These are two questions and two answers:
Question 1:
<span>A
quadratic equation is shown below: 3x^2 − 15x + 20 = 0 Part A: Describe
the solution(s) to the equation by just determining the radicand. Show
your work.
Answer: </span><span>The negative value of the radicand means that the equation does not have real solutions.
Explanation:
1) With radicand the statement means the disciminant of the quadratic function.
2) The discriminant is: b² - 4ac, where a, b, and c are the coefficients of the quadratic equation: ax² + bx + c
3) Then, for 3x² - 15x + 20, a = 3, b = - 15, and c = 20
and the discriminant (radicand) is: (-15)² - 4(3)(20) = 225 - 240 = - 15.
4) The negative value of the radicand means that the equation does not have real solutions.
Question 2:
Part B: Solve 3x^2 + 5x − 8 = 0 by using an appropriate
method. Show the steps of your work, and explain why you chose the
method used.
Answer: </span> two solutions x = 1 and x = - 8/3x
Explanation:
1) I choose factoring (you may use the quadratic formula if you prefer)
2) Factoring
Given: 3x² + 5x − 8 = 0
Make 5x = 8x - 3x: 3x² + 8x - 3x - 8 = 0
Group: (3x² - 3x) + (8x - 8) = 0
Common factors for each group: 3x(x -1) + 8(x - 1) = 0
Coomon factor x - 1: (x - 1) (3x + 8) = 0
The two solutions are for each factor equal to zero:
x - 1 = 0 ⇒ x = 1
3x + 8 = 0 ⇒ x = -8/3
Those are the two solutions. x = 1 and x = - 8/3
Answer:
Step-by-step explanation:
Midsegment is the half the length of the parallel side of the triangle.
So parallel sides of the triangles, which are the side of the base of the pyramid are:
Answer:
False
Step-by-step explanation:
If the conditional statement was if you have a dog then you have a pet, that would be the biconditional statement. The conditional is true but since not both parts of the biconditional are true, it is false.
2/4=1/2 two is 2 times greater than one so what is two times greater that 64?
64x2=128
Answer:
A - 0%
B- 50%
C- 50%
D- 100%
Step-by-step explanation:
Cystic fibrosis is inherited in an autosomal recessive form, meaning that a person has to inherit two abnormal genes for the disease to manifest. In the case of this question, one parent is a gene carrier, so his genotype is Aa, while the other does not have the cystic fibrosis gene, so AA.
Performing the cross of Aa x AA, we can see that:
a.) The probability of a child would have cystic fibrosis is 0%, since the disease is recessive and to be affected it should receive a recessive gene from each parent.
b.) The probability of a child would be a carrier is 50%, as 50% of the crossing phenotypes are Aa.
c.) The probability of a child would not have cystic fibrosis and is not a carrier is 50%, as 50% of the child's genotype is AA.
d.) The probability of a child would be healthy is 100%, as of all possible phenotypes none is affected.