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kifflom [539]
3 years ago
11

Please explain this to me

Mathematics
2 answers:
grigory [225]3 years ago
7 0
First, you can convert them to improper fractions (3/2 and 19/6). then, you multiply the numerators (3•19=57). after that, you multiply the denominators (2•6=12). you are left with 57/12. now you just need to change it back to a mixed number. 12•4=48, 57-48=9. The answer is 4 9/12. This can be simplified to 4 3/4.
KengaRu [80]3 years ago
6 0

\huge\text{Hey there!}

\large\boxed{\rm{1\dfrac{1}{2}\times3\dfrac{1}{6}}}\\\\\large\boxed{\rm{\rightarrow \dfrac{3}{2}\times\dfrac{19}{6}}}\\\\\large\boxed{\rightarrow\dfrac{3\times19}{2\times6}}\\\large\boxed{\rightarrow\dfrac{57}{12}}\\\\\large\boxed{\rightarrow \dfrac{57}{12}\approx 4 \dfrac{3}{4}}\\\\\huge\boxed{\mathsf{Therefore, your\ answer\ is: \dfrac{57}{12} \ or\ 4\dfrac{3}{4}}}\huge\checkmark

\large\text{Good luck on your assignment \& enjoy your day!}

~\frak{Amphitrite1040:)}

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Answer:

C. 5/7

Step-by-step explanation:

8 0
2 years ago
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What is the x-intercept of -2x 5y = 20 ? question 1 options: 10 4 -4 -10?
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You mean -2x+5y=20 if it was that's how you do it
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-2x/-2=20/-2
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7 0
3 years ago
Determine whether the integral is convergent or divergent. ∫[infinity] 2 e^−1/x / x^2 dx : O Convergent O divergent If it is con
monitta

Let f(x)=e^{-1/x}. Then f'(x)=\frac1{x^2}e^{-1/x}>0 for all x\ge2, so f is strictly increasing. As x\to\infty, e^{-1/x}\to e^0=1, so f is bounded above by 1. This is to say,

e^{-1/x}

and the integral of \frac1{x^2} converges over the same domain, so this integral must also converge by comparison.

We have, by setting y=-\frac1x,

\displaystyle\int_2^\infty\frac{e^{-1/x}}{x^2}\,\mathrm dx=\int_{-1/2}^0e^y\,\mathrm dy=e^0-e^{-1/2}=1-\frac1{\sqrt e}

8 0
3 years ago
Find the inverse of the function. Show work. <br> g(x)= - 3/x-2 +2
4vir4ik [10]

Answer:

\displaystyle g^{-1}(x)=\frac{-7+2x}{x-2}

Step-by-step explanation:

We are given the function:

\displaystyle g(x)=-\frac{3}{x-2}+2

Let's find the inverse of g.

Call y=g(x):

\displaystyle y=-\frac{3}{x-2}+2

We need to solve for x. Multiply both sides by x-2 to eliminate denominators:

y(x-2)=-3+2(x-2)

Operate:

yx-2y=-3+2x-4

Collect the x's to the left side and the rest to the right side of the equation:

yx-2x=-3-4+2y

Factor the left side and operate on the right side:

x(y-2)=-7+2y

Solve for x:

\displaystyle x=\frac{-7+2y}{y-2}

Interchange variables:

\displaystyle y=\frac{-7+2x}{x-2}

Call y as the inverse function:

\boxed{\displaystyle g^{-1}(x)=\frac{-7+2x}{x-2}}

5 0
3 years ago
Does the table represent y as a function of x? Justify your answer.​
Otrada [13]

Answer:

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Step-by-step explanation:

As per definition of the function, every x- value (input) has exactly one corresponding y-value (output)

<u>We can see on the table:</u>

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  • x = 3.45 and y = 3.36

Two same input values with different outputs.

So this not a function

6 0
3 years ago
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