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2 years ago

5

Hi. I need help wih this question (see image). Please show workings.

1 answer:

Kipish [7]2 years ago

4 0

Answer:

$\frac{dy}{dx}$ = 12x³ - 6x² - 7

Step-by-step explanation:

Differentiate each term using the power rule

$\frac{d}{dx}$ (a $x^{n}$ ) = na $x^{n-1}$

Given

y = 3 $x^{4}$ - 2x³ - 7x + 5, then

$\frac{dy}{dx}$ = 4.3 $x^{4-1}$ - 3.2 $x^{3-1}$ - 7 $x^{1-1}$ + 0

= 12x³ - 6x² - 7

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What would be the answer to g+3=17

andrew-mc [135]

Answer:

g = 14

Step-by-step explanation:

g + 3 = 17

minus 3 from both sides

g = 14

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2 years ago

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Step-by-step explanation:

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2 years ago

What divides a line segment into two congruent segments?

Nutka1998 [239]

Answer:

segment bisector

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3 years ago

2m-nx=x+4<br> Solve for x

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3 years ago

Suppose that the Celsius temperature at the point (x, y) in the xy-plane is T(x, y) = x sin 2y and that distance in the xy-plane

liraira [26]

Missing information:

How fast is the temperature experienced by the particle changing in degrees Celsius per meter at the point

$P = (\frac{1}{2}, \frac{\sqrt 3}{2})$

Answer:

$Rate = 0.935042^\circ /cm$

Step-by-step explanation:

Given

$P = (\frac{1}{2}, \frac{\sqrt 3}{2})$

$T(x,y) =x\sin2y$

$r = 1m$

$v = 2m/s$

Express the given point P as a unit tangent vector:

$P = (\frac{1}{2}, \frac{\sqrt 3}{2})$

$u = \frac{\sqrt 3}{2}i - \frac{1}{2}j$

Next, find the gradient of P and T using: $\triangle T = \nabla T * u$

Where

$\nabla T|_{(\frac{1}{2}, \frac{\sqrt 3}{2})} = (sin \sqrt 3)i + (cos \sqrt 3)j$

So: the gradient becomes:

$\triangle T = \nabla T * u$

$\triangle T = [(sin \sqrt 3)i + (cos \sqrt 3)j] * [\frac{\sqrt 3}{2}i - \frac{1}{2}j]$

By vector multiplication, we have:

$\triangle T = (sin \sqrt 3)* \frac{\sqrt 3}{2} - (cos \sqrt 3) \frac{1}{2}$

$\triangle T = 0.9870 * 0.8660 - (-0.1606 * 0.5)$

$\triangle T = 0.9870 * 0.8660 +0.1606 * 0.5$

$\triangle T = 0.935042$

Hence, the rate is:

$Rate = \triangle T = 0.935042^\circ /cm$

3 0

2 years ago