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lozanna [386]
3 years ago
8

The raidus of circle p the larger circle is 7 times the radius of circle 0 the smaller circle. What is the ratio of the area of

circle to the area of circle p?
Your choices are: 1 to 7, 7 to 1, 49 to 1 , 1 to 49
plz explain
Mathematics
1 answer:
DaniilM [7]3 years ago
8 0
Since we know that the radius of circle p is 7 times that of the radius of circle o we can say that radius of circle p equals 7x and the radius of circle o is x.
using the area formula we can say that area of circle p equals 
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End of semester help
aliya0001 [1]

Answer:

Step-by-step explanation:

A= 1/2 bh

10.98  = 1/2b (3.6)

10.98=b/2 (3.6)

10.98=b (1.8)

10.98/1.8=b

6.1=b

6 0
3 years ago
Solve for n.<br> n + 1 = 4(n – 8)<br> n=1<br> n=8<br> n=11<br> U n = 16
Alika [10]

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3 0
3 years ago
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Two is equal to 4 less than 3 times a certain number. Find the number.
Allushta [10]

Answer:

2

Step-by-step explanation:

2=3x-4

6=3x

x=2

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3 years ago
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Given: ΔABC. If ∠B = 35°, a = 4.5 cm, and b = 6.0 cm, what is the measurement of ∠A to the nearest tenth of a degree?
soldier1979 [14.2K]
We know that

the law of sines established 
a/sin A=b/sin B--------sin A=[a*sin B]/b
a=4.5 cm
b=6 cm
B=35°
so
sin A=[4.5*sin 35°]/6-----> sin A=0.43018
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the answer is the option
<span>C) 25.5° </span>


6 0
3 years ago
25 POINTS
uysha [10]

Answer:

a reflection over the x-axis and then a 90 degree clockwise rotation about the origin

Step-by-step explanation:

Lets suppose triangle JKL has the vertices on the points as follows:

J: (-1,0)

K: (0,0)

L: (0,1)

This gives us a triangle in the second quadrant with the 90 degrees corner on the origin. It says that this is then transformed by performing a 90 degree clockwise rotation about the origin and then a reflection over the y-axis. If we rotate it 90 degrees clockwise we end up with:

J: (0,1) , K: (0,0), L: (1,0)

Then we reflect it across the y-axis and get:

J: (0,1), K:(0,0), L: (-1,0)


Now we go through each answer and look for the one that ends up in the second quadrant;

If we do a reflection over the y-axis and then a 90 degree clockwise rotation about the origin we end up in the fourth quadrant.

If we do a reflection over the x-axis and then a 90 degree counterclockwise rotation about the origin we also end up in the fourth quadrant.

If we do a reflection over the x-axis and then a reflection over the y-axis we also end up in the fourth quadrant.

The third answer is the only one that yields a transformation which leads back to the original position.

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