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3 years ago

The function f(x) = x2 + 10x – 3 written in vertex form is f(x) = (x + 5)2 – 28. What are the coordinates of the vertex?

1 answer:

4 0

Vertex form uses the following formula:

$y = a(x-h)^2 + k$

Identify the values of h and k in your function:

$(x+5)^2 - 28$
$h = -5, k = -28$

The vertex is found with the following point:

$(h,k)$
$\text{Vertex: (-5, -28)}$

The vertex will be found at (-5, -28).

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Calculate the Laplace transforms of the following from the definition. 1. y = t^2. y = t^3

Likurg_2 [28]

Answer:

1) $L(y)=\frac{2}{s^{3}}$

2) $L(y)=\frac{6}{s^{4}}$

Step-by-step explanation:

To find : Calculate the Laplace transforms of the following from the definition ?

Solution :

We know that,

Laplace transforms of $t^n$ is given by,

$L(t^n)=\frac{n!}{s^{n+1}}$

1) $y=t^2$

Laplace of y,

$L(y)=L(t^2)$ here n=2

$L(y)=\frac{2!}{s^{2+1}}$

$L(y)=\frac{2}{s^{3}}$

2) $y=t^3$

Laplace of y,

$L(y)=L(t^3)$ here n=3

$L(y)=\frac{3!}{s^{3+1}}$

$L(y)=\frac{3\times 2}{s^{4}}$

$L(y)=\frac{6}{s^{4}}$

5 0

3 years ago

Could someone help me, please?

k0ka [10]

We can rewrite the expression under the radical as

$\dfrac{81}{16}a^8b^{12}c^{16}=\left(\dfrac32a^2b^3c^4\right)^4$

then taking the fourth root, we get

$\sqrt[4]{\left(\dfrac32a^2b^3c^4\right)^4}=\left|\dfrac32a^2b^3c^4\right|$

Why the absolute value? It's for the same reason that

$\sqrt{x^2}=|x|$

since both $(-x)^2$ and $x^2$ return the same number $x^2$ , and $|x|$ captures both possibilities. From here, we have

The absolute values disappear on all but the $b$ term because all of $\dfrac32$ , $a^2$ and $c^4$ are positive, while $b^3$ could potentially be negative. So we end up with