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2 years ago

9

(x 2 + 2x - 1)(x 2 + 2x + 5) find product

1 answer:

tankabanditka [31]2 years ago

7 0

Answer:

x^4+4x^3+8x^2+8x-5

Step-by-step explanation:

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Find the range of the function f(x)=(x-2)^2+2

kari74 [83]

The function is in vertex form of a quadratic function.
Vertex form is f(x)=a(x-h)^2+k
The vertex is (h, k)

In this case, our vertex is (2, 2). The a value is positive which means that the graph points up. Since the graph points up, the vertex is the minimum, the lowest point of the graph. The range refers to y-values, so the range of the function is y/y>2. It can also be written as 2≤y<span>≤positive infinity.</span>

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First-order linear differential equations

kkurt [141]

Answer:

$(1)\ logy\ =\ -sint\ +\ c$

$(2)\ log(y+\dfrac{1}{2})\ =\ t^2\ +\ c$

Step-by-step explanation:

1. Given differential equation is

$\dfrac{dy}{dt}+ycost = 0$

$=>\ \dfrac{dy}{dt}\ =\ -ycost$

$=>\ \dfrac{dy}{y}\ =\ -cost dt$

On integrating both sides, we will have

$\int{\dfrac{dy}{y}}\ =\ \int{-cost\ dt}$

$=>\ logy\ =\ -sint\ +\ c$

Hence, the solution of given differential equation can be given by

$logy\ =\ -sint\ +\ c.$

2. Given differential equation,

$\dfrac{dy}{dt}\ -\ 2ty\ =\ t$

$=>\ \dfrac{dy}{dt}\ =\ t\ +\ 2ty$

$=>\ \dfrac{dy}{dt}\ =\ 2t(y+\dfrac{1}{2})$

$=>\ \dfrac{dy}{(y+\dfrac{1}{2})}\ =\ 2t dt$

On integrating both sides, we will have

$\int{\dfrac{dy}{(y+\dfrac{1}{2})}}\ =\ \int{2t dt}$

$=>\ log(y+\dfrac{1}{2})\ =\ 2.\dfrac{t^2}{2}\ + c$

$=>\ log(y+\dfrac{1}{2})\ =\ t^2\ +\ c$

Hence, the solution of given differential equation is

$log(y+\dfrac{1}{2})\ =\ t^2\ +\ c$

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2 years ago

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mr Goodwill [35]

Answer:

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Step-by-step explanation:

I did the test and got this right.

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Answer:

Step-by-step explanation:8

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2 years ago

Some body plis help me

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Answer:

im not positive but I think the answer is 7x+5

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