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Salsk061 [2.6K]
2 years ago
15

Is there any video related to this or can you explain how to do it

Mathematics
1 answer:
uranmaximum [27]2 years ago
4 0

Angles ECD  and CEF add to 180

40+140 = 180

So that means we have EF parallel to CD (due to the same side interior angle theorem)

--------------

Angles BCE and ECD combine to 30+40 = 70, which is congruent to angle ABC = 70 as well.

In other words, this shows angle ABC = angle BCD. Both of these angles are alternate interior angles. Since they're congruent, they lead to AB being parallel to CD.

--------------

So far we have

AB || CD

CD || EF

Using the transitive property, we can then link the two statements to say AB || EF. Think of a chain where CD is the common link. We go from AB to CD, then from CD to EF. So we can just take a single path from AB to EF.

It's like saying "P --> Q and Q --> R, therefore P --> R"

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slega [8]
1. 30 in
2. 40 in
I hope this helped
7 0
2 years ago
Help! open the attachment below!
gregori [183]

Answer:

yes it is

Step-by-step explanation:X=side lenghths 2/3

                                            Y=perimeter 12/18

2 time 6=12 and 3 times 6=18 and it is paraticly telling you

2/6 and 3/6 i beleve i need a lil more info

6 0
2 years ago
1.
photoshop1234 [79]

Answer:

0.923

Step-by-step explanation:

There are 13 total marbles. (12 Purple Marbles + One Blue Marble)

The total events are 13 and the number of preferred events is 12.

\frac{12}{13}= 0.92307692307 ≈ 0.923

So there should be a 0.923 chance that a purple marble is picked.

3 0
2 years ago
Read 2 more answers
PLEASE HELP!!!! The problem is in the photo below.
Ratling [72]
Answer :36 explanation: (39x39)-(15-15)=1296 then square root that and get 36
7 0
2 years ago
21. In 4 + In(4x - 15) = In(5x + 19)
Alexxx [7]

Answer:

x=-30;\quad \:I\ne \:0

Step-by-step explanation:

In\cdot \:4+In\left(4x-15\right)=In\left(5x+19\right)

\:4+In\left(4x-15\right):\quad -11nI+4nxI\\In\cdot \:4+In\left(4x-15\right)\\=4nI+nI\left(4x-15\right)\\\\\:In\left(4x-15\right):\quad 4nxI-15nI\\\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b-c\right)=ab-ac\\a=In,\:b=4x,\:c=15\\=In\cdot \:4x-In\cdot \:15\\=4nxI-15nI\\=In\cdot \:4+4nxI-15nI\\\mathrm{Simplify}\:In\cdot \:4+4nxI-15nI:\quad -11nI+4nxI\\In\cdot \:4+4nxI-15nI\\\mathrm{Group\:like\:terms}\\=4nI-15nI+4nxI\\\mathrm{Add\:similar\:elements:}\:4nI-15nI=-11nI\\=-11nI+4nxI\\

\mathrm{Expand\:}In\left(5x+19\right):\quad 5nxI+19nI\\In\left(5x+19\right)\\=nI\left(5x+19\right)\\\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b+c\right)=ab+ac\\a=In,\:b=5x,\:c=19\\=In\cdot \:5x+In\cdot \:19\\=5nxI+19nI\\\\-11nI+4nxI=5nxI+19nI\\\\\mathrm{Add\:}11nI\mathrm{\:to\:both\:sides}\\-11nI+4nxI+11nI=5nxI+19nI+11nI\\Simplify\\4nxI=5nxI+30nI\\\mathrm{Subtract\:}5nxI\mathrm{\:from\:both\:sides}\\4nxI-5nxI=5nxI+30nI-5nxI\\\mathrm{Simplify}\\-nxI=30nI\\

\mathrm{Divide\:both\:sides\:by\:}-nI;\quad \:I\ne \:0\\\frac{-nxI}{-nI}=\frac{30nI}{-nI};\quad \:I\ne \:0\\\mathrm{Simplify}\\\frac{-nxI}{-nI}=\frac{30nI}{-nI}\\\mathrm{Simplify\:}\frac{-nxI}{-nI}:\quad x\\\frac{-nxI}{-nI}\\\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{-b}=\frac{a}{b}\\=\frac{nxI}{nI}\\\mathrm{Cancel\:the\:common\:factor:}\:n\\=\frac{xI}{I}\\\mathrm{Cancel\:the\:common\:factor:}\:I\\=x\\\mathrm{Simplify\:}\frac{30nI}{-nI}:\quad -30\\\mathrm{Apply\:the\:fraction\:rule}:

\quad \frac{a}{-b}=-\frac{a}{b}\\\mathrm{Cancel\:the\:common\:factor:}\:n\\=\frac{30I}{I}\\\mathrm{Cancel\:the\:common\:factor:}\:I\\=-30\\x=-30;\quad \:I\ne \:0

3 0
3 years ago
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