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vova2212 [387]
3 years ago
8

Simplify 18÷3x2+1 please

Mathematics
2 answers:
telo118 [61]3 years ago
7 0

Answer:

the answer is gonna be

6x^2+1

Ganezh [65]3 years ago
5 0

\huge\textsf{Hey there!}

\huge\textsf{Just do PEMDAS}\\\\\\\huge\textsf{Parentheses}\\\\\huge\textsf{Exponents}\\\\\huge\textsf{Multiplication}\\\\\huge\textsf{Division}\\\\\huge\textsf{Addition}\\\\\huge\textsf{Subtraction}

\large\textsf{18 }\div\large\textsf{ 3 }\times\large\textsf{ 2 + 1}\\\\\\\large\textsf{18 }\div\large\textsf{ 3 = \bf 6}\\\\\\\large\textsf{6 }\times\large\textsf{ 2 + 1}\\\\\\\large\textsf{6 }\times\large\textsf{ 2 = \bf 12}\\\\\\\large\textsf{12 + 1}\\\\\\\large\textsf{= \bf 13}\\\\\\\boxed{\boxed{\huge\textsf{Answer: \bf 13}}}\huge\checkmark

\huge\textsf{Or you could if your equation is: }\\\\\large\textsf{18}\div\large\textsf{3x}\mathsf{^2}\large\textsf{ + 1}

\large\textsf{18}\div\large\textsf{3x}\mathsf{^2}\large\textsf{ + 1}\\\\\large\textsf{18}\div\large\textsf{3x}\mathsf{^2}\large\textsf{ = \bf 6x}\mathsf{\bf ^2}\\\\\large\textsf{= \bf 6x}\mathsf{\bf ^2  \ + }\large\textsf{ \bf 1}\\\\\\\\\\\boxed{\boxed{\large\textsf{Answer:  \bf 6x}\mathsf{\bf ^2}\large\textsf{ \bf + 1}}}\huge\checkmark

\large\textsf{Good luck on your assignment and enjoy your day!}

~\frak{Amphitrite1040:)}

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Find the equation of the line that is perpendicular to y=5 through the point (-4,-6)
Nikolay [14]

Answer:

x=-4

Step-by-step explanation:

we know that

The given equation y=5 is a horizontal line (is parallel to the x-axis)

The slope of the given line is equal to zero

A perpendicular line to the given line is a vertical line (parallel to the y-axis)

so

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4 0
3 years ago
B) Let g(x) =x/2sqrt(36-x^2)+18sin^-1(x/6)<br><br> Find g'(x) =
jolli1 [7]

I suppose you mean

g(x) = \dfrac x{2\sqrt{36-x^2}} + 18\sin^{-1}\left(\dfrac x6\right)

Differentiate one term at a time.

Rewrite the first term as

\dfrac x{2\sqrt{36-x^2}} = \dfrac12 x(36-x^2)^{-1/2}

Then the product rule says

\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 x' (36-x^2)^{-1/2} + \dfrac12 x \left((36-x^2)^{-1/2}\right)'

Then with the power and chain rules,

\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12\left(-\dfrac12\right) x (36-x^2)^{-3/2}(36-x^2)' \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} - \dfrac14 x (36-x^2)^{-3/2} (-2x) \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12 x^2 (36-x^2)^{-3/2}

Simplify this a bit by factoring out \frac12 (36-x^2)^{-3/2} :

\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-3/2} \left((36-x^2) + x^2\right) = 18 (36-x^2)^{-3/2}

For the second term, recall that

\left(\sin^{-1}(x)\right)' = \dfrac1{\sqrt{1-x^2}}

Then by the chain rule,

\left(18\sin^{-1}\left(\dfrac x6\right)\right)' = 18 \left(\sin^{-1}\left(\dfrac x6\right)\right)' \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac x6\right)'}{\sqrt{1 - \left(\frac x6\right)^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac16\right)}{\sqrt{1 - \frac{x^2}{36}}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{3}{\frac16\sqrt{36 - x^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18}{\sqrt{36 - x^2}} = 18 (36-x^2)^{-1/2}

So we have

g'(x) = 18 (36-x^2)^{-3/2} + 18 (36-x^2)^{-1/2}

and we can simplify this by factoring out 18(36-x^2)^{-3/2} to end up with

g'(x) = 18(36-x^2)^{-3/2} \left(1 + (36-x^2)\right) = \boxed{18 (36 - x^2)^{-3/2} (37-x^2)}

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2 years ago
you make two cylinders using 8.5-by-11-inch pieces of paper. One has a height of 8.5 inches, and the other has a height of 11 in
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The second cylinder is taller correct? 
8 in < 11 in 

If the height is also larger, the base is most likely larger too.

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