Answer:
Area = 140 sq. ft
Step-by-step explanation:
Divide the figure into two shapes: rectangle and triangle.
Solve the area of the triangle first:
A = 1/2 bh
A = 1/2 (6)(4)
A = 1/2 (24)
A = 12
Solve the area of rectangle:
A = lw
A = (8)(16)
A = 128
Area of figure = area of triangle + area of rectangle
Area of figure = 12 + 128
Area = 140 sq. ft
Answer:
Wow!!! you in college or some that $h!t looks hard
Step-by-step explanation:
Can u help me plz?
Answer:
B
Step-by-step explanation:
To select the correct equation, check to see that the c term of is 2 since the y-intercept is (0,2). This means only A and B are options since C and D have -2.
For A and B, substitute the other two points (-2,8) and (1,5) into the equations and see if it hods true.
This is not 8 and is not true.
This is true. This is the solution.
3rd choice ~ Multiply 4 by the difference of 10 and 3, then add 5
Answer:
A polynomial is prime if it can not be factored into polynomials of lower degree also with integer coefficients.
For example, the first option:
x^3 + b*x^2 can be rewritten as:
(x - 0)*(x^2 + b*x)
So it is not prime.
The second option:
x^2 -4x - 12
Because here we can factor this into:
(x + 2)*(x - 6) = x^2 - 6x + 2*x - 12 = x^2 - 4x - 12
Now, the third option is a two variable polynomial, here the degree is equal to the sum of the degrees of both variables.
x^4 + 8*x*y^3
(x - 0)*(x^3 + 8*y^3)
So each side has a lower degree than the original polynomial, then it is not prime.
4th option:
x^2 - b^3
This can be written as:
(x + b^(3/2))*(x - b^(3/2))
Now, here we have a problem.
If for example, b = 1, this would not be a prime.
because 1^(3/2) = 1.
But if b^(3/2) is not an integer, then we can not factorize the initial polynomial into lower degree polynomials with only integer coefficients, then we can not be 100% sure that this is not a prime polynomial, then this is the correct option.