Any help please if possible Thank you ❤❤❤

Mathematics

1 answer:

creativ13 [48]3 years ago

5 0

155 - 2⁷ = 27 =3³
155 - 2⁷ = 3³
2⁷ + 3³ =155

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consider the graph of the following qudratic equation. y= -x^2 -10x + 24.What is the y-value of the vertex?

3241004551 [841]

Given that the quadratic equation is $y=-x^{2}-10 x+24$

We need to determine the y - value of the vertex.

The x - value of the vertex:

The x - value of the vertex can be determined using the formula,

$x=-\frac{b}{2 a}$

where $a=-1, b=-10, c=24$

Substituting these values, we get;

$x=-\frac{(-10)}{2(-1)}$

Simplifying the terms, we get;

$x=-\frac{-10}{-2}$

$x=-5$

Thus, the x - value of the vertex is -5.

The y - value of the vertex:

The y - value of the vertex can be determined by substituting the x - value of the vertex ( x = -5) in the equation $y=-x^{2}-10 x+24$

Thus, we get;

$y=-(-5)^{2}-10(-5)+24$

Simplifying the values, we have;

$y=-25+50+24$

$y=49$

Thus, the y - value of the vertex is 49.

8 0

2 years ago

Use chain rule to find dw/dt w=ln(x^2 + y^2 + z^2)^(1/2) , x=sint, y=cost, z=tant

son4ous [18]

The chain rule states that

$\dfrac{\mathrm dw}{\mathrm dt}=\dfrac{\partial w}{\partial x}\dfrac{\mathrm dx}{\mathrm dt}+\dfrac{\partial w}{\partial y}\dfrac{\mathrm dy}{\mathrm dt}+\dfrac{\partial w}{\partial z}\dfrac{\mathrm dz}{\mathrm dt}$

Since $\ln(x^2+y^2+z^2)^{1/2}=\dfrac12\ln(x^2+y^2+z^2)$ , you have

$\dfrac{\partial w}{\partial x}=\dfrac x{x^2+y^2+z^2}$
$\dfrac{\partial w}{\partial x}=\dfrac y{x^2+y^2+z^2}$
$\dfrac{\partial w}{\partial z}=\dfrac z{x^2+y^2+z^2}$

and

$\dfrac{\mathrm dx}{\mathrm dt}=\cos t$
$\dfrac{\mathrm dy}{\mathrm dt}=-\sin t$
$\dfrac{\mathrm dz}{\mathrm dt}=\sec^2t$

Also, since $x^2+y^2+z^2=\sin^2t+\cos^2t+\tan^2t=1+\tan^2t=\sec^2t$ , the derivative is

$\dfrac{\mathrm dw}{\mathrm dt}=\dfrac{\sin t\cos t}{\sec^2t}-\dfrac{\sin t\cos t}{\sec^2t}+\dfrac{\tan t\sec^2t}{\sec^2t}$
$\dfrac{\mathrm dw}{\mathrm dt}=\tan t$