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mamaluj [8]
2 years ago
11

Every quadrilateral with opposite angles supplementary can be inscribed in a circle.TrueFalseIf a triangle is inscribed in a cir

cle, the center of the circle is called the circumcenter.TrueFalse
Mathematics
1 answer:
solniwko [45]2 years ago
7 0

Explanation

The Inscribed Quadrilateral Theorem states that a quadrilateral can be inscribed in a circle if and only if the opposite angles of the quadrilateral are supplementary

Answer 1: True

Also, Given a triangle, the circumscribed circle is the circle that passes through all three vertices of the triangle. The center of the circumscribed circle is the circumcenter of the triangle, the point where the perpendicular bisectors of the sides meet.

Answer 2: True

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Answer:

Option C.  No, because the two populations from which the samples are selected do not appear to have equal variances.

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The variance measures how far a set of (random) numbers are spread out from their average value.

The fact that the younger adults show diversity in their brain activity while the older adults produce similar activities show that there are no equal variances in the two populations from which the samples are selected.

Therefore it would not be valid for Dr. Park to use the independent-measures t-test to test whether the brain activity of younger adults is different from that of older adults during a visual recognition task

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4 years ago
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What is the surface area of a cylinder with a diameter of 8 centimeters and a height of 12 centimeters?
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The formula for this is A=2 \pi rh+2 \pi  r^{2}.  If the diameter is 8 the radius is 4, so filling in accordingly we have A=2 \pi (4)(12)+2 \pi (16) and A=96 \pi +32 \pi which is A=128 \pi and when you multiply in 3.14 you get that A = 401.92 or B above.
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4 years ago
What is the value of Fraction 1 over 2x3 + 3.4y when x = 3 and y = 4? 15.85 18.1 27.1 31.6
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Step-by-step explanation:

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3 years ago
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L(x,y,z,\lambda)=10x+10y+2z+\lambda(5x^2+5y^2+2z^2-42)

L_x=10+10\lambda x=0\implies1+\lambda x=0

L_y=10+10\lambda y=0\implies1+\lambda y=0

L_z=2+4\lambda z=0\implies1+2\lambda z=0

L_\lambda=5x^2+5y^2+2z^2-42=0

\begin{cases}L_x=0\\L_y=0\\L_z=0\end{cases}\implies1+\lambda x=1+\lambda y=1+2\lambda z=0\implies x=y=2z

5x^2+5y^2+2z^2=5(2z)^2+5(2z)^2+2z^2=42z^2=42\implies z^2=1

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There are two critical points, at which we have

f\left(2,2,1\right)=42\text{ (a maximum value)}

f\left(-2,-2,-1\right)=-42\text{ (a minimum value)}

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