(x+2y)(x-5y)
do foil method:
x^2-5xy+2xy-10y^2
simplify x^2-3xy-10y^2,
so d is correct.
Answer: in statistics deviation is the difference between the value of one number in a series of numbers and the average value of all numbers in the series ( i hope this helped)
Step-by-step explanation:
Answer:
1 standard deviation
Step-by-step explanation:
Given that :
IQs as measured by the Wechsler Adult Intelligence Scale are approximately Normal with:
Mean = 100
Standard deviation = 15
About 84% of people will have IQs below 115 because:
84% = 0.84
From z - table, 0.84 corresponds to a z- score of about 0.995
0.995 = 1. 00
The z-score gives the number of standard deviations by which a raw score is above or below the average or mean value.
Raw score = 115
Mean = 100
Standard deviation (sd) = 15
Mean + number of sd's
100 + 1(sd)
100 + 1(15) = 100 + 15 = 115
Answer:
(14,-1)
Step-by-step explanation:
okay so first you need to use the "x=4y+10" in the first equation.
Rewrite as
3(4y+10) +11y=7
Distribute the 3(4y+10)
Rewrite as
12y+30
You'll have the equation
12y +30 +11y= 7
Subtract the 7 from both sides
12y+30+11y-7= 0
Combine like terms
23y+23=0
the best thing to do here is divide the equation by 23.
23/23 = 1
you'll have
y + 1 = 0
this simplifies to
y= -1
use this y value to solve for x
the easiest way to do this is start with the equation
x = 4y +10
Substitute the y for -1
x = -4(-1) +10
x = 4 +10
x = 14
the solution point is
(14,-1)
Answer:
A quadratic equation can be written as:
a*x^2 + b*x + c = 0.
where a, b and c are real numbers.
The solutions of this equation can be found by the equation:
Where the determinant is D = b^2 - 4*a*c.
Now, if D>0
we have the square root of a positive number, which will be equal to a real number.
√D = R
then the solutions are:
Where each sign of R is a different solution for the equation.
If D< 0, we have the square root of a negative number, then we have a complex component:
√D = i*R
We have two complex solutions.
If D = 0
√0 = 0
then:
We have only one real solution (or two equal solutions, depending on how you see it)
