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mafiozo [28]
2 years ago
7

(7m+6) (6m^2+4m+4) I really suck at math

Mathematics
2 answers:
lions [1.4K]2 years ago
5 0

Answer:

<h2>42 {m}^{3}  + 64 {m}^{2}  + 52m + 24</h2>

Step-by-step explanation:

(7m + 6)(6 {m}^{2}  + 4m + 4)

= 7m(6 {m}^{2}  + 4m + 4) + 6(6 {m}^{2}  + 4m + 4)

= 42 {m}^{3}  + 28 {m}^{2}  + 28m + 36 {m}^{2}  + 24m + 24

= 42 {m}^{3}  + 64 {m}^{2}  + 52m + 24( ans)

Ostrovityanka [42]2 years ago
3 0

\large\mathfrak{{\pmb{\underline{\orange{Your\:answer }}{\orange{:}}}}}

:⟼  (7m + 6)(6 {m}^{2}  + 4m + 4)

:⟼  7m \: (6  {m}^{2}  + 4m + 4) + 6 \: (6 {m}^{2}  + 4m + 4)

:⟼  42 {m}^{3}  + 28 {m}^{2}  + 28m + 36 {m}^{2}  +24  m  + 24

:⟼  42 {m}^{3}  + 28 {m}^{2}  + 36 {m}^{2}  +  28m + 24m + 24

:⟼  42 {m}^{3}  + 64 {m}^{2}  + 52m + 24

:⟼  2(21 {m}^{3}  + 32 {m}^{2}  + 26m + 12)

\sf \bf {\boxed {\mathbb {HAPPY\:LEARNING.}}}

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Step-by-step explanation:

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Answer:

<u><em></em></u>

  • <u><em>Event A: 1/35</em></u>
  • <u><em>Event B: 1/840</em></u>

<u><em></em></u>

Explanation:

<u>Event A</u>

For the event A, the order of the first 4 acts does not matter.

The number of different four acts taken from a set of seven acts, when the order does not matter, is calculated using the concept of combinations.

Thus, the number of ways that the first <em>four acts</em> can be scheduled is:

          C(m,n)=\dfrac{m!}{n!(m-n)!}

         C(7,4)=\dfrac{7!}{4!(7-4)!}=\dfrac{7!}{4!(3)!}=35

And<em> the number of ways that four acts is the singer, the juggler, the guitarist, and the violinist, in any order</em>, is 1: C(4,4).

Therefore the<em> probability of Event A</em> is:

           P(A)=1/35

Event B

Now the order matters. The difference between combinations and permutations is ordering. When the order matters you need to use permutations.

The number of ways in which <em>four acts </em>can be scheculed when the order matters is:

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From top to bottom:

A, J, E, B, I, C, D, G, F, H

See below for more clarification.

Step-by-step explanation:

We are given that AB is parallel to CD, XY is the perpendicular bisector of AB, and E is the midpoint of XY. And we want to prove that ΔAEB ≅ ΔDEC.

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In a plane, if a transveral is perpendicular to one of the two parallel lines, then it is perpendicular to the other.

3) m∠AXE = 90°, m∠DYE = 90°.

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5) XE ≅ YE

Definition of a midpoint.

6) ∠A ≅ ∠D.

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7) ΔAEX ≅ ΔDEY

AAS Triangle Congruence*

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8) AE ≅ DE

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9) ∠AEB ≅ ∠DEC

Vertical Angles Theorem

10) ΔAEB ≅ ΔDEC

ASA Triangle Congruence**

(**∠A ≅ ∠D, AE ≅ DE, and ∠AEB ≅ ∠DEC)

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