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4 years ago

If f(x)=3^2 and g(x)=4x^3+1 , what is the degree of (f*g)(x)?

1 answer:

N76 [4]4 years ago

5 0

I'm assuming you meant to say f(x) = 3x^2. If that assumption is correct, then the degree is found by multiplying the leading terms from both f(x) and g(x). The leading terms are 3x^2 and 4x^3

3x^2*4x^3 = (3*4)*(x^2*x^3) = 12x^(2+3) = 12x^5

The exponent of that result is 5, so the degree of (f*g)(x) is 5

Answer: 5

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2(x 4) = 35 − (7 − 4x)

Rus_ich [418]

2(x4) = 35 - (7-4x)
+4x
2(x8)= 28
/2 /2
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6 0

3 years ago

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ad-work [718]

Answer:

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Step-by-step explanation:

6 0

3 years ago

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Arisa [49]

It should be C., is that what the graph looks like?

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4 years ago

Which of the following is equivalent to tan2θcos(2θ) for all values of θ for which tan2θcos(2θ) is defined?

Aloiza [94]

Answer:

2sin²θ - tan²θ

Step-by-step explanation:

Given

tan²θcos(2θ)

Required

Simplify

We start by simplifying cos(2θ)

cos(2θ) = cos(θ+θ)

From Cosine formula

cos(A+A) = cosAcosA - sinAsinA

cos(A+A) = cos²A - sin²A

By comparison

cos(2θ) = cos(θ+θ)

cos(2θ) = cos²θ - sin²θ ----- equation 1

Recall that cos²θ + sin²θ = 1

Make sin²θ the subject of formula

sin²θ = 1 - cos²θ

Substitute sin²θ = 1 - cos²θ in equation 1

cos(2θ) = cos²θ - (1 - cos²θ)

cos(2θ) = cos²θ - 1 +cos²θ

cos(2θ) = cos²θ + cos²θ - 1

cos(2θ) = 2cos²θ - 1

Substitute 2cos²θ - 1 for cos(2θ) in the given question

tan²θcos(2θ) becomes

tan²θ(2cos²θ - 1)

Open brackets

2cos²θtan²θ - tan²θ

------------------------

Simplify tan²θ

tan²θ = (tanθ)²

Recall that tanθ = sinθ/cosθ

So, we have

tan²θ = (sinθ/cosθ)²

tan²θ = sin²θ/cos²θ

------------------------

Substitute sin²θ/cos²θ for tan²θ

2cos²θtan²θ - tan²θ becomes

2cos²θ(sin²θ/cos²θ) - tan²θ

Open bracket (cos²θ will cancel out cos²θ) to give

2(sin²θ) - tan²θ

2sin²θ - tan²θ

Hence, the simplification of tan²θcos(2θ) is 2sin²θ - tan²θ

Option E is correct

7 0

3 years ago

A. Solve the differential equation <img src="https://tex.z-dn.net/?f=y%27%3D2x%20%5Csqrt%7B1-y%5E2%7D%20" id="TexFormula1" title

kirill [66]

$y' = \frac{dy}{dx}$

seperable differential equations will have the form
$\frac{dy}{dx} = F(x) G(y)$

what you do from here is isolate all the y terms on one side and all the X terms on the other
$\frac{dy}{G(y)} = F(x) dx$
just divided G(y) to both sides and multiply dx to both sides

then integrate both sides
$\int \frac{1}{G(y)} dy = \int F(x) dx

$

once you integrate, you will have a constant. use the initial value condition to solve for the constant, then try to isolate x or y if the question asks for it

In your problem,
$G(y) = \sqrt{1-y^2}

F(x) = 2x$

so all you need to integrate is
$\int \frac{1}{\sqrt{1-y^2}} dy = \int 2x dx$

5 0

4 years ago