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Illusion [34]
3 years ago
14

You and a friend went to the movies.You spent half your money on tickets. Then you spent 1/4 of the money on popcorn, $2 on cand

y, and $3 on a soda. If you have $1 left, hiw much money did u take to the movies?
Mathematics
1 answer:
bagirrra123 [75]3 years ago
3 0

Answer:

We took $ 24 to the movies.

Step-by-step explanation:

Let total money be

On movies = Half of the money =\frac{x}{2}

On popcorn =  \frac{x}{4}

on candy = $2

On soda = $3

total money spent =  \frac{x}{2} +\frac{x}{4}+ 2+3+1

         x   =  \frac{x}{2} +\frac{x}{4}+ 2+3+1

x- \frac{x}{2} -\frac{x}{4} = 6

\frac{x}{1} -\frac{x}{2} - \frac{x}{4}  =6

    \frac{4x-2x-x}{4}  = 6\\          \frac{x}{4} =6\\x =24

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Vlad1618 [11]

Answer:

A f(x) is a negative number

Step-by-step explanation:

Lets see what happens

f(x) = 1/x

Let x get smaller and smaller from the negative side

x = -1/2

f(-1/2) = 1/ (-1/2) = -2

x = -1/3

f(-1/3) = 1/ (-1/3) = -3

x = -1/10

f(-1/10) = 1/-1/10 = -10

What is happening

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4 years ago
10 to the zero power times 2 to the 1 power​
bezimeni [28]

the answer is two :D

8 0
3 years ago
Imagine the United States only had two postage stamps, a 3 cent stamp and a 7 cent stamp. If you can put any number of stamps on
cricket20 [7]

Answer:

  • not possible:  1, 2, 4, 5, 8, 11
  • possible: all others

Step-by-step explanation:

Obviously, all multiples of 3 and 7 can be paid.

1 cent can be added by adding a 7-cent stamp and taking away two 3-cent stamps. 2 cents can be added by adding three 3-cent stamps and taking away one 7-cent stamp. Hence any value more than 7 +2(3) = 13 cents can always be accommodated.

Amounts that are impossible:

  1, 2, 4, 5, 8, 11 cents

All other positive integer amounts are possible.

7 0
4 years ago
Someone please tell me what the cube root of x to the power of 3 is?
defon

Answer:

x

Step-by-step explanation:

The cube root of x to the power of 3 is

( \sqrt[3]{x} )³ = ( x^{\frac{1}{3} } )³ = x

or

\sqrt[3]{x^3} = x

5 0
3 years ago
<img src="https://tex.z-dn.net/?f=%5Clim_%7Bh%20%5Cto%20%5C116%7D%20x-16%2F%5Csqrt%7Bx%7D%20-4" id="TexFormula1" title="\lim_{h
Evgen [1.6K]

Answer:

\displaystyle    8

Step-by-step explanation:

we would like to compute the following limit

\displaystyle \lim_{x \to 16} \left( \frac{x - 16}{ \sqrt{x}  - 4}  \right)

if we substitute 16 directly we'd end up

\displaystyle = \frac{16 - 16}{ \sqrt{16}  - 4}

\displaystyle = \frac{0}{ 0}

which isn't a good answer now notice that we have a square root on the denominator so we can rationalise the denominator to do so multiply the expression by √x+4/√x+4 which yields:

\displaystyle \lim_{x \to 16} \left( \frac{x - 16}{ \sqrt{x}  - 4} \times  \frac{ \sqrt{x} +  4 }{ \sqrt{x} + 4 }   \right)

simplify which yields:

\displaystyle \lim_{x \to 16} \left( \frac{(x - 16)( \sqrt{x}  + 4)}{ x  - 16}  \right)

we can reduce fraction so that yields:

\displaystyle \lim_{x \to 16} \left( \frac{ \cancel{(x - 16)}( \sqrt{x}  + 4)}{  \cancel{x  - 16} } \right)

\displaystyle  \lim _{x \to 16} \left(  \sqrt{x }   + 4\right)

now it's safe enough to substitute 16 thus

substitute:

\displaystyle =   \sqrt{16}   + 4

simplify square root:

\displaystyle  =  4   + 4

simplify addition:

\displaystyle  =  8

hence,

\displaystyle \lim_{x \to 16} \left( \frac{x - 16}{ \sqrt{x}  - 4}  \right)  = 8

6 0
3 years ago
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