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3 years ago

Any suggestions on how to solve this? thanks

Mathematics

1 answer:

amm18123 years ago

8 0

Answer: x = 4

Step-by-step explanation:

You basically want to solve for the angle A. Do arcsin of 1/3, which is approximately 19.47 degrees. In this triangle, we are missing the adjacent and opposite, but we do have the angle luckily. Lets use SOH-CAH-TOA.

$sin(theta) = \frac{opposite}{hypotenuse}$

We can see here that doing sine will give us the length of the opposite

$sin19.47 = \frac{x}{12}$ . Do sine of 19.47, then multiply by 12. I got 4

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Help me pleasee , timed

solong [7]

Answer:

The second alternative is correct

Step-by-step explanation:

We have been given the expression;

$(x^{27}y)^{\frac{1}{3}}$

The above expression can be re-written as;

$(x^{27})^{\frac{1}{3}}*y^{\frac{1}{3}}\\\\(x^{27})^{\frac{1}{3}}=x^{27*\frac{1}{3}}=x^{9}$

On the other hand;

$y^{\frac{1}{3}}=\sqrt[3]{y}$

Therefore, we have;

$x^{9}\sqrt[3]{y}$

6 0

2 years ago

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A recipe for chili calls for 5 pounds of ground beef for 18 people. How many pounds of ground beef are needed for 50 people

tatyana61 [14]

The answer is a repeating decimal of 13.888... round accordingly.

Set up the proportion 5/18 = x/50 18x=250 Solve for x

4 0

3 years ago

What is 372 rounded to the nearest 100

Xelga [282]

372

The number 3 is in the hundreds place.

Look to the number immediately to the right.

We know that 7 is bigger than 3.

So, we add 1 to 3 and the rest become 0.

Answer: 400

7 0

3 years ago

Read 2 more answers

System of equations

nalin [4]

Answer:

The solution of the system of equations is the point (-3,0)

Step-by-step explanation:

we have

$-3x+y=9$ -----> equation A

$2x-y=-6$ -----> equation B

Adds equation A and equation B

$-3x+y=9\\2x-y=-6\\------\\-3x+2x=9-6\\-x=3\\x=-3$

Find the value of y

$-3(-3)+y=9$

$9+y=9$

$y=9-9=0$

The solution of the system of equations is the point (-3,0)

4 0

2 years ago

Rewrite the expression as a single power with a negative exponent

goblinko [34]

$~\hspace{7em}\textit{negative exponents} \\\\ a^{-n} \implies \cfrac{1}{a^n} ~\hspace{4.5em} a^n\implies \cfrac{1}{a^{-n}} ~\hspace{4.5em} \cfrac{a^n}{a^m}\implies a^na^{-m}\implies a^{n-m} \\\\[-0.35em] ~\dotfill\\\\ \left(\cfrac{1}{3}\cdot \cfrac{1}{3}\cdot \cfrac{1}{3}\cdot \cfrac{1}{3} \right)^2\implies \left[ \left( \cfrac{1}{3} \right)^4 \right]^2\implies \left( \cfrac{1^4}{3^4} \right)^2 \\\\\\ \left( \cfrac{1^{4\cdot 2}}{3^{4\cdot 2}} \right) \implies \cfrac{1}{3^8} \implies 3^{-8}$

let's recall that

$\begin{array}{llll} 1^1&=&1\\ 1^{10}&=&1\\ 1^{1,000}&=&1\\ 1^{1,000,000,000}&=&1\\ 1^{1,000,000,000,000}&=&1\\ \end{array}$

8 0

2 years ago

Any suggestions on how to solve this? thanks ​

Any suggestions on how to solve this? thanks