<u>The x and y coordinates of the origin is equal to (0, 0), which is the point of intersection of the x-axis and y-axis. The x and y coordinates of a point on the x-axis is of the form (a, 0), and here the y coordinate of the point is equal to zero.</u>
L.S.A.=1/2pl where p represents the perimeter of the base and l the slant height.
L.S.A. = ½(4*32)(24)
L.S.A. = ½ (128)(24)
L.S.A. =1/2(3072)
L.S.A. = 1536 cm
Find the reciprocal of the second fraction (flip numerator and denominator's spots) and multiply the two fractions.
Answer:
Therefore the required polynomial is
M(x)=0.83(x³+4x²+16x+64)
Step-by-step explanation:
Given that M is a polynomial of degree 3.
So, it has three zeros.
Let the polynomial be
M(x) =a(x-p)(x-q)(x-r)
The two zeros of the polynomial are -4 and 4i.
Since 4i is a complex number. Then the conjugate of 4i is also a zero of the polynomial i.e -4i.
Then,
M(x)= a{x-(-4)}(x-4i){x-(-4i)}
=a(x+4)(x-4i)(x+4i)
=a(x+4){x²-(4i)²} [ applying the formula (a+b)(a-b)=a²-b²]
=a(x+4)(x²-16i²)
=a(x+4)(x²+16) [∵i² = -1]
=a(x³+4x²+16x+64)
Again given that M(0)= 53.12 . Putting x=0 in the polynomial
53.12 =a(0+4.0+16.0+64)
=0.83
Therefore the required polynomial is
M(x)=0.83(x³+4x²+16x+64)
What is the solution set of x2 + y2 = 26 and x − y = 6? A. {(5, -1), (-5, 1)} B. {(1, 5), (5, 1)} C. {(-1, 5), (1, -5)} D. {(5,
Rus_ich [418]
He two equations given are
x^2 + y^2 = 26
And
x - y = 6
x = y +6
Putting the value of x from the second equation to the first equation, we get
x^2 + y^2 = 26
(y + 6) ^2 + y^2 = 26
y^2 + 12y + 36 + y^2 = 26
2y^2 + 12y + 36 - 26 = 0
2y^2 + 12y + 10 = 0
y^2 + 6y + 5 = 0
y^2 + y + 5y + 5 = 0
y(y + 1) + 5 ( y + 1) = 0
(y + 1)(y + 5) = 0
Then
y + 1 = 0
y = -1
so x - y = 6
x + 1 = 6
x = 5
Or
y + 5 = 0
y = - 5
Again x = 1
So the solutions would be (-1, 5), (1 , -5). The correct option is option "C".
