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1 year ago

How many licks does it take to get to the center of a tootsie pop ?

Mathematics

1 answer:

Bess [88]1 year ago

7 0

Answer:

411

Step-by-step explanation:

Yet another study by the University of Michigan concluded that it takes 411 licks to get to the center of a Tootsie Pop. A 1996 study by undergraduate students at Swarthmore College concluded that it takes a median of 144 licks (range 70–222) to get to the center of a Tootsie Pop.

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What value of x makes the equation 3^(x-5)=27^(x+4) true?

SVEN [57.7K]

Answer:

The attached file contains the solution to the question

7 0

1 year ago

When three times the sum of a number and seven is increased by ten, the result is four. What is the number ?

densk [106]

Answer:

x=9 :))

Step-by-step explanation:

First set up your equation.

"3 times" is multiplication

"THE SUM of A NUMBER and seven" that is parenthesis because of the multiplication before it

"is increased by ten" that is an addition at the end

Now we have 3(x+7)+10=4

Solve your equation!

Distribute the 3 now we have 3x+21+10=4

Add like terms

now it should look like this... 3x+31=4

Subtract 31 from both sides and it looks like...

3x=27

Divide by the x-value

x=9

4 0

3 years ago

If I have 2 quarters 12 dimes and 23 pennies how many more pennies will I need in order to have a toldal of $2.00

umka2103 [35]

Answer:

You will need 7 more pennies.

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

Which impression is equivalent to - 3/4 / -(<br> 8 / 12)

Makovka662 [10]

I’m not sure if you can

3 0

2 years ago

If A+B+C=<img src="https://tex.z-dn.net/?f=%5Cpi" id="TexFormula1" title="\pi" alt="\pi" align="absmiddle" class="latex-formula"

seraphim [82]

Answer:

$a + b + c = \pi \\ = > c= \pi - a - b \\ < = > \tan(c) = \tan(\pi - a - b) = -\tan(a + b)$

Step-by-step explanation:

we have:

$\tan(a) + \tan(b) + \tan(c) \\ = \tan(a) + \tan(b) - \tan(a + b) \\ = \tan( a) + \tan(b) - \frac{ \tan(a) + \tan(b) }{1 - \tan(a) \tan(b) } \\ = \frac{ ( \tan(a) + \tan(b) ) \tan(a) \tan(b) }{ \tan(a) \tan(b) - 1 } (1)$

we also have:

$\tan(a) \tan(b) \tan(c) \\ = - \tan(a) \tan(b) \tan(a + b) \\ = \frac{ -(\tan( a ) + \tan(b) ) \tan(a) \tan(b) }{1 - \tan(a) \tan(b) } \\ = \frac{( \tan(a) + \tan(b)) \tan(a) \tan(b) }{ \tan(a) \tan(b) - 1 } (2)$

from (1)(2) => proven

5 0

2 years ago