There are nine 'different' slots we can fill in for each value. Thus, we get 9! different ways, including repetition.
Now, we need to account for repetition. This can be visualised as separate letters as below:
s₁ u c₁ c₂ e₁ s₂ s₃ e₂ s₄
We can arrange them like this, and we'd never know the difference:
s₁ u c₂ c₁ e₂ s₃ s₂ e₁ s₄
s₄ u c₁ c₂ e₁ s₁ s₃ e₂ s₂
Thus, we have overcounted by a factor of 2! (c's can change in 2! ways), 2! (e's can change in 2! ways), and 4! (s' can change in 4! ways).
Therefore, our final number of permutations becomes:
<em>Greetings from Brasil... </em>
Check Attachment!!!
According to the statement of the question, we have:
a/b = 3/4
<em>Note that according to the attached figure, </em><em>b > a</em><em>, then the correct proportion will be </em><em>a</em><em> to </em><em>3</em><em>, as well as </em><em>b</em><em> to </em><em>4</em><em>, because </em><em>b > a</em><em> and </em><em>4 > 3</em>
Isolating a and b:
a = 3b/4
b = 4a/3
From Pythagoras:
h² = a² + b² - as a = 3b/4 and b = 4a/3, so
<h3>b = 4h/5</h3><h3>a = 6h/5</h3>
For Area:
A = b.a/2
as b = 4h/5 and a = 6h/5, so
<h3>A = 12h/25</h3>
Answer:
-1+3i
Step-by-step explanation:
Sqr root of 25 and 36 are 5 and 6 respectively
Sqr root of -25 and -4 are 5i and 2i respectively
Combine like with like
5-6=-1
5i-2i=3i
-1+3i
0=-\frac{233}{4}
sides are not equal
Answer:
100 Candy Bars
Step-by-step explanation:
Hey!
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Ayden wants to sell a total of $150 worth of candy bars for his school fundraiser. Each candy bar costs $1.50. How many candy bars will he have to sell in order to meet his goal?
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We can create an algebraic Equation to solve this.
<u>- We know that Each Candy Bar costs $1.50.</u>
<u>-We multiply the Amount of Candy Bars with the price. </u>
<u>-We know the profit should be $150</u>
⇒ 1.50x = 150
-<u>Divide</u> by 1.50 on both sides
⇒ 1.50x/1.50 = 150/1.50
-<u>Simplify</u>
⇒ x = 100
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<em>Hope I Helped, Feel free to ask any questions to clarify :)
</em>
<em>
Have a great day!
</em>
<em>
More Love, More Peace, Less Hate.
</em>
<em> -Aadi x</em>