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DanielleElmas [232]
3 years ago
6

(2t-3)^3, can you help me?, please?

Mathematics
2 answers:
grandymaker [24]3 years ago
6 0

Answer:

8t^3-36t^2+54t-27

Step-by-step explanation:

(2t-3)^3

(2t-3)(2t-3)(2t-3)

Multiply the first two terms

(4t^2 -6t -6t+9)(2t-3)

Combine like terms

( 4t^2 -12t +9) ( 2t-3)

Multiply

2t*( 4t^2 -12t +9) -3 ( 4t^2 -12t +9)

Distribute

8t^3 -24t^2 +18t-12t^2 +36t-27

Combine like terms

8t^3-36t^2+54t-27

MissTica3 years ago
6 0

Answer:

8t³ - 36t² + 54t - 27

Step-by-step explanation:

(2t - 3)³

(2t - 3)(2t - 3)(2t - 3)

FOIL the first two binomials - first - outer - inner - last

(2t - 3)(2t - 3)

= 4t² - 6t - 6t + 9

= 4t² - 12t + 9

multiply this polynomial by the last binomial

(4t² - 12t + 9)(2t - 3)

distribute 2t

2t • 4t² = 8t³

2t • -12t = -24t²

2t • 9 = 18t

distribute -3

-3 • 4t² = -12t²

-3 • -12t = 36t

-3 • 9 = -27

combine

8t³ - 24t² + 18t - 12t² + 36t - 27

simplify

8t³ - 36t² + 54t - 27

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Step-by-step explanation:

Here, we want to get the coordinates of the image after dilating the pre-image by a scale factor of 4

What we have to do here is to multiply each of the coordinate on the pre-image by 4

We have this as;

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3 years ago
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Varvara68 [4.7K]

Answer:

1.) can not be found

2.)V=11403.98

3.)

Step-by-step explanation:

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r = Radius

H = height

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5 0
3 years ago
Can someone please help me with this problem?? **It's high-school geometry.
jeka94

Hello!

Answer:

\huge\boxed{59.04 units}

To solve, we will need to use Right-Triangle Trigonometry:

Begin by solving for angles ∠S and ∠R using tangent (tan = opp/adj)

tan ∠S = a / (1/2b)

tan ∠S = 3√5 / 14

tan ∠S ≈ 0.479

arctan 0.479  = m∠S (inverse)

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Use cosine to solve for the hypotenuse, or the missing side-length:

cos ∠S = 14 / x

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Both triangles are congruent, so we can go ahead and find the perimeter of the figure:

RS + RQ + QS = 28 + 15.52 + 15.52 = 59.04 units.

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3 years ago
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