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Gennadij [26K]
3 years ago
6

Sarah has a wall hanging in the shape of a parallelogram. What is the height of the wall hanging if its area and base are 300 sq

uare centimeters and 20 centimeters respectively?
Mathematics
1 answer:
Anastasy [175]3 years ago
4 0
Use the formula
Area= base times height.
You're given that area =300cm^2 and base=20 cm.
Plug these into the equation to get 300=20 x height. Now, you just need to solve for height. To do this, you need to undo the multiplication of b x h by dividing the area by the base.
So, h = 300/20
Simplify for an answer of 15.
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Find the average rate of change for the given function over the indicated values of x. If necessary, round your final answer to
Pavlova-9 [17]

Answer:

The average rate of change of the function in this interval is of 18.

Step-by-step explanation:

The average rate of change of a function f(x) in an interval from a to b is given by:

A = \frac{f(b) - f(a)}{b - a}

In this question:

f(x) = x^2 + 6x

Where x goes from 5 to 7.

This means that b = 7, a = 5. So

f(7) = 7^2 + 6(7) = 49 + 42 = 91

f(5) = 5^2 + 6(5) = 25 + 30 = 55

The rate of change is:

A = \frac{f(7) - f(5)}{7 - 5} = \frac{91 - 55}{2} = 18

The average rate of change of the function in this interval is of 18.

8 0
3 years ago
1. A bag contains rubber bands with lengths that are normally distributed with a mean of 6 cm of length, and a standard deviatio
romanna [79]
Roughly 1.7 percent of the bands are shorter than 3cm. We calculate the z score of the data point in standard distribution. By definition of z score, we use score minus mean divided by standard deviation. z=(3-6)/1.5=-2. A z score of -2 corresponds to approximately 1.7%, in other words, roughly 1.7 percent of data is less than 3cm.
4 0
3 years ago
Is anybody else here to help me ??​
Akimi4 [234]

Answer:

\cot(x)+\cot(\frac{\pi}{2}-x)

\cot(x)+\tan(x)

\frac{\cos(x)}{\sin(x)}+\frac{\sin(x)}{\cos(x)}

\frac{1}{\sin(x)}(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})

\csc(x)(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})

\csc(x)[\frac{\cos(x)\cos(x)}{\cos(x)}+\sin(x)\frac{sin(x)}{\cos(x)}]

\csc(x)[\frac{\cos(x)\cos(x)+\sin(x)\sin(x)}{\cos(x)}]

\csc(x)[\frac{\cos^2(x)+\sin^2(x)}{\cos(x)}]

\csc(x)[\frac{1}{\cos(x)}]

\csc(x)[\sec(x)]

\csc(x)[\csc(\frac{\pi}{2}-x)]

\csc(x)\csc(\frac{\pi}{2}-x)

Step-by-step explanation:

I'm going to use x instead of \theta because it is less characters for me to type.

I'm going to start with the left hand side and see if I can turn it into the right hand side.

\cot(x)+\cot(\frac{\pi}{2}-x)

I'm going to use a cofunction identity for the 2nd term.

This is the identity: \tan(x)=\cot(\frac{\pi}{2}-x) I'm going to use there.

\cot(x)+\tan(x)

I'm going to rewrite this in terms of \sin(x) and \cos(x) because I prefer to work in those terms. My objective here is to some how write this sum as a product.

I'm going to first use these quotient identities: \frac{\cos(x)}{\sin(x)}=\cot(x) and \frac{\sin(x)}{\cos(x)}=\tan(x)

So we have:

\frac{\cos(x)}{\sin(x)}+\frac{\sin(x)}{\cos(x)}

I'm going to factor out \frac{1}{\sin(x)} because if I do that I will have the \csc(x) factor I see on the right by the reciprocal identity:

\csc(x)=\frac{1}{\sin(x)}

\frac{1}{\sin(x)}(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})

\csc(x)(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})

Now I need to somehow show right right factor of this is equal to the right factor of the right hand side.

That is, I need to show \cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)} is equal to \csc(\frac{\pi}{2}-x).

So since I want one term I'm going to write as a single fraction first:

\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)}

Find a common denominator which is \cos(x):

\frac{\cos(x)\cos(x)}{\cos(x)}+\sin(x)\frac{sin(x)}{\cos(x)}

\frac{\cos(x)\cos(x)+\sin(x)\sin(x)}{\cos(x)}

\frac{\cos^2(x)+\sin^2(x)}{\cos(x)}

By  the Pythagorean Identity \cos^2(x)+\sin^2(x)=1 I can rewrite the top as 1:

\frac{1}{\cos(x)}

By the quotient identity \sec(x)=\frac{1}{\cos(x)}, I can rewrite this as:

\sec(x)

By the cofunction identity \sec(x)=\csc(x)=(\frac{\pi}{2}-x), we have the second factor of the right hand side:

\csc(\frac{\pi}{2}-x)

Let's just do it all together without all the words now:

\cot(x)+\cot(\frac{\pi}{2}-x)

\cot(x)+\tan(x)

\frac{\cos(x)}{\sin(x)}+\frac{\sin(x)}{\cos(x)}

\frac{1}{\sin(x)}(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})

\csc(x)(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})

\csc(x)[\frac{\cos(x)\cos(x)}{\cos(x)}+\sin(x)\frac{sin(x)}{\cos(x)}]

\csc(x)[\frac{\cos(x)\cos(x)+\sin(x)\sin(x)}{\cos(x)}]

\csc(x)[\frac{\cos^2(x)+\sin^2(x)}{\cos(x)}]

\csc(x)[\frac{1}{\cos(x)}]

\csc(x)[\sec(x)]

\csc(x)[\csc(\frac{\pi}{2}-x)]

\csc(x)\csc(\frac{\pi}{2}-x)

7 0
3 years ago
Write the equation of the line in slope intercept from that: Has a slope of m= −2 and goes through (1, 2). please help
Lisa [10]

Answer:

y=-2x+4

Step-by-step explanation:

5 0
3 years ago
Xavier writes the expression y+7x2\5 which expression is equivalent to the one xavier writes
kenny6666 [7]

Answer:

bhkhbkb bkhjgjvbhbhhhtfdik gfjkbfg

Step-by-step explanation:

8 0
2 years ago
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