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svlad2 [7]
3 years ago
14

A recipe calls for 2 2/3 teaspoons of salt.you can only find 3 of your measuring spoons:a 1/2 teaspoon,a 1/8 teaspoon,and a 1/6

teaspoon
A- what measuring spoon(s)would you use to measure the salt
B-how many scoops of each measuring spoon would you need
Mathematics
1 answer:
anyanavicka [17]3 years ago
7 0
U would measure it with the teaspoon and the 1/6 teaspoon
you would need 2 use the 1/2 teaspoon 4 times to get 2 teaspoons and the `1/6 teaspoon 4 times to get 2/3.

1/6+1/6=2/6*2=4/6 =2/3

hope this helps
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59  is D 
because  with the point (-3,7) you substitute it into the equation, making it: 7=4x+b.  solve for b.  then you have y=4x+19.     work out the algebra in the possible choices and whatever equals y=4x+19 will be the answer. in this case, its D.


60 is C
same as above, you do the algebra of the equation. bring the one over after doing distribution with the 4 and voila!


61 is A
a relatively easy one, all you do is the the slope -4 where m goes, and 3 where b goes. y= -4x+3


62 is C.
this one requires more work.
chose one of the points, in this case (2,7) and put them into the equation. 
but wait, you need a slope! 
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voila!



63 is  C.   y= 1/2x+3

64 is B. (3, -5)

66 is B. negative. the line  goes \ ( not / which is positive)

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70 is C.  2/3.   as before, remember we can but the points into this equation and have (6-4)/(3-0) which = 2/3


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74 can't read 
 
75 can't read.

 
4 0
3 years ago
Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n
aliya0001 [1]

Answer:

f(x)=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}

Step-by-step explanation:

The Maclaurin series of a function f(x) is the Taylor series of the function of the series around zero which is given by

f(x)=f(0)+f^{\prime}(0)x+f^{\prime \prime}(0)\dfrac{x^2}{2!}+ ...+f^{(n)}(0)\dfrac{x^n}{n!}+...

We first compute the n-th derivative of f(x)=\ln(1+2x), note that

f^{\prime}(x)= 2 \cdot (1+2x)^{-1}\\f^{\prime \prime}(x)= 2^2\cdot (-1) \cdot (1+2x)^{-2}\\f^{\prime \prime}(x)= 2^3\cdot (-1)^2\cdot 2 \cdot (1+2x)^{-3}\\...\\\\f^{n}(x)= 2^n\cdot (-1)^{(n-1)}\cdot (n-1)! \cdot (1+2x)^{-n}\\

Now, if we compute the n-th derivative at 0 we get

f(0)=\ln(1+2\cdot 0)=\ln(1)=0\\\\f^{\prime}(0)=2 \cdot 1 =2\\\\f^{(2)}(0)=2^{2}\cdot(-1)\\\\f^{(3)}(0)=2^{3}\cdot (-1)^2\cdot 2\\\\...\\\\f^{(n)}(0)=2^n\cdot(-1)^{(n-1)}\cdot (n-1)!

and so the Maclaurin series for f(x)=ln(1+2x) is given by

f(x)=0+2x-2^2\dfrac{x^2}{2!}+2^3\cdot 2! \dfrac{x^3}{3!}+...+(-1)^{(n-1)}(n-1)!\cdot 2^n\dfrac{x^n}{n!}+...\\\\= 0 + 2x -2^2  \dfrac{x^2}{2!}+2^3\dfrac{x^3}{3!}+...+(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}+...\\\\=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^n\dfrac{x^n}{n}

3 0
3 years ago
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