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4 years ago

Consider the sequence 10, 2..

Mathematics

1 answer:

romanna [79]4 years ago

8 0

Answer: see below

Step-by-step explanation:

1) 10, 2, 1.2, 1.12, 1.112, 1.1112

$t_n=\dfrac{t_{n-1}}{10}+1$

t₁ = 10

t₂ = 10/10 + 1 = 2

t₃ = 2/10 + 1 = 1.2

t₄ = 1.2/10 + 1 = 1.12

t₅ = 1.12/10 + 1 = 1.112

t₆ = 1.112/10 + 1 = 1.1112

2) 10, 2, -2, -4, -5, -5.5, ...

$t_n=\dfrac{t_{n-1}}{2}-3$

t₁ = 10

t₂ = 10/2 - 3 = 2

t₃ = 2/2 - 3 = -2

t₄ = -2/2 - 3 = -4

t₅ = -4/2 - 3 = -5

t₆ = -5/2 - 3 = -5.5

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Jenny runs a daycare center and charges $60 per day for each child. There are currently 50 children enrolled at the daycare. At

Nutka1998 [239]

Answer:

Step-by-step explanation:

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3 years ago

If bc=24 and ad=5y-1, find y

s2008m [1.1K]

Bc = 24
Ad = 5y-1

5 times what is 25?
5 x 5 = 25
25 - 1 = 24

Therefore Y = 5
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4 0

4 years ago

Which of the following describes the transformation of g (x) = 3 (2) Superscript negative x Baseline + 2 from the parent functio

Tanzania [10]

Option d: reflect across the y-axis, stretch the graph vertically by a factor of 3, shift 2 units up.

Step-by-step explanation:

The parent function is $f(x)=2^{x}$

The transformation function is $g(x)=3(2)^{-x} +2$

In the transformed function, the function is added +2, which shifts the graph by 2 units up.

Also, the function is multiplied by 3, which stretches the function $f(x)=2^{x}$ vertically by a factor of 3 units.

The variable x is multiplied by -1, such that the function reflects across y-axis.

Thus, the correct answer is option d.

The graph is attached below which shows the parent function and the transformed function.

The transformation function is reflect across the y-axis, stretch the graph vertically by a factor of 3, shift 2 units up.

8 0

3 years ago

Read 2 more answers

Identify the standard form of the equation by completing the square.

OLEGan [10]

Answer:

$\dfrac{(x-1)^2}{9}-\dfrac{(y-2)^2}{4}=1$

Step-by-step explanation:

Given equation:

$4x^2-9y^2-8x+36y-68=0$

This is an equation for a horizontal hyperbola.

To complete the square for a hyperbola

Arrange the equation so all the terms with variables are on the left side and the constant is on the right side.

$\implies 4x^2-8x-9y^2+36y=68$

Factor out the coefficient of the x² term and the y² term.

$\implies 4(x^2-2x)-9(y^2-4y)=68$

Add the square of half the coefficient of x and y inside the parentheses of the left side, and add the distributed values to the right side:

$\implies 4\left(x^2-2x+\left(\dfrac{-2}{2}\right)^2\right)-9\left(y^2-4y+\left(\dfrac{-4}{2}\right)^2\right)=68+4\left(\dfrac{-2}{2}\right)^2-9\left(\dfrac{-4}{2}\right)^2$

$\implies 4\left(x^2-2x+1\right)-9\left(y^2-4y+4\right)=36$

Factor the two perfect trinomials on the left side:

$\implies 4(x-1)^2-9(y-2)^2=36$

Divide both sides by the number of the right side so the right side equals 1:

$\implies \dfrac{4(x-1)^2}{36}-\dfrac{9(y-2)^2}{36}=\dfrac{36}{36}$

Simplify:

$\implies \dfrac{(x-1)^2}{9}-\dfrac{(y-2)^2}{4}=1$

Therefore, this is the standard equation for a horizontal hyperbola with:

center = (1, 2)
vertices = (-2, 2) and (4, 2)
co-vertices = (1, 0) and (1, 4)
$\textsf{Asymptotes}: \quad y = -\dfrac{2}{3}x+\dfrac{8}{3} \textsf{ and }y=\dfrac{2}{3}x+\dfrac{4}{3}$
$\textsf{Foci}: \quad (1-\sqrt{13}, 2) \textsf{ and }(1+\sqrt{13}, 2)$

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2 years ago

Factorization of 60x^2+48x+24

OLga [1]

Answer:

hope it helps uh..............

5 0

3 years ago