All categories

1 year ago

Match the trigonometry formula to its correct ratio. Reduce the ratio if necessary.

Mathematics

2 answers:

Mandarinka [93]1 year ago

5 0

Step-by-step explanation:

mark me as brilliant plzzz

Diano4ka-milaya [45]1 year ago

4 0

Answer:

4/3 = tan A

4/5 = sin A

3/5 = cos A

Step-by-step explanation:

Define the adjacent, hypotenuse, and opposite lengths in the triangle.

Hypotenuse: 20

Opposite: 18

Adjacent: 12

Tan, sin, and cos formulas:

Tan A = opposite/adjacent
Sin A = opposite/hypotenuse
Cos A = adjacent/hypotenuse

I think it should be pretty easy to do the rest by simplifying the fractions and then getting the answer!

Have a lovely rest of your day/night, and good luck with your assignments! ♡

You might be interested in

If number eight is underlined what is the position

pshichka [43]

Is there any numbers to answer the question?

5 0

3 years ago

Find the missing side (X) round to the nearest hundredth

nata0808 [166]

Answer:

$\displaystyle 18,28 ≈ x$

Step-by-step explanation:

$\displaystyle \frac{x}{11} = csc\:37 \hookrightarrow 11csc\:37 = x \hookrightarrow 18,278041552... = x \\ \\ 18,28 \approx x$

$\displaystyle \frac{11}{x} = sin\:37 \hookrightarrow xsin\:37 = 11 \hookrightarrow \frac{11}{sin\:37} = x \hookrightarrow 18,278041552... = x \\ \\ 18,28 \approx x$

Information on trigonometric ratios

$\displaystyle \frac{OPPOCITE}{HYPOTENUSE} = sin\:θ \\ \frac{ADJACENT}{HYPOTENUSE} = cos\:θ \\ \frac{OPPOCITE}{ADJACENT} = tan\:θ \\ \frac{HYPOTENUSE}{ADJACENT} = sec\:θ \\ \frac{HYPOTENUSE}{OPPOCITE} = csc\:θ \\ \frac{ADJACENT}{OPPOCITE} = cot\:θ$

I am joyous to assist you at any time.

7 0

2 years ago

The function is y=6x.<br><br> Find the horizontal asymptote of the curve.

Lynna [10]

Answer:

y=-6x/square root (4x^2-160

Step-by-step explanation:

hope this helps please mark me brainliest

6 0

2 years ago

Read 2 more answers

A chemist wants to make 100 liters of a 44% acid solution. She has solutions that are 20% acid and 60% acid.

alexgriva [62]

the first solution is 20% acid, and say we'll be using "x" liters, so how many liters of just acid are in it? well 20% of "x" or namely 0.2x. Likewise for the 60% acid solution, if we had "y" liters of it, the amount of only acid in it is 0.6y.

$\begin{array}{lcccl} &\stackrel{solution}{quantity}&\stackrel{\textit{\% of }}{amount}&\stackrel{\textit{liters of }}{amount}\\ \cline{2-4}&\\ \textit{1st solution}&x&0.20&0.2x\\ \textit{2nd solution}&y&0.60&0.6y\\ \cline{2-4}&\\ mixture&100&0.44&44 \end{array}~\hfill \begin{cases} x+y=100\\\\ 0.2x+0.6y=44 \end{cases}$

$x+y=100\implies y=100-x~\hfill \stackrel{\textit{substituting on the 2nd equation}}{0.2x+0.6(100-x)=44} \\\\\\ 0.2x+60-0.6x=44\implies -0.4x+60=44\implies -0.4x=-16 \\\\\\ x=\cfrac{-16}{-0.4}\implies \boxed{x=40}~\hfill \boxed{\stackrel{100-40}{y=60}}$