All categories

4 years ago

You went to Mexico for vacation and came back with 3000 pesos. How much money is that worth in U.S. dollars?

Mathematics

1 answer:

jeka57 [31]4 years ago

7 0

161.11 us dollars is 3000 pesos

You might be interested in

I have to turn this in by tonight

Artist 52 [7]

Answer:

x = -11

Step-by-step explanation:

Convert fraction into negative exponential form: $(36^-^1)^x^-^4=(216^-^1)^x^+^1\\$

Simplify using exponent power rule: $(a^n)^m=a^n^m:36^-^x^+^4=216^-^x^-^1$

Convert both sides of the equation into terms with same base: $(6^2)^-^x^+^4=(6^3)^-^x^-^1$

Simplify using exponent power rule:

$(a^n)^m=a^n^m:6^-^2^x^+^8=6^-^3^x^-^3$

Based on the given conditions, corresponding exponents are equal:

$-2x=8=-3x-3$

Rearrange unknown terms to the left side of the equation:

$-2x+3x=-3-8$

Combine like terms: $x=-3-8$

Calculate the sum or difference: $x = -11$

Thurs, x = -11

[RevyBreeze]

5 0

2 years ago

4a. In a book store, for every 9 books sold, 13 magazines were sold. If they had a 572 sales, how many of the sales were for boo

topjm [15]

Answer:

234 books.

Step-by-step explanation:

You need to set up a proportion. Be careful to note what it looks like and the explanation that follows.

9 books 13 magazines

======= = ===========

x 572 - x Cross multiply

What you want is the number of book sales from the proportion.

However all you have is the ratio of books to magazines. So you have to set up the total sales to magazines, which is a bit awkward. 572 - x is the number of magazine sales.

9*(572 - x) = 13x Remove the brackets on the left.

5148- 9x = 13x Add 9x to both sides.

5148 = 13x + 9x Transpose and add the right.

22x = 5148 Divide by 22

x = 5148/22

x = 234

234 books were sold.

5 0

3 years ago

Find cot and cos <br> If sec = -3 and sin 0 > 0

Natali5045456 [20]

Answer:

Second answer

Step-by-step explanation:

We are given $\displaystyle \large{\sec \theta = -3}$ and $\displaystyle \large{\sin \theta > 0}$ . What we have to find are $\displaystyle \large{\cot \theta}$ and $\displaystyle \large{\cos \theta}$ .

First, convert $\displaystyle \large{\sec \theta}$ to $\displaystyle \large{\frac{1}{\cos \theta}}$ via trigonometric identity. That gives us a new equation in form of $\displaystyle \large{\cos \theta}$ :

$\displaystyle \large{\frac{1}{\cos \theta} = -3}$

Multiply $\displaystyle \large{\cos \theta}$ both sides to get rid of the denominator.

$\displaystyle \large{\frac{1}{\cos \theta} \cdot \cos \theta = -3 \cos \theta}\\\displaystyle \large{1=-3 \cos \theta}$

Then divide both sides by -3 to get $\displaystyle \large{\cos \theta}$ .

Hence, $\displaystyle \large{\boxed{\cos \theta = - \frac{1}{3}}}$

__________________________________________________________

Next, to find $\displaystyle \large{\cot \theta}$ , convert it to $\displaystyle \large{\frac{1}{\tan \theta}}$ via trigonometric identity. Then we have to convert $\displaystyle \large{\tan \theta}$ to $\displaystyle \large{\frac{\sin \theta}{\cos \theta}}$ via another trigonometric identity. That gives us:

$\displaystyle \large{\frac{1}{\frac{\sin \theta}{\cos \theta}}}\\\displaystyle \large{\frac{\cos \theta}{\sin \theta}$

It seems that we do not know what $\displaystyle \large{\sin \theta}$ is but we can find it by using the identity $\displaystyle \large{\sin \theta = \sqrt{1-\cos ^2 \theta}}$ for $\displaystyle \large{\sin \theta > 0}$ .

From $\displaystyle \large{\cos \theta = -\frac{1}{3}}$ then $\displaystyle \large{\cos ^2 \theta = \frac{1}{9}}$ .

Therefore:

$\displaystyle \large{\sin \theta=\sqrt{1-\frac{1}{9}}}\\\displaystyle \large{\sin \theta = \sqrt{\frac{9}{9}-\frac{1}{9}}}\\\displaystyle \large{\sin \theta = \sqrt{\frac{8}{9}}}$

Then use the surd property to evaluate the square root.

Hence, $\displaystyle \large{\boxed{\sin \theta=\frac{2\sqrt{2}}{3}}}$

Now that we know what $\displaystyle \large{\sin \theta}$ is. We can evaluate $\displaystyle \large{\frac{\cos \theta}{\sin \theta}}$ which is another form or identity of $\displaystyle \large{\cot \theta}$ .

From the boxed values of $\displaystyle \large{\cos \theta}$ and $\displaystyle \large{\sin \theta}$ :-

$\displaystyle \large{\cot \theta = \frac{\cos \theta}{\sin \theta}}\\\displaystyle \large{\cot \theta = \frac{-\frac{1}{3}}{\frac{2\sqrt{2}}{3}}}\\\displaystyle \large{\cot \theta=-\frac{1}{3} \cdot \frac{3}{2\sqrt{2}}}\\\displaystyle \large{\cot \theta=-\frac{1}{2\sqrt{2}}$

Then rationalize the value by multiplying both numerator and denominator with the denominator.

$\displaystyle \large{\cot \theta = -\frac{1 \cdot 2\sqrt{2}}{2\sqrt{2} \cdot 2\sqrt{2}}}\\\displaystyle \large{\cot \theta = -\frac{2\sqrt{2}}{8}}\\\displaystyle \large{\cot \theta = -\frac{\sqrt{2}}{4}}$

Hence, $\displaystyle \large{\boxed{\cot \theta = -\frac{\sqrt{2}}{4}}}$

Therefore, the second choice is the answer.

__________________________________________________________

Summary

Trigonometric Identity

$\displaystyle \large{\sec \theta = \frac{1}{\cos \theta}}\\ \displaystyle \large{\cot \theta = \frac{1}{\tan \theta} = \frac{\cos \theta}{\sin \theta}}\\ \displaystyle \large{\sin \theta = \sqrt{1-\cos ^2 \theta} \ \ \ (\sin \theta > 0)}\\ \displaystyle \large{\tan \theta = \frac{\sin \theta}{\cos \theta}}$