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DochEvi [55]
3 years ago
9

Solve for x in terms of r, s and t : s=2x+t/r

Mathematics
1 answer:
Anna007 [38]3 years ago
4 0

For this case we have the following equation:

s = \frac {2x + t} {r}

We must clear the value of the variable "x" as a function of r, s and t:

If we multiply by "r" on both sides of the equation we have:

rs =2x + t

If we subtract "t" on both sides of the equation we have:

rs-t = 2x

If we divide by "2" on both sides of the equation we have:

x = \frac {rs-t} {2}

Thus, the value of the variable "x" is:

x = \frac {rs-t} {2}

Answer:

x = \frac {rs-t} {2}

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Answer:

I think they would be 7 4/6 & 5 2/8

Step-by-step explanation:

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Find the limit of the function by using direct substitution.
serg [7]

Answer:

Option a.

\lim_{x \to \frac{\pi}{2}}(3e)^{xcosx}=1

Step-by-step explanation:

You have the following limit:

\lim_{x \to \frac{\pi}{2}{(3e)^{xcosx}

The method of direct substitution consists of substituting the value of \frac{\pi}{2} in the function and simplifying the expression obtained.

We then use this method to solve the limit by doing x=\frac{\pi}{2}

Therefore:

\lim_{x \to \frac{\pi}{2}}{(3e)^{xcosx} = \lim_{x\to \frac{\pi}{2}}{(3e)^{\frac{\pi}{2}cos(\frac{\pi}{2})}

cos(\frac{\pi}{2})=0\\

By definition, any number raised to exponent 0 is equal to 1

So

\lim_{x\to \frac{\pi}{2}}{(3e)^{\frac{\pi}{2}cos(\frac{\pi}{2})} = \lim_{x\to \frac{\pi}{2}}{(3e)^{\frac{\pi}{2}(0)}\\\\

\lim_{x\to \frac{\pi}{2}}{(3e)^{0}} = 1

Finally

\lim_{x \to \frac{\pi}{2}}(3e)^{xcosx}=1

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3 years ago
5×100+9×1+41/100 as a decimal
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Answer:

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5X100=500  and 9X1=9 and 41/100=0.41 Then put them back into the equation

5X100+9X1+41/100=

 500  +  9  +  0.41  =  Now add.  509.41

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3 years ago
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