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Katyanochek1 [597]
1 year ago
15

Use the euclidean algorithm to find integers $x$ and $y$ such that $164x + 37y = 1,$ with the smallest possible positive value o

f $x$. state your answer as a list with $x$ first and $y$ second, separated by a comma.
Mathematics
1 answer:
Advocard [28]1 year ago
8 0

Well, let's use the Euclidean algorithm:

164 = 4×37 + 16

37 = 2×16 + 5

16 = 3×5 + 1

Then working backwards,

1 = 16 - 3×5

1 = 16 - 3×(37 - 2×16) = 7×16 - 3×37

1 = 7×(164 - 4×37) - 3×37 = 7×164 - 31×37

so that x = 7 and y = -31.

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In the middle of a park there is a fenced off area in the shape of the fence says is a square. If the area of the fenced off por
pashok25 [27]

Answer:

Step-by-step explanation:

The formula for the area of a square is A = x^2, where x is the length of one side.  Equivalently, x^2 = A.

Here x^2 = 225 yd^2.  Taking the positive square root of both sides, x = 15 yd.

Thus, the perimeter of the square area enclosed by the fence is

P = 4(15 yd) = 60 yd

3 0
3 years ago
What is (3x^3)^3 in factored form? Help
Sergio [31]

Answer:

This can't really be factored, but it can be expanded slightly.

(3x³)³ = 27x⁹

4 0
3 years ago
One solution to a quadratic function, h, is given.
Margaret [11]

Answer:

B. The other solution to function h is -4-i\,7.

Step-by-step explanation:

From Algebra, we remember that quadratic functions of the form a\cdot x^{2}+b\cdot x + c has solutions of the form:

x = \frac{-b\pm \sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}, where both are complex if and only if \sqrt{b^{2}-4\cdot a \cdot c}< 0.

Hence, if one solution of the quadratic function is -4+i\,7, then the other solution is -4-i\,7. The correct answer is B.

7 0
3 years ago
How many triangles can be constructed with sides measuring 7 cm, 6 cm, and 9 cm? more than one one none
REY [17]

Answer:

none

Step-by-step explanation:


3 0
3 years ago
Figure ABCD is graphed on a coordinate plane.
Phoenix [80]
1. BC is 1+3=4 (units)

2. AD is 3+5= 8 (units)

3. Draw the altitude CH from point C, as shown in the figure.

4. Then triangle CHD is a right triangle, with hypothenuse CD, and sides HD and CH.

5. CD^2=HD^2+CH^2

CD^2=2^2+4^2=4+16=20

CD= \sqrt{20}= \sqrt{4*5}= \sqrt{4} *  \sqrt{5}=2 \sqrt{5}= 4.47

6. AB=CD=4.47 because the trapezoid is isosceles.

7. P= BC+AD+AB+CD=4+8+4.47+4.47=20.9 (units)

3 0
3 years ago
Read 2 more answers
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