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3 years ago

(X^4-81)/x^3+3x^2+9x+27

Mathematics

1 answer:

Iteru [2.4K]3 years ago

6 0

Answer:

x-3

Step-by-step explanation:

(x^2-9)(x^2-9)/x^2*(x+3)+9(x+3)

(x+3)*(x^2+9)/(x+3)*(x^2+9) cross multiply

(x^2+9)*(x+3)/(x^2+9) cross multiply

x+3 final anser

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B(n)=1 (-2)^n-1<br> What is the 4th term in the sequence

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Answer:

The fourth term of the sequence B(4) =-8

The sequence is 1,-2,4,-8...

Step-by-step explanation:

<u><em>Explanation:-</em></u>

Given that the sequence

B(n) = 1(-2)ⁿ⁻¹

Put n=1

B(1) = 1(-2)¹⁻¹ = 2⁰ =1

Put n=2

B(2) = 1(-2)²⁻¹ = (-2)¹ =-2

Put n=3

B(3) =1(-2)³⁻¹ = (-2)² =4

Put n=4

B(4) = 1(-2)⁴⁻¹ =(-2)³ =-8

The sequence

1,-2,4,-8...

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To make a trail mix you must have 4 cups of pretzels and 2 cups of bread chips.What is the ratio of pretzels to bread chips?

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4:2 which can be converted to 2:1

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<h2>✒️VALUE</h2>

$\\ \quad \begin{array}{c} \qquad \bold{Distance \: \green{ Formula:}}\qquad\\ \\ \boldsymbol{ \tt d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}} \end{array}\\ \begin{array}{l} \\ 1.)\: \bold{Given:}\: \begin{cases}\tt D(- 5,6), E(2.-1),\textsf{ and }F(x,0) \\ \tt DF = EF \end{cases} \\ \\ \qquad\bold{Required:}\:\textsf{ value of }x \\ \\ \qquad \textsf{Solving for }x, \\ \\ \tt \qquad DF = EF \\ \\ \implies\small \tt{\sqrt{(x -(- 5))^2 + (0 - 6)^2} = \sqrt{(x - 2)^2 + (0 - (-1))^2}} \\ \\ \implies\tt\sqrt{(x + 5)^2 + 36 } = \sqrt{(x - 2)^2 + 1 } \\ \\ \textsf{Squaring both sides yields} \\ \\ \implies\tt (x + 5)^2 + 36 = (x - 2)^2 + 1 \\ \\ \implies\tt x^2 + 10x + 25 + 36 = x^2 - 4x + 4 + 1 \\ \\ \implies \tt x^2 + 10x + 61 = x^2 - 4x + 5 \\ \\ \implies\tt10x + 4x = 5 - 61 \\ \\ \implies\tt14x = -56 \\ \\ \implies \red{\boxed{\tt x = -4}}\end{array} \\ \\ \\ \\\begin{array}{l} \\ 2.)\: \bold{Given:}\: \begin{cases}\tt P(6,-1), Q(-4,-3),\textsf{ and }R(0,y) \\ \tt PR = QR \end{cases} \\ \\ \bold{Required:}\:\textsf{ value of }y \\ \\ \qquad\textsf{Solving for }y, \\ \\ \qquad\tt PR = QR \\ \\ \implies \tt\small{\sqrt{(0 - 6)^2 + (y - (-1))^2} = \sqrt{(0 - (-4))^2 + (y - (-3))^2}} \\ \\ \implies\tt\sqrt{36 + (y + 1)^2} = \sqrt{16 + (y + 3)^2 } \\ \\ \textsf{Squaring both sides yields} \\ \\ \implies \tt \: 36 + (y + 1)^2 = 16 + (y + 3)^2 \\ \\ \implies\tt 36 + y^2 + 2y + 1 = 16 + y^2 + 6y + 9 \\ \\ \implies \tt \: y^2 + 2y + 37 = y^2 + 6y + 25 \\ \\ \implies \tt \: 2y - 6y = 25 - 37 \\ \\ \implies \tt -4y = -12 \\ \\ \implies\red{\boxed{ \tt y = 3}} \end{array} \\ \\ \\ \begin{array}{l} \\ 3.)\: \bold{Given:}\: \begin{cases}\: A(4,5), B(-3,2),\textsf{ and }C(x,0) \\ \: AC = BC \end{cases} \\ \\ \bold{Required:}\:\textsf{ value of }x \\ \\ \qquad\textsf{Solving for }x, \\ \\ \qquad\tt AC = BC \\ \\ \implies\tt\small{\sqrt{(x - 4)^2 + (0 - 5)^2} = \sqrt{(x - (-3))^2 + (0 - 2)^2}} \\ \\ \implies\tt\sqrt{(x - 4)^2 + 25} = \sqrt{(x + 3)^2 + 4} \\ \\ \textsf{Squaring both sides yields} \\ \\ \implies\tt\:(x - 4)^2 + 25 = (x + 3)^2 + 4 \\ \\ \implies\tt\:x^2 - 8x + 16 + 25 = x^2 + 6x + 9 + 4 \\ \\ \implies\tt\:x^2 - 8x + 41 = x^2 + 6x + 13 \\ \\ \implies\tt-8x - 6x = 13 - 41 \\ \\\implies\tt -14x = -28 \\ \\ \implies\red{\boxed{\tt\:x = 2}} \end{array}$

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7 0

2 years ago