Answer:
You can go for 1x9, 2x8, 3x7, 4x6, 5x5.
5x5 is the best as it builds a square, which maximizes space.
20*2.50 = $50.
Answer:
The triangle does not exist because sin(A)/a can not be equal to sin(B)/b
Step-by-step explanation:
we know that
step 1
Find the measure of angle B
Applying the law of sines

we have




substitute




Remember that the value of sine can not be greater than 1
therefore
The triangle does not exist because sin(A)/a can not be equal to sin(B)/b
Step-by-step explanation:
I'll do one for you.
Using the formula for turning exponents into radicals
![(b) {}^{ \frac{x}{y} } = \sqrt[y]{b} {}^{x}](https://tex.z-dn.net/?f=%28b%29%20%7B%7D%5E%7B%20%5Cfrac%7Bx%7D%7By%7D%20%7D%20%20%3D%20%20%20%20%5Csqrt%5By%5D%7Bb%7D%20%7B%7D%5E%7Bx%7D%20%20)
where b is the base
This means that the numerator in the exponet form becomes the power under the radical in radical form and
the denominator in exponet form becomes the nth root in radical form.
For example 5,

That becomes in radical form
![\sqrt[5]{5 {}^{ - 3} }](https://tex.z-dn.net/?f=%20%5Csqrt%5B5%5D%7B5%20%7B%7D%5E%7B%20-%203%7D%20%7D%20)
or if you want to write it using positive exponents
![\frac{1}{ \sqrt[5]{5 {}^{3} } }](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B%20%5Csqrt%5B5%5D%7B5%20%7B%7D%5E%7B3%7D%20%7D%20%7D%20)
I'll do one more for you
For example 6,

That becomes in radical form
![\sqrt[3]{11 {}^{4} }](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B11%20%7B%7D%5E%7B4%7D%20%7D%20)
It depends on what did you mean by saying perfect square. If I've understood it correctly, I can help you with a part of your problem. The squares of mod <span>9</span><span> are </span><span><span>1</span><span>,4,7</span></span><span> which are came from </span><span><span>1,2,</span><span>4.</span></span><span> </span>Addition of the given numbers are 2,3,5,6, 8, which are exactly the part of your problem. This number, which is not shown as squares Mod 9, and thus doesn't appear as a sum of digits of a perfect square. I hope you will find it helpful.