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konstantin123 [22]
2 years ago
5

Is this wrong?

Mathematics
1 answer:
sammy [17]2 years ago
4 0

Step-by-step explanation:

The last step is a little wrong. Lemme show you:

x(x + 2) + (x + 2)

Imagine there is an invisible '1' infront of (x + 2) like this:

= x(x + 2) + 1(x + 2)

Now we can combine the like terms together:

= (x + 1)(x + 2)

This is the correct factorization.

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Answer:

  1.  9 < s < 17

  2.  5 < MN < 19

  3.  AD > BD

Step-by-step explanation:

1. The triangle inequality tells you the sum of any two sides of a triangle must exceed the length of the other side. (Some versions say, "must be not less than ..." rather than "must exceed.") In practice, this means two things:

  • the sum of the shortest two sides is greater than the length of the longest side
  • the length of any side lies between the sum and the difference of the other two sides

Here, we can use the latter fact to write the desired inequality. The difference of the given sides is 13 -4 = 9; their sum is 13 +4 = 17. The third side must lie between 9 and 17. If that side length is designated "s", then ...

  9 < s < 17

(If you don't mind a "triangle" that looks like a line segment, you can use ≤ instead of <.)

__

2. Same as (1) using different numbers.

  12 -7 < MN < 12 +7

  5 < MN < 19

__

3. Side CD is congruent to itself, and side CA is shown congruent to side CB. This means the requirements of the Hinge Theorem are met. That theorem tells you the longer side is opposite the greater angle:

  AD > BD

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A right square pyramid with base edges of length $8\sqrt{2}$ units each and slant edges of length 10 units each is cut by a plan
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Answer:

32 unit^{3}

Step-by-step explanation:

Given:

  • The slant length 10 units
  • A right square pyramid with base edges of length 8\sqrt{2}

Now we use  Pythagoras to get the slant height in the middle of each triangle:

\sqrt{10^{2 - (4\sqrt{2} ^{2} }) } = \sqrt{100 - 32} = \sqrt{68}  units

One again, you can use Pythagoras again to get the perpendicular height of the entire pyramid.

\sqrt{68-(4\sqrt{2} ^{2} )} = \sqrt{68 - 32} = 6 units.

Because slant edges of length 10 units each is cut by a plane that is parallel to its base and 3 units above its base. So we have the other dementions of the small right square pyramid:

  • The height 3 units
  • A right square pyramid with base edges of length 4\sqrt{2}

So the volume of it is:

V  = 1/3 *3* 4\sqrt{2}

= 32 unit^{3}

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