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galben [10]
3 years ago
5

i need help please i have alot of problems on math and i was hoping you all can help me and explain it to me

Mathematics
1 answer:
gizmo_the_mogwai [7]3 years ago
8 0
You want to break the shape apart
the square would be 2•2 so 4
the triangle on the right would be .5•2•6 so 6
the triangle on the left would be .5•2•2 so 2
and 6+4+2 is 12
so 12 units ^2
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3 numbers that are less than 2 and greater then -1
Viefleur [7K]

Three numbers less than 2, but greater than -1 are:

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4 0
3 years ago
Complete the equation of the line through [1,4] and [2,2]
son4ous [18]

First find the slope of the line so change in y over change in x so 4-2/1-2 which is 2/-1 so -2 is your slope then plug that as well as a point from above to find your y intercept so y=mx+b 2=-2(2)+ b then you have 2=-4+b then add 4 to each side and you find b is six then you make ur equation from your slope and y int(b) y=-2x+6
6 0
3 years ago
Which is equivalent to 8X- Y + 3X
Zarrin [17]

Answer:

11x-y or 5x+y.

Step-by-step explanation: I hope this helps, I really don't know what the question means, since it not that specific.

6 0
3 years ago
Mai received a $70 gift card for a coffee store. She used it in buying coffee that cost $8.48 per pound. After buying the coffee
Yuri [45]
Since she had $35.48 on her card.

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4 0
3 years ago
Read 2 more answers
*In ∆ABC, on the extension of the side BC , draw a line segment CD ≅ CA . Draw the segment AD . The line segment CE is the angle
Leya [2.2K]

CE ⊥ CF because m∠ECA + m∠FCA = 90°

Step-by-step explanation:

In ∆ABC

  • On the extension of the side BC , draw a line segment CD ≅ CA
  • Draw the segment AD
  • The line segment CE is the angle bisector of ∠ACB
  • The line segment CF is the median towards AD in ∆ ACD

We want to prove that CF ⊥ CE

Look to the attached figure

In Δ ABC

∵ CE is the bisector of angle ACB

∴ ∠ACE ≅ ∠BCE

In Δ ACD

∵ CA = CD

∴ Δ ACD is an isosceles triangle

∵ AD is the median towards AD

- In any isosceles triangle the median from a vertex to its opposite

  side bisects this vertex

∴ AD bisects ∠ACD

∴ ∠ACF ≅ ∠DCF

∵ BCD is a straight segment

∵ CE , CA , CF are drawn from point C

∴ m∠BCE + m∠ACE + m∠ACF + m∠DCF = 180°

∵ m∠ACE ≅ m∠BCE

∵ m∠ACF ≅ m∠DCF

- Replace m∠BCE by m∠ACE and m∠DCF by m∠ACF

∴ m∠ACE + m∠ACE + m∠ACF + m∠ACF = 180°

∴ 2 m∠ACE + 2 m∠ACF = 180°

- Divide all terms by 2

∴ m∠ACE + m∠ACF = 90°

∴ EC ⊥ CF

CE ⊥ CF because m∠ECA + m∠FCA = 90°

Learn more:

You can learn more about perpendicular lines in brainly.com/question/11223427

#LearnwithBrainly

7 0
3 years ago
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