What is the perimeter of a if it has an area of 81 centimetre squares?
Answers
Related Questions
Solve the system of equations and choose the correct answer from the list of options.d + e = 15 –d + e = –5 Label the ordered pair as (d, e). (4 points) Select one: a. (0, 0) b. (10, –5) c. (5, 10) d. (10, 5)
Simplify. 10.5 • (–3) ÷ 10 • (–2) ÷ 0.5 A. –12.6 B. –3.15 C. 12.6 D. 3.15
PLEASEE HURRY!!!!!!!! Tom determines that the system of equations below has two solutions, one of which is located at the vertex of the parabola.Equation 1: (x – 3)2 = y – 4Equation 2: y = -x + bIn order for Tom’s thinking to be correct, which qualifications must be met? A: b must equal 7 and a second solution to the system must be located at the point (2, 5).B: b must equal 1 and a second solution to the system must be located at the point (4, 5).C: b must equal 7 and a second solution to the system must be located at the point (1, 8).D: b must equal 1 and a second solution to the system must be located at the point (3, 4).
What is the sum of three fours and four threes
Answers
Answers
2.y=4x^3+3x^2+2
3.y=2x+3+8/x
4.y=x^3-9x^2-21x+11
5.y=9x^2/3-2x+5
Please provide full working out, thanks.
Answers
y=x²+2x
we have to calculate the first derived
y´=2x+2
we equal to "0" the first derived and find out the value of "X"
2x+2=0
x=-2/2=-1
we have to calculate de second derived
y´´=2>0 ⇒we have a minimun at x=-1
we calculate y
y=(-1)²+2(-1)=1-2=-1
Answer: we have a minimum at (-1,-1)
2)
y=4x³+3x²+2
we have to calculate the first derived
y´=12x²+6x
we equalize to "0" the first derived and find out the value of "X"
12x²+6x=0
6x(2x+1)=0
6x=0 ⇒x=0
2x+1=0 ⇒x=-1/2
we have to calculate de second derived
y´´=24x+6
y``(0)=24(0)+6=6 >0 ⇒we have a minimun at x=0
y``(-1/2)=24(-1/2)+6=-12+6=-6 ⇒we have a maximum at x=-1/2
we calculate y
if x=0, y=2
if x=-1/2; y=4(-1/2)³+3(-1/2)²+2=9/4.
we have to equalize the second derived to "0" and find out the value of "x"
24x+6=0
x=-6/24=-1/4; in x=-1/4 we have an inflection point.
y=4(-1/4)³+3(-1/4)²+2=31/16
Answer: we have a minimum at (0,2), a maximum at (-1/2, 9/4) and a inflection point at (-1/4, 31/16).
3.
y=2x+3+8/x
y=(2x²+3x+8)/ x
we have to calculate the first derived
y´= [(4x+3)x-(2x²+3x+8)] / x²=(4x²+3x-2x²-3x-8) / x²=(2x²-8)/x²
we equal to "0" the first derived and find out the value of "X"
(2x²-8) / x²=0
2x²-8=0
x=⁺₋2
we have to calculate de second derived
y´´=[(4x)x²-2x(2x²-8)] / x⁴=(4x³-4x³+16x)/x⁴)=16/x³
y``(-2)=16/(-2)³=-2>0 ⇒ at x=-2 exist a maximum
y´´(2)=16/(2)³=2<0 ⇒ at x=2 exist a minimum
we calculate y
y(-2)=-4+3-4=-5
y(2)=4+3+4=11
Answer: we have a maximum at (-2,-5 ) and a minimun at (2,11)
4)
y=x³-9x²-21x+11
we have to calculate the first derived
y´=3x²-18x-21
we equal to "0" the first derived and find out the value of "X"
3x²-18x-21=0
x²-6x-7=0
x=[6⁺₋√(36+28)]/2=(6⁺₋8)/2
x₁=-1
x₂=7
we have to calculate de second derived
y´´=6x-18
y``(-1)=-6-18=-24<0 ⇒at x=-1 exist a maximum
y´´(7)=42-18=24>0 ⇒ at x=7 exist a minimum.
we calculate y
y(-1)=-1-9+21+11=22
y(7)=343-441-147+11=-234
We equalize the second derive to 0, and find out the value of "x"
6x-18=0
x=3 in x=3 exist an inflection point
y=27-81-63+11=-106
Answer: we have a minimum at (7,-234), a maximum at (-1,22) and an inflection point at (3,-106).
5)
y=9x²/³-2x+5
we have to calculate the first derived
y´=6x⁻¹/³-2
we equal to "0" the first derived and find out the value of "X"
6x⁻¹/³-2=0
x⁻¹/³=1/3
x¹/³=3
x=27
we have to calculate de second derived
y´´=-2x^(-4/3)
y´´(27)=-0.024<0 ⇒ at x=27 exist a maximum
we calculate y
y=32
Answer: we have a maximum at (27,32)
Answers
Answer: 7
Step-by-step explanation:
See attached picture
Detailed explanation please, I want to be able to understand this thoroughly.
Diagram is attached, thanks in advance!
Answers
Answer:
Given : BRDG is a kite that is inscribed in a circle,
With BR = RD and BG = DG
To prove : RG is a diameter
Proof:
Since, RG is the major diagonal of the kite BRDG,
By the property of kite,
∠ RBG = ∠ RDG
Also, BRDG is a cyclic quadrilateral,
Therefore, By the property of cyclic quadrilateral,
∠ RBG + ∠ RDG = 180°
⇒ ∠ RBG + ∠ RBG = 180°
⇒ 2∠ RBG = 180°
⇒ ∠ RBG = 90°
⇒ ∠ RDG = 90°
Since, Angle subtended by a diameter or semicircle on any point of circle is right angle.
⇒ RG is the diameter of the circle.
Hence, proved.