Hi can someone please help me?

1 answer:

4 0

Both can be right because if the function goes up, turns and goes down, between x = -2 and x = 2, it can happen that f(-2) = f(2) and then the average rate of change is [f(2) - f(-2)] / [2-(-2)] which is 0/4 = 0.

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What are the solutions of the equation x4 – 5x2 - 36 = 0? Use factoring to solve.

balandron [24]

Answer:

x = -2i and x = 2i

x = 3 and x = - 3

Step-by-step explanation:

x^4 - 5x^2 - 36 = 0

let u = x^2 and substitute

we have:

u^2 - 5u - 36 = 0

(u - 9)(u + 4) = 0

u = 9 and u = - 4

substitute back for u

x^2 = 9

x = 3 and x = - 3

x^2 = -4 -- i am going to assume you've learned imaginary numbers; otherwise, you cannot take the square root of a negative number.

x = -2i and x = 2i

5 0

3 years ago

The jar only contains nickels and dimes altogether there are 10 coins total of the value is 90 cents how many nickels and how ma

lilavasa [31]

Write a system of equations and solve.

x + y = 10
5x + 10y = 90

Multiply and combine equations

-5x - 5y = -50
5x + 10y = 90
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5y = 40
y = 8
x = 2

There are 2 nickels and 8 dimes in the jar. Hope this helps! :)

3 0

3 years ago

Finding an Equation of a Tangent Line In Exercise, find an equation of the tangent line to the graph of the function at the give

neonofarm [45]

Answer:

Equation of tangent:

$y = xe^{x}$

At point (1,0):

y = 2.713

Step-by-step explanation:

The equation of tangent line to the function can be calculated by taking the first derivative.

We have,

$y = xe^{x}-e^{x}\\\frac{dy}{dx}=\frac{d}{dx}[ xe^{x} ]-\frac{d}{dx} [e^{x}]\\$

Applying Product Rule:

d/dx [u.v] = (d/dx u) . (v) + (u) . (d/dx v)

Therefore,

$\frac{dy}{dx}=\frac{d}{dx}(x) . e^{x}+x .\frac{d}{dx}(e^{x})- \frac{d}{dx}(e^{x})\\\\\frac{dy}{dx}=(1)(e^{x})+(x)(e^{x})-e^{x}\\ \frac{dy}{dx}=xe^{x}\\\\$