ΔAOB is a right angled triangle. Therefore the Pythagorean Theorem applies in this situation.
θ is the angle from a standard position of the line OA
The length of the y component is √(1-0)2 +(-3-(-3))2] =√(12+ 02) = 1 A(-3,1) to B(-3,0) which is opposite
Then the length of the x-component is √[(-3-0)2 +(0-0)2] = √(9+0)= 3 B(-3,0) to O(0,0) which is adjacent
The length of vector OA is √[(-3-0)2 + (1-0)2] = √(9+1) = √(10) A(-3,1) to O(0,0) which is the hypotenuse of the triangle
θ = 180 - α
sinθ = sin(180-α) = opposite/hypotenuse = 1/√10
cosθ = adjacent/hypotenuse = -3/√10
tanθ = opposite/adjacent = 1/-3 = -1/3
α= arcsin(1/√10) ≈ 18
θ =180 -18 ≈162
Step-by-step explanation:
if the two integers are x and y
xy=-10
x+y=-9
x+(-10/x)=-9
x2+9x-10=0
(x-1)(x+10)=0
x=1 or x=-10
if x=1, y=-10
if x=-10, y=1
so the two integers are 1 and -10
Any number ends with 0, 1, 2, 3, 4, 5, 6, 7, 8 or 9
Given any number. The square of this number is the last digit of square of the original numbers units digit.
For example
23*23 ends with 9 (3*3=9)
149*149 ends with 1 (9*9=81)
2564*2564 ends with 6 (4*4=16)
and so on
so all the possible unit digits of a square number are {0, 1, 4, 5, 6, 9}
because:
0*0= 0 ; 1*1=1; 2*2=4; 3*3=9; 4*4=16; 5*5=25, 6*6=36; 7*7=49; 8*8=64, 9*9=81
Thus, the probability that the square of a number selected from any set of numbers being 7, is 0.
Answer: 0
Answer:
9/16
Step-by-step explanation:
i took the test ;)
Answer:
A) x = 3 or -1
B) x = -7
C)x = -7
Step-by-step explanation:
A) x² + 2x + 1 = 2x² - 2
Rearranging, we have;
2x² - x² - 2x - 2 - 1 = 0
x² - 2x - 3 = 0
Using quadratic formula, we have;
x = [-(-2) ± √((-2)² - 4(1 × -3))]/(2 × 1)
x = (2 ± √16)/2
x = (2 + 4)/2 or (2 - 4)/2
x = 6/2 or -2/2
x = 3 or -1
B) ((x + 2)/3) - 2/15 = (x - 2)/5
Multiply through by 15 to get;
5(x + 2) - 2 = 3(x - 2)
5x + 10 - 2 = 3x - 6
5x - 3x = -6 - 10 + 2
2x = -14
x = -14/2
x = -7
C) log(2x + 3) = 2log x
From log derivations, 2 log x is same as log x²
Thus;
log(2x + 3) = logx²
Log will cancel out to give;
2x + 3 = x²
x² - 2x - 3 = 0
Using quadratic formula, we have;
x = [-(-2) ± √((-2)² - 4(1 × -3))]/(2 × 1)
x = (2 ± √16)/2
x = (2 + 4)/2 or (2 - 4)/2
x = 6/2 or -2/2
x = 3 or -1
