a.This set of vectors are not basis for vector space for two-dimentional space R2 due to high number of vectors (3). It means three vector is two much to span 2-dimentional space.
b.This set of vectors are not basis for vector space for three-dimentional space R3 due to small number of vectors (2). It means two vector can't span three-dimentional space.
Answer:
3.74637674E20
Step-by-step explanation:
Answer:
AB + BC = AC Segment Addition Postulate
(x + 10) + (x + 14) = 22 Substitution
2x + 24 = 22 Simplify (added like terms)
2x = -2 Subtraction Property of Equality
x = -1 Divison Property of Equality
Answer: x = -1
Step-by-step explanation:
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Use the rational root theorem
a0=16, an=1
The dividers of a0: 1,2,4,8,16 The dividers of an:1
Therefore check the following rational numbers: +- 1,2,4,8,16/1
4/1 is a root of the expression, so factor out x-4,
compute (x^4+13x^3-64x^2-20x+16)/(x-4) to get the rest of the equation:
x^3+17x^2+4x-4
=(x-4)(x^3+17x^2+4x-4)/x-4
which will leave you with answer B
x^3+17x^2+4x-4
9514 1404 393
Answer:
(a) 4/3
(b) y -3 = 4/3(x -1)
(c) y -3 = -3/4(x -1)
(d) r = 5
Step-by-step explanation:
a) The slope is given by the slope formula:
m = (y2 -y1)/(x2 -x1)
m = (7 -3)/(4 -1) = 4/3
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b) The radius is normal to the circle. The point-slope form of the equation for a line can be useful here:
y -k = m(x -h) . . . . . line with slope m through point (h, k)
For slope 4/3, the line through point (1, 3) will have the equation ...
y -3 = 4/3(x -1) . . . . point-slope equation of the normal
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c) The tangent is perpendicular to the radius. It will have a slope that is the opposite reciprocal of the slope of the radius: -1/(4/3) = -3/4.
y -3 = -3/4(x -1) . . . . point-slope equation of the tangent
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d) The radius can be found from the distance formula.
d = √((x2 -x1)² +(y2 -y1)²)
d = √((4 -1)² +(7 -3)²) = √(3² +4²) = √25 = 5
The radius of the circle is 5.