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Viktor [21]
3 years ago
13

Question 2

Mathematics
1 answer:
slamgirl [31]3 years ago
6 0

Answer:

37/117

Step-by-step explanation:

There are 117 gumballs and 37 are green.

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Which simplified fraction is equal to 0.53 with the three repeating?
topjm [15]

Answer:

24/46 = 0.52173913

8/15 = 0.53333333

42/90 = 0.46666666

5/9 = 0.55555555

answer is 8/15

8 0
3 years ago
What is the answer to this math problem
pishuonlain [190]

Answer:

Domain: c)

Range: d)

Step-by-step explanation:

The domain means all of the possible x-values. To find the domain, find the leftmost and rightmost value of x. They are 0 and 8.5.

The range means all of the possible y-values. To find the range, find the lowest and higher values for y. They are 0 and 12.5.

"All numbers" means x or y can be any number, decimal decimal numbers. "Whole numbers" mean x or y can only be numbers that don't have decimals.

Both x and y can be decimal numbers because the graph is not broken up. If it could only be whole numbers, the graph would be dots.

4 0
3 years ago
What is the range and domain of f(x)=60-0.18x
Gnom [1K]

The domain is always the value of x, but there is no value of x shown..

3 0
3 years ago
Drag each tile to the correct location. Not all tiles will be used.
Novosadov [1.4K]

Answer:

25:64

Step-by-step explanation:

surface : 4×pi×R^2

R1 = 5x

R2 =8x

4piR1^2 / 4piR2^2 = R1^2/R2^2 = 5x^2/8x^2 = 25x^2/64x^2 = 25/64

4 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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