Drag the correct steps into order to evaluate 21+ m 7 for m=28.

In a right triangle, angle is an acute angle and sin angle = 4/9. Evaluate the other five trigonometric functions of angle.

kkurt [141]

Answer:

Part 1) $cos(A)=\frac{\sqrt{65}}{9}$

Part 2) $tan(A)=4\frac{\sqrt{65}}{65}$

Part 3) $cot(A)=\frac{\sqrt{65}}{4}$

Part 4) $sec(A)=9\frac{\sqrt{65}}{65}$

Part 5) $csc(A)=\frac{9}{4}$

Step-by-step explanation:

Let

A------> the angle

we have that

$sin(A)=\frac{4}{9}$

step 1

Find the cos(A)

we know that

$cos^{2}(A)+sin^{2}(A)=1$

substitute the value of sin(A)

$cos^{2}(A)+(\frac{4}{9})^{2}=1$

$cos^{2}(A)=1-(\frac{4}{9})^{2}$

$cos^{2}(A)=1-(\frac{16}{81})$

$cos^{2}(A)=(\frac{65}{81})$

$cos(A)=\frac{\sqrt{65}}{9}$

step 2

Find the tan(A)

we know that

$tan(A)=\frac{sin(A)}{cos(A)}$

substitute the values

$tan(A)=\frac{\frac{4}{9}}{\frac{\sqrt{65}}{9}}$

$tan(A)=\frac{4}{\sqrt{65}}$

$tan(A)=4\frac{\sqrt{65}}{65}$

step 3

Find the cot(A)

we know that

$cot(A)=\frac{1}{tan(A)}$

we have

$tan(A)=\frac{4}{\sqrt{65}}$

substitute

$cot(A)=\frac{1}{\frac{4}{\sqrt{65}}}$

$cot(A)=\frac{\sqrt{65}}{4}$

step 4

Find the sec(A)

we know that

$sec(A)=\frac{1}{cos(A)}$

we have

$cos(A)=\frac{\sqrt{65}}{9}$

substitute

$sec(A)=\frac{1}{\frac{\sqrt{65}}{9}}$

$sec(A)=\frac{9}{\sqrt{65}}$

$sec(A)=9\frac{\sqrt{65}}{65}$

step 5

Find the csc(A)

we know that

$csc(A)=\frac{1}{sin(A)}$

we have

$sin(A)=\frac{4}{9}$

substitute the value

$csc(A)=\frac{1}{\frac{4}{9}}$

$csc(A)=\frac{9}{4}$

5 0

2 years ago

The lateral surface area of the wrapper for an ice cream cone is 47.1 cm squared.If the lateral height of the ice cream cone is

Simora [160]

To determine the radius of the cone, we make use of the lateral surface area of the cone. The lateral surface of an object is the area of all the sides of a certain object. For a cone, it is the product of pi, the radius and the slant height (lateral height). We calculate as follows:
Lateral Surface Area = πrh
47.1 = πr(6)
r = 2.5 cm

5 0

3 years ago

Help me please im beggingggg

icang [17]

Answer: 96 sq ft.

Step-by-step explanation:

First, we will find the area of the rectangular sides. Since the triangle is an equilateral triangle, all the rectangles will be the same size.

A = L * W

A = (10 ft) * (3 ft)

A = 30 ft²

30 ft² * 3 = 90 ft²

Next, we will find the area of the two triangles.

A = $\frac{b*h}{2}$

A = $\frac{3*2}{2}=\frac{6}{2} =3$

3 ft + 3 ft = 6 ft²

Lastly, we will add them together for the total surface area.

90 ft² + 6 ft² = 96 ft²

4 0

1 year ago

12a. Shan has 32 books, including 12 books by his favorite author. Express as a decimal the number of books by his favorite auth

LenaWriter [7]

12a. First expressing as a fraction, Shan has 12/32 books from his favorite author, so:

12/32=3/8=0.375 books from his favorite author

12b. Because the decimal 0.375 isn't a periodical decimal, so it would be a terminating decimal.

5 0

1 year ago

20 points best answer gets brainliest

lana66690 [7]

Note: The second image should be titled "Data Set #2"

It seems like you've already answered a part of this question yourself, but let's get into the details.

<h3>Parts A and B: 5 Number Summary</h3>

The five values the questions ask you to find the minimum, maximum, median, and the first and third quartiles for both of the data sets. These data are frequently called the five number summary of a data set, and we can use them to create a box plot of our data. The meanings of maximum and minimum are pretty obvious - they're just the biggest and smallest values in the set - but the median and the first and third quartiles all refer to different "middles" in the set.

The median is the "middle value" of a set of ordered data. When we have an odd number of data points, the median is simply the middle number of the set, but when we have an even number, as is the case with these two data sets, we have to find the number halfway between the two middle values. In data set 1, that number is

$\dfrac{16+25}{2}=\dfrac{41}{2}=20.5$

In data set 2, it's the number halfway between 8 and 10, which is 9.

The median splits any set of data into two parts: all the data points smaller than the median, and all of those larger than the median. In data set one, it's the two subsets {1, 4, 9, 16} and {25, 36, 49, 64}. The median of the smaller set gives us the first quartile, and the median of the larger one gives us the third quartile.

Why "first" and "third" quartile? Where are the second and fourth ones? While they don't go be the titles officially, those values are already part of our five number summary:

First quartile
Second quartile (the median)
Third quartile
Fourth quartile (the maximum)

<h3>Part C: Range, spread, and box plots</h3>

To get a visual for how our data is spread out, we can visualize our five-number-summary with a box plot. I've created a box plot for each of the data sets in the first two image uploads. The little nubs on the far ends, sometimes called the "whiskers" of the plot, are the minimum and maximum of the data set; the "box" represents the interquartile range of the data: all the values between the first and third quartile of the data; and the notch going down through the box is the median of the data.

We can see at a glance that data set 1 spans a far greater range of values that data set 2, and that its data points tend to be more concentrated in the lower values. Data set 2, by contrast, is much more uniform; its median lies right in the center of its range, and the "box" is centered similarly along it.

Comparing the medians of two data sets, especially those with the same number of values, can give us valuable information as to how much "larger" or "smaller" one set is than the other, but we need to bring in the other numbers in the five-number summary for a better picture about how that data is spread out.

<h3>Part D: Histograms vs. Box Plots: Which one is better?</h3>

There's no correct answer to this, because each type of graph gives us insight into different aspect of a data set.

A box and whisker plot is great for understanding:

The range of a set of data
Its spread
Its center

While a histogram can reveal:

How and where values are concentrated
Gaps and outliers

The histogram of data set 1, set to constant intervals of 7 units, shows us that many of the values at the lower end, and get more spread out as we go further - the empty patches become more frequent as we continue to the right, suggesting that our values will become more sparse as they get larger.

Contrast that with data set 2, which has a totally flat, uniform distribution when viewed at a constant interval of 4 units. The box plot and histogram work in tandem to give us a visual, quantitative picture of our data which we can use to make informed conclusions about it.

6 0

2 years ago