our linear term is 16, so half of 16 is 8 annd 8 squared is 64.  HOWEVER, on the left side, that - \frac{1}{16} is still hanging out in front, which means that when we add in 64, we are actually adding in - \frac{1}{16} *64 which is -4.  Now here's what we have: - \frac{1}{16} ( x^{2} +16x+64)=-2-4which simplifies to - \frac{1}{16}( x^{2} +16x+64)=-6.  Creating the perfect square binomial on the left was the point of this (to give us our vertex), so when we do that we have - \frac{1}{16} (x+8) ^{2} =-6.  Now just for simplicity, we will take baby steps.  Move the -6 back over by addition and set it back equal to y: - \frac{1}{16}(x+8) ^{2}+6=y.  Now we will work on getting into standard form.  Move the 6 back over by the y (baby steps, remember) to get - \frac{1}{16} (x+8) ^{2} =y-6.  Multiply both sides by -16 to get our "p" on the right: (x+8) ^{2} =-16(y-6).  We need to use our "4p" part of the standard form to find the p, which is the distance from the vertex to the focus. 4p=-16, and p = -4.  That means that the focus is 4 units below the vertex.  Let's figure out what the vertex is.  From our equation, the vertex is ( -8, 6), and since this is an upside-down opening parabola, the focus will be aligned with the x-coordinate of the vertex.  So our focus lies 4 units below 6 (6 is the y coordinate of the vertex which indicates up and down movement), so our focus has coordinates of (-8, 2), the first choice above.  Told you it was a tough one!  These conics are quite challenging!
7 0
3 years ago
The end points of one diagonal of a rhombus are (-5,2) and (1,6). If the coordinates of the 3rd vertexare (-6,10), what are the
ad-work [718]
Look at the picture.

E\left(\dfrac{-5+1}{2};\ \dfrac{2+6}{2}\right)\to E\left(\dfrac{-4}{2};\ \dfrac{8}{2}\right)\to E(-2;\ 4)

\vec{BE}=[-2-(-6);\ 4-10]=[4;\ -6]

D(x;\ y)\\\\\vec{ED}=[x-(-2);\ y-4]=[x+2;\ y-4]\\\\\vec{ED}=\vec{BE}\ therefore\\\\ \ [x+2;\ y-4]=[4;\ -6]\to x+2=4\ and\ y-4=-6\\\\x=2\ and\ y=-2\\\\Answer:\ \boxed{A.\ (2;\ -2)}

8 0
3 years ago
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