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Dominik [7]
3 years ago
12

Help please!!! Find the point in the first quadrant where the line y=4x intersects a circle of radius 4 centered at the origin.

Mathematics
2 answers:
lianna [129]3 years ago
8 0

Answer:

Step-by-step explanation:

X^2+Y^2=16

X^2+4X^2=16

5X^2=16

X^2=16/5

X=+- SQRT(16/5)

X=+- 4SQRT(5)

DISREGARD THE -4SQRT(5) BECAUSE WE ARE DEALING WITH THE FIRST QUADRANT

Firdavs [7]3 years ago
6 0

Answer:

the point is : (1 ; 4 )

Step-by-step explanation:

an equation for this circle is : x²+y² = 16

you have the system : x²+y² = 16 ....(1)

                                      y = 4x .....(2)

Substitute y = 4x into (1) and solve for x :

x² + (4x)² = 16

x² +16x² = 16

17x² = 16

x² = 16/17

x = 4/√17   or x = - 4/√17 ( refused) because the point in the first quadrant where  : x ≥0

 x = 4/√17 ≅ 0.970

exact values or give decimal values accurate to at least 3 decimal places

is : x =1

Substitute  x = 1 into (2) : y =4    the point is : (1 ; 4 )

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VladimirAG [237]

Answer:

  A) 24 ft³

Step-by-step explanation:

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_____

<em>Comment on the triangle</em>

You know the 3-4-5 triangle is a right triangle, because you have seen those numbers many times in relation to the Pythagorean theorem. If you don't know that, then you can check to see by using the Pythagorean theorem:

  5² = 3² +4²

  25 = 9 + 16 . . . . . true; so the triangle is a right triangle

In any event, there are other means for finding the area of the triangle based on the lengths of its sides. You will get the same area result.

Using Heron's formula, s = (a+b+c)/2 = (3+4+5)/2 = 6.

  A = √(s(s -a)(s -b)(s -c)) = √(6(3)(2)(1)) = √36 = 6 . . . . area of base triangle

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4 years ago
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