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Yuri [45]
3 years ago
13

Help with Algebra questions? I will Mark brainliest!

Mathematics
2 answers:
kondaur [170]3 years ago
8 0

Answer:

Step-by-step explanation:

1)  First graph is correct! Because clearly, the roots are visible, and it's a 3rd degree polynomial, so it can't be bottom twos, cuz they're parabolas.

2) Roots are at -6, -2 and 3. (If you need further assistance on this, let me know!) Hence, It is the last graph!

3) Roots are at -4, -2 and 2, (Again, if you need assistance, let me know!)

Hence, it is the first one!

Hope this cleared things up for you!

Anastaziya [24]3 years ago
5 0

Answer:

1) The degree of the polynomial function f(x) is 3, because it has three roots when f(x) = 0. Remember that the degree of a polynomial is dictated by the number of roots it has. (first graph)

2) The graph that represents g(x)=x^{3}+5x^{2}  -12x-36 is attached. There you can observe that the roots of this functoin are x=-6, x=-2 and x=3. The right choice is the graph placed at the bottom-right side.

3) The graph of the function f(x)=x^{3}+4x^{2}  -4x-16 is attached. Notice that its roots are x=-4, x=-2 and x=2. Therefore, the right choice is the graph placed at the top-left side.

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Kobotan [32]

Answer: 18 square ft.

Step-by-step explanation:

Surface area is length*width. 1*1= 1*6= 6

since there are three cubes, it is 6*3= 18 square ft total.

4 0
3 years ago
11. Give two examples of each type of number: real, natural, integers, rational and irrational.
zaharov [31]

Answer:

real:√2, 3,-1, 1/2 etc

natural: 1,2,3,4,5,6.....(0 is not included)

integers:. .............-4,-3,-2,-1,0,1,2,3,4........ etc

rational: nos. which are in p/q form

:1/2,3/4,4/9 etc

irrational: nos. which cannot be written in p/q form

: √2,√3... etc

irrational: √2, √3, √5, √11, √21, π(Pi)

7 0
3 years ago
*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

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3 years ago
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Is there a picture of the question?
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