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circles and radius
The radius is the distance between the center of the circle and the circumference. Half of the diameter is the radius. Hence, $Radius= \dfrac{diameter}{2}$ In terms of area, $radius= \sqrt{\dfrac{area}{\Pi}}$ In terms of perimeter, $radius = \dfrac{perimeter}{2\Pi}$
The radius is the distance between the center of the circle and the circumference. Half of the diameter is the radius. Hence,
In terms of area,
In terms of perimeter,
See less20 + 3x – 15 + x = 27
$20+3x-15+x=27$ Collect like terms $3x+x=27-20+15$ $4x=22$ Divide both sides by 4 $x=11/2$ $5\tfrac{1}{2}$
Collect like terms
Divide both sides by 4
See lessWhat do you mean by algebra?
According to Oxford advance learners dictionary, algebra means the part of mathematics in which letters and other general symbols( Latin or Greek) are used to represent numbers and quantities in formulae and equations. When these symbols are combined with numbers and arithmetic operations, it becomeRead more
According to Oxford advance learners dictionary, algebra means the part of mathematics in which letters and other general symbols( Latin or Greek) are used to represent numbers and quantities in formulae and equations. When these symbols are combined with numbers and arithmetic operations, it becomes an algebraic expression.
See lessTry to solve what is the value of X in x / 2 = 3
$x/2=3$ Cross multiply $x=3×2$ $x=6$
Cross multiply
See less2 ( x + 5 ) – 7 = 3 ( x – 2)
$2(x+5)-7=3(x-2)$ Open up the brackets $2x+10-7=3x-6$ $2x+3=3x-6$ Collect like terms $3x-2x=3+6$ $x=9$
Open up the brackets
Collect like terms
See lessAlgebra question
$-x^{2}-(k+7)x-8=-(x-2)(x-4)$ open up the brackets $-x^{2}-kx-7x-8=-(x^{2}-6x+8)$ $-x^{2}-kx-7x-8=-x^{2}+6x-8$ collect like terms $-x^{2}-kx-7x-8+x^{2}-6x+8=0$ $-kx-7x-6x=0$ $-kx-13x=0$ $kx=-13x$ $\therefore k=-13$
See lessopen up the brackets
collect like terms
Write the equation of the line with
$m=3; x_{1}=2;y_{1}=5$ $y-y_{1}=m(x-x_{1})$ $y-5=3(x-2)$ $y-5=3(x-2)$ open up the bracket $y-5=3x-6$ add $5$ to both sides $y=3x-6+5$ $y=3x-1$
to both sides
See lessopen up the bracket
add
Write the equation of the line with
$m=-2; x_{1}=1;y_{1}=1$ $y-y_{1}=m(x-x_{1})$ $y-1=-2(x-1)$ $y-1=-2(x-1)$ open up the bracket $y-1=-2x+2$ add $1$ to both sides $y=-2x+2+1$ $y=-2x+3$
to both sides
See lessopen up the bracket
add
$m=5; x_{1}=1;y_{1}=-3$ $y-y_{1}=m(x-x_{1})$ $y--3=5(x-1)$ $y+3=5(x-1)$ open up the bracket $y+3=5x-5$ Subtract $3$ from both sides $y=5x-5-3$ $y=5x-8$
from both sides
See lessopen up the bracket
Subtract
Write the equation of the line with
$m=\dfrac{1}{3}; x_{1}=0;y_{1}=-3$ $y-y_{1}=m(x-x_{1})$ $y--3=\dfrac{1}{3}(x-0)$ $y+3=\dfrac{1}{3}(x)$ open up the bracket $y+3=\dfrac{1}{3}x$ Subtract $3$ from both sides $y=\dfrac{1}{3}x-3$
from both sides
See lessopen up the bracket
Subtract