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2 years ago

What are 4 terms of a polynomial

Mathematics

1 answer:

velikii [3]2 years ago

3 0

Monomial binomial trinomial and quadratic

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Please help if you want brianleist!! :) uhm don't mind update- NO LINkS child.. Ty!

GrogVix [38]

12 since there are 12 x’s:)

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2 years ago

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user100 [1]

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2 years ago

Hester sells televisions. She earns a fixed amount for each television and an additional $ 15 if the buyer gets an extended war

tiny-mole [99]

Answer:

$y =mx +bx$

Where y is the total amount earned, m the amount for each extended warranty and b the fixed cost and x the total amount of TVs

For this case the value of x = 16 since we have 16 Tvs with extended warranties and we can do this:

$1200 = 15*16 +16b$

and solving for b we got:

$b= \frac{1200-15*16}{16}= 60$

And then we can conclude that she earns 60 for each TV

Step-by-step explanation:

For this case we can set a linear model like this:

$y =mx +bx$

Where y is the total amount earned, m the amount for each extended warranty and b the fixed cost and x the total amount of TVs

For this case the value of x = 16 since we have 16 Tvs with extended warranties and we can do this:

$1200 = 15*16 +16b$

and solving for b we got:

$b= \frac{1200-15*16}{16}= 60$

And then we can conclude that she earns 60 for each TV

8 0

3 years ago

Please give me the correct answer.Please only answer if you're good at math.

lina2011 [118]

The rate of change is 2 euro per hour. Please mark me brainliest :)

4 0

3 years ago

Find gradient xe^y + 4 ln y = x² at (1, 1)

cricket20 [7]

$xe^y+4\ln y=x^2$

Differentiate both sides with respect to x, assuming y = y(x).

$\dfrac{\mathrm d(xe^y+4\ln y)}{\mathrm dx}=\dfrac{\mathrm d(x^2)}{\mathrm dx}$

$\dfrac{\mathrm d(xe^y)}{\mathrm dx}+\dfrac{\mathrm d(4\ln y)}{\mathrm dx}=2x$

$\dfrac{\mathrm d(x)}{\mathrm dx}e^y+x\dfrac{\mathrm d(e^y)}{\mathrm dx}+\dfrac4y\dfrac{\mathrm dy}{\mathrm dx}=2x$

$e^y+xe^y\dfrac{\mathrm dy}{\mathrm dx}+\dfrac4y\dfrac{\mathrm dy}{\mathrm dx}=2x$

Solve for dy/dx :

$e^y+\left(xe^y+\dfrac4y\right)\dfrac{\mathrm dy}{\mathrm dx}=2x$

$\left(xe^y+\dfrac4y\right)\dfrac{\mathrm dy}{\mathrm dx}=2x-e^y$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2x-e^y}{xe^y+\frac4y}$

If y ≠ 0, we can write

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2xy-ye^y}{xye^y+4}$

At the point (1, 1), the derivative is

$\dfrac{\mathrm dy}{\mathrm dx}\bigg|_{x=1,y=1}=\boxed{\dfrac{2-e}{e+4}}$

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3 years ago