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3 years ago

What is (x+3) (2x squared+x-3) polynomials

Mathematics

1 answer:

Whitepunk [10]3 years ago

7 0

I'm pretty sure it's 4x+6x²-9

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Kelly runs a restaurant that sells two types of desserts. Kelly knows that the restaurant must make at least 25 dozen and at mos

mash [69]

Answer: Whats is the question asking???

Step-by-step explanation:

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3 years ago

I am having trouble with these two questions . 1) -2+x= -x+2 2) 2+3x=x+7

olga2289 [7]

Answer:

1)x=2. 2)x=5/2

Step-by-step explanation:

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3 years ago

Doughnuts at one shop cost £1.20 each.

charle [14.2K]

Answer: 30p

Step-by-step explanation:

Cost of donuts in the first shop £1.20

Cost at the second shop at 25% higher

= 1.20 x 1.25 = 1.50

Difference in price

= 1.50 - 1.20 = 0.30

3 0

3 years ago

Mike is a car salesman. He earns $200 for every car he sells plus a 3% commission. If Mike sells 3 cars in one week for a total

Serhud [2]

For every car Mike earns------------------ >$200 plus 3% commission

3 cars----------------- > $200*3+$32,343*0.03=600+970.29=$1570.29

Total earnings for a week is $1570.29

6 0

3 years ago

Read 2 more answers

Need help with Calculus 1 inverse trig functions

lidiya [134]

Answer:

$\displaystyle y'=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{2+2x^2+2x\sqrt{1+x^2}}$

Step-by-step explanation:

The Derivative of a Function

The derivative of f, also known as the instantaneous rate of change, or the slope of the tangent line to the graph of f, can be computed by the definition formula

$\displaystyle f'(x)=\lim\limits_{\Delta x \rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$

There are tables where the derivative of all known functions are provided for an easy calculation of specific functions.

The derivative of the inverse tangent is given as

$\displaystyle (tan^{-1}u)'=\frac{u'}{1+u^2}$

Where u is a function of x as provided:

$y=3tan^{-1}(x+\sqrt{1+x^2})$

If we set

$u=(x+\sqrt{1+x^2})$

Then

$\displaystyle u'=1+\frac{2x}{2\sqrt{1+x^2}}$

$\displaystyle u'=1+\frac{x}{\sqrt{1+x^2}}$

Taking the derivative of y

$y'=3[tan^{-1}(x+\sqrt{1+x^2})]'$

Using the change of variables

$\displaystyle y'=3[tan^{-1}u]'=3\frac{u'}{1+u^2}$

$\displaystyle y'=3\frac{u'}{1+u^2}=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{1+(x+\sqrt{1+x^2})^2}$

Operating

$\displaystyle y'=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{1+x^2+2x\sqrt{1+x^2}+1+x^2}$

$\boxed{\displaystyle y'=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{2+2x^2+2x\sqrt{1+x^2}}}$

8 0

3 years ago