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kirill [66]
3 years ago
13

Thomas paid a one-time registration

Mathematics
2 answers:
Mkey [24]3 years ago
3 0

Answer:

There is no error

Step-by-step explanation:

He stated with 15 15x and made 10 for each y=15+10

ziro4ka [17]3 years ago
3 0

Step-by-step explanation:

y= 25 this is the answer because the other person jdhddhrjrjr

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The answer to this is A.) 8/17

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Solve this please <br> 6x = 1,620
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6 x 270= 1,620
I hope this helped
8 0
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Read 2 more answers
I need help please Asap
s2008m [1.1K]

Answer:

Step-by-step explanation:

1.

Equation one:

x = -5, x = -1 (Both are real)

Equation two:

No real solutions

Equation three:

x = -3 (Real)

Equation four:

No real solutions

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The easiest way to figure out if an equation has real solutions is to factor it. If it is factorable, then it has real solutions. If it isn't, then it doesn't have real solutions.

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1 year ago
27. Stuart wants to play a trick on his friend by hiding his pencil in the box of tissues shown below. If
Alex17521 [72]

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5 0
3 years ago
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:

<u>The Derivative of a Function</u>

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
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