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IRINA_888 [86]
3 years ago
9

What is the center of the circle with this equation is (x+2)²+(y-4)²=41?

Mathematics
1 answer:
OLga [1]3 years ago
5 0

Answer:

gkutrersfghjkdt

Step-by-step explanation:

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Why is it important to know the different properties of three-dimensional figures
IgorLugansk [536]

A two-dimensional shape has length and width. A three-dimensional solid shape also has depth. Three-dimensional shapes, by their nature, have an inside and an outside, separated by a surface. All physical items, things you can touch, are three-dimensional.



4 0
4 years ago
a fan is marked up 40% on the original price. The original was $20. What is the new price of the fan before sales tax?
Iteru [2.4K]
100/40×20.
100/40=2•5
2•5×20=50
the answer is 50
6 0
3 years ago
Given that a=25, b=43 and c=17 solve triangle ABC
Montano1993 [528]
The given lengths cannot form a triangle. They do not meet the requirements of the triangle inequality.

17 + 25 < 43
The triangle inequality requires each side be shorter than the sum of the other two.

3 0
3 years ago
A piece of wire measuring 20 feet is attached to a telephone pole as a guy wire. The distance along the ground from the bottom o
dem82 [27]

Answer: Height at which the wire is attached to the pole is 12 feet.

Explanation:

Since we have given that

Length of the wire = 20 feet

Let the height at which wire is attached to the pole be h

and distance along the ground from the bottom of the pole to the end of the wire be x+4

Now, it forms  a right angle triangle so, we can apply "Pythagorus theorem".

H^2=B^2+P^2\\\\20^2=x^2+(x+4)^2\\\\400=x^2+x^2+8x+16\\\\400=2x^2+8x+16\\\\400=2(x^2+4x+8)\\\\200=x^2+4x+8\\\\x^2+4x-192=0\\\\x^2+16x-12x-192=0\\\\x(x+16)-12(x+16)=0\\\\(x+16)(x-12)=0\\\\x=-16\ and\ 12

But height cant be negative so, height will be 12 feet.

Hence, height at which the wire is attached to the pole is 12 feet.


8 0
3 years ago
HELP I WILL GIVE 5 POINTS TO WHOM EVER CAN ANSWER THESE LAST 2 QUESTIONS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
vodka [1.7K]
Group and factor
undistribute then undistribute again
remember
ab+ac=a(b+c)
this is important



6d^4+4d^3-6d^2-4d
undistribute 2d
2d(3d^3+2d^2-3d-2)
group insides
2d[(3d^3+2d^2)+(-3d-2)]
undistribute
2d[(d^2)(3d+2)+(-1)(3d+2)]
undistribute the (3d+2) part
(2d)(d^2-1)(3d+2)
factor that difference of  2 perfect squares
(2d)(d-1)(d+1)(3d+2)






77.
group
(45z^3+20z^2)+(9z+4)
factor
(5z^2)(9z+4)+(1)(9z+4)
undistribuet (9z+4)
(5z^2+1)(9z+4)
7 0
3 years ago
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