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lozanna [386]
3 years ago
13

-34 -2p = 8(8 - 2p)​

Mathematics
1 answer:
Lera25 [3.4K]3 years ago
8 0

Answer:

p = 7

Step-by-step explanation:

Look at the picture hope this helps!

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Find the area of the rectangle that has the vertices (-2, 0). (0-4). (2.0), and (0,4), respectively.
steposvetlana [31]

Answer:

Step-by-step explanation:

....... im not sure

5 0
3 years ago
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Parker wants to ride his bicycle 34 miles this week. He has already ridden 10 miles. If he rides for 5 more days, how many miles
vazorg [7]

Answer: He rode 4.8 miles each day.

Variable: x = Constant distance distance traveledd each day.

Equation : 10+5x=34

Step-by-step explanation:

Total distance needs to travel = 34 miles

Distance already traveled = 10 miles

He will ride bicycle for 5  more days.

Let x = Constant distance distance traveledd each day.

Total distance = 10+5x

i.e.

10+5x=34\\\\\Rightarrow\ 5x=24\\\\\Rightarrow\ x=4.8

Hence, he rode 4.8 miles each day.

7 0
3 years ago
Could someone help me with this?
umka2103 [35]
---------------------------------------------
Find area of one triangle
---------------------------------------------
Area of triangle = 1/2 x base x height
Area of triangle = 1/2 x 9 x 11.75
Area of triangle = 52.875 ft²

---------------------------------------------
Find area of two triangles
---------------------------------------------
Area of 2 triangles = 52.875 x 2 
Area of 2 triangles = 105.75 ft²

---------------------------------------------
Find the cost of the flower beds
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1 ft² = $4.25
105.75 ft² = 105.75 x 4.25
105.75 ft² = $449.44 (nearest hundredth)

---------------------------------------------
Answer: $449.44
---------------------------------------------
8 0
3 years ago
I need help which one is it?
Natalka [10]
Answer:
Alternate interior
8 0
3 years ago
Read 2 more answers
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
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