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julia-pushkina [17]
4 years ago
14

I have two names, i can make a good tiling pattern. I am symmetrical in 2 ways, all my corners are the same size and 2 pairs of

equal sides
what am i?
Mathematics
1 answer:
garri49 [273]4 years ago
7 0

Rectangle.

A rectangle can also be called a quadrilateral, the two length sides are the same and the two height sides are the same. All the angles are 90°, and tiling patterns need to be something that tessellates, and rectangles tessellate. 
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3 years ago
True or false? Any matrix plus the identity matrix is the original matrix.
lapo4ka [179]
The answer is a true
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Find the derivative with respect to x of the integral from 2 to x^3 of ln(x^2)dx
strojnjashka [21]

The derivative follows from the fundamental theorem of calculus:

\displaystyle\frac{\mathrm d}{\mathrm dx}\int_c^{g(x)}f(t)\,\mathrm dt=f(g(x))\dfrac{\mathrm dg}{\mathrm dx}

where c is any constant in the domain of f.

We have

g(x)=x^3\implies\dfrac{\mathrm dg}{\mathrm dx}=3x^2

so

\displaystyle\frac{\mathrm d}{\mathrm dx}\int_2^{x^3}\ln(t^2)\,\mathrm dt=3x^2\ln((x^3)^2)=18x^2\ln x

(applying the property \ln a^b=b\ln a)

5 0
3 years ago
What is the value of y?<br><br> Enter your answer, as an exact value, in the box.
OLga [1]
We know the bottom triangle is a 45, 45, 90 triangle, so the hypotenuse is √2 times the value of the legs:

(√2)(√2)
=√4
=2

Now, we can use this to solve for y. The top triangle is a 30, 60, 90 triangle. The side we found above is the side across from the 30 degree angle. The side opposite the 60 degree angle is √3 times the side across the 30 degree angle. Therefore, we can solve for y by multiplying 2 by √3

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7 0
4 years ago
I’m confused on how to find x. Can someone help me on this worksheet please ASAP!?
igor_vitrenko [27]

9514 1404 393

Answer:

  1. x=2 (work is correct)
  2. x=4
  3. x=44
  4. x=4
  5. x=any number
  6. impossible; no such number

Step-by-step explanation:

1. The equation is ...

  3x +5 = 11

  3x = 11 -5 = 6

  x = 6/3 = 2 . . . . matches the work shown

__

2. The equation is ...

  (x +2)·5 = 30

  x +2 = 30/5 = 6

  x = 6 -2 = 4

__

3. The equation is ...

  (x/2) +2 = 24 . . . . matches the work shown

  x/2 = 24 -2 = 22

 x = 22·2 = 44

__

4. The equation is ...

  2x +8 = x +12

  2x = x + 12 -8 = x +4

  2x -x = x -x +4

  x = 4

__

5. The equation is ...

  2x +4 -x -3 = x +1

  x +1 = x +1 . . . . . true for any value of x

__

6. The equation is ...

  (3x +6) -2x = x +3

  x +6 = x +3 . . . . . not true for any number

_____

<em>Additional comment</em>

The work shown is correct. You find x by writing and solving an appropriate equation. To solve the equation, undo the steps that are done to x. Undo addition by adding the opposite. Undo multiplication by diving by the multiplier. It often helps to collect terms before you start the "undo" steps.

For x on both sides of the equation (as in problems 4–6), you can subtract the term with the lowest coefficient. (Of course, you must subtract it from both sides of the equation.) In problem 5, if you do that, you get 1=1, which is true for any value of x. In problem 6, if you do that, you get 6=3, a false equation, indicating no value of x will make it true.

5 0
3 years ago
Read 2 more answers
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