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love history [14]
3 years ago
12

Find the surface area of the pyramid

Mathematics
1 answer:
cricket20 [7]3 years ago
7 0

Answer:

32√3

Step-by-step explanation:

I hope this helps... if not i dont know then

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Marissa counted 29 coins in her change purse which contained d dimes and q quarters. The value of the coins is $5.00. How many d
SVETLANKA909090 [29]
15 dimes 14 quarters

8 0
3 years ago
Read 2 more answers
Type the correct answer in the box.
statuscvo [17]

Step-by-step explanation:

V should be written as (1/3) pi r^2 h

V = (1/3) pi r^2 h multiply by 3

3V = pi r^2 h Divide by pi

3V/ pi = r^2 h Divide by r^2

3V / (pi *r^2 ) = h

4 0
3 years ago
Ayuda en este cripto-dichos
Cerrena [4.2K]
Ooo I don’t speak that language
6 0
3 years ago
Can someone please teach me how to do this? You can please do like 2 questions and please explain exactly how you got the answer
spayn [35]

All of these questions require one thing: trigonometric functions.

There are 3 main trigonometric functions, which can only be used on right triangles: sine, cosine, and tangent.

Sine = opposite / hypotenuse

Cosine = adjacent / hypotenuse

Tangent = opposite / adjacent

When trying to figure out what function to use, we always start by looking from the angle. Take problem a, for example. Looking from angle E, of which the value is not given, we have the side opposite and the side adjacent. Therefore, we should use the tangent function.

---The hypotenuse is always the longest side of the triangle. It is never considered the opposite or adjacent side.

Let's set up our function with the given information from problem a.

tan(x) = 9.7 / 5.2

---The tangent of an unknown angle is equal to the quotient of the opposite side and the adjacent side.

Now, solving for the value of x will require a calculator. We'll need to use what's called an inverse trigonometric function. Most calculators have these directly above the regular trigonometric functions, and the inverse functions are accessed using a "second" key.

---Ensure that your calculator is in degrees, not radians!

x = tan^-1(9.7 / 5.2)

x = 61.805 = 62 degrees

Next, let's take a look at problem b. This time, we're solving for a side length instead of an angle. But, we're still going to start by looking from our angle.


Looking from the 38 degree angle, we are given the adjacent side and an unknown hypotenuse. Therefore, we should use the cosine function.

cosine(38) = 53.1 / r

---The cosine of a 38 degree angle is equal to the quotient of 53.1 and an unknown hypotenuse, r.

Use your algebra skills to isolate the variable r.

r * cosine(38) = 53.1

r = 53.1 / cosine(38)

---From here, all you need to do is plug this into your calculator. Since we are solving for a side length (and given an angle), we are just using the regular trigonometric function buttons on the calculator.

r = 67.385 = 67.4 units

Hope this helps!

4 0
2 years ago
Suppose that you had the following data set. 500 200 250 275 300 Suppose that the value 500 was a typo, and it was suppose to be
hodyreva [135]

Answer:

\bar X_B = \frac{\sum_{i=1}^5 X_i}{5} =\frac{500+200+250+275+300}{5}=\frac{1525}{5}=305

s_B = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(500-305)^2 +(200-305)^2 +(250-305)^2 +(275-305)^2 +(300-305)^2)}{5-1}} = 115.108

\bar X_A = \frac{\sum_{i=1}^5 X_i}{5} =\frac{-500+200+250+275+300}{5}=\frac{525}{5}=105

s_A = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(-500-105)^2 +(200-105)^2 +(250-105)^2 +(275-105)^2 +(300-105)^2)}{5-1}} = 340.221

The absolute difference is:

Abs = |340.221-115.108|= 225.113

If we find the % of change respect the before case we have this:

\% Change = \frac{|340.221-115.108|}{115.108} *100 = 195.57\%

So then is a big change.

Step-by-step explanation:

The subindex B is for the before case and the subindex A is for the after case

Before case (with 500)

For this case we have the following dataset:

500 200 250 275 300

We can calculate the mean with the following formula:

\bar X_B = \frac{\sum_{i=1}^5 X_i}{5} =\frac{500+200+250+275+300}{5}=\frac{1525}{5}=305

And the sample deviation with the following formula:

s_B = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(500-305)^2 +(200-305)^2 +(250-305)^2 +(275-305)^2 +(300-305)^2)}{5-1}} = 115.108

After case (With -500 instead of 500)

For this case we have the following dataset:

-500 200 250 275 300

We can calculate the mean with the following formula:

\bar X_A = \frac{\sum_{i=1}^5 X_i}{5} =\frac{-500+200+250+275+300}{5}=\frac{525}{5}=105

And the sample deviation with the following formula:

s_A = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(-500-105)^2 +(200-105)^2 +(250-105)^2 +(275-105)^2 +(300-105)^2)}{5-1}} = 340.221

And as we can see we have a significant change between the two values for the two cases.

The absolute difference is:

Abs = |340.221-115.108|= 225.113

If we find the % of change respect the before case we have this:

\% Change = \frac{|340.221-115.108|}{115.108} *100 = 195.57\%

So then is a big change.

8 0
3 years ago
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