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3 years ago

Shakia buys a bunk bed orginally priced at $119.99 at a discounted price of $71.99. What is the discount percent?

Mathematics

1 answer:

Ainat [17]3 years ago

4 0

Answer:

40%

Step-by-step explanation:

119.99 - 71.99 = 48.00 is 40%

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Which is greater 52 + (-77) or 52 - (-77)

Kay [80]

52 + (-77) =

52 - 77 = -25

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

52 - (-77) =

52 + 77 = 129

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

your answer is: 52 - (-77)

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3 years ago

Please help with this question!

ZanzabumX [31]

S > less than or equal to 130
d < greater than or equal to $500

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2 years ago

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Brent created a mural, 4/5 of which he painted in shades of blue. Three eighths of the blue area is sky blue. What portion of th

Snezhnost [94]

Answer: $\frac{3}{10}$ of the mural is sky blue.

Step-by-step explanation:

Let x be the full portion of mural.

Since, $\frac{4}{5}$ of mural is painted in shades of blue.

Therefore, the blue portion= $\frac{4}{5}x$

Also, Three eighths of the blue area is sky blue.

Therefore, the portion of the mural is sky blue= $\frac{3}{8}\times\frac{4}{5}x$

The portion of the mural is sky blue= $\frac{12}{40}x$

Simplify, by dividing numerator and denominator by 4, we get

The portion of the mural is sky blue= $\frac{3}{10}x$

5 0

2 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

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

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

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julia-pushkina [17]

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