2 years ago

James wants to purchase a set of headphones.

Mathematics

1 answer:

Fudgin [204]2 years ago

6 0

52
Wants 20 charters to replie

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QUICK I NEED HELP!!

natta225 [31]

7 (6) + 14!!! And this needs to be 20 characters long so I’m adding this

4 0

3 years ago

Ming rolls a number cube, tosses a coin, and chooses a card from two cards marked a and b. if an even number and heads appears,

Zarrin [17]

B I think I am petty sure

7 0

2 years ago

How do I go about solving (27x^3/8y^9)^5/3? And what is the role of the numerator and denominator?

MrRissso [65]

$\left( \frac{27x^3}{8y^9}\right)^ \frac{5}{3} \\\\\\ =\left( \frac{(3x)^3}{(2y^3)^3}\right)^ \frac{5}{3} \\\\\\ = \frac{(3x)^{3 \times \frac{5}{3} }}{(2y^3)^{3 \times \frac{5}{3} }} \\\\\\ =\frac{(3x)^5}{(2y^3)^{5 }} \\\\\\ =\frac{243x^5}{32y^{15}}$

Now, If the exponent was negative like you asked....

$\left( \frac{27x^3}{8y^9}\right)^ {-\frac{5}{3}} \\\\\\ =\left( \frac{8y^9}{27x^3}\right)^ {\frac{5}{3}}\\\\\\ =\left( \frac{(2y^3)^3}{(3x)^3}\right)^ \frac{5}{3} \\\\\\ = \frac{(2y^3)^{3 \times \frac{5}{3} }}{(3x)^{3 \times \frac{5}{3} }} \\\\\\ =\frac{(2y^3)^{5 }}{(3x)^5} \\\\\\ =\frac{32y^{15}}{243x^5}$

5 0

2 years ago

A woman has 7 keys. only one of them will open her door. when seh arrives at the door there is no sufficient light for her to fi

Gala2k [10]

Since she has seven key the probability that the first one will open the door is
$\frac{1}{7}$
if it takes three attempts to open the door,the probability is
$\frac{1}{3}$
if the three attempts would not work
then the probability that the fourth attempt will open the door is
$\frac{1}{4}$
hope this helps

7 0

3 years ago

900 is 4 1/2% of what?

Svet_ta [14]

Simple, first change the 4.5% into 0.045

Thus,

900=0.045*n

900/0.045=0.045n/0.045

20,000=n

Thus, your answer is 20,000.

3 0

2 years ago

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