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2 years ago

Whats the difference of squares

Mathematics

1 answer:

stepan [7]2 years ago

6 0

An example of the difference of squares would be: a^2-b^2 = (a+b)(a-b)

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What is the equation, in standard form, of a parabola that contains the following points? (-2,18),(0,4),(4,24)

nydimaria [60]

Y=ax² + bx + c and A(-2,18), B(0.4), C(4,24) given

1)Let's determine 1st the value of c in using the pair of B
4= a(0)² + b(0) + c , c=4 and y=ax²+bx+4

2) Use the pair of A: 18 = 4a -2b + 4===> 4a-2b=14

3) Use C pair: 24 = 16a+4b+4 ==> 16a+4b=20

4) solve the system equation in 2) and 3) and you'll find a=2 and b=-3

Hence the equation is y =2x²-3x+4

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2 years ago

Mason and Nora decided to swim across the river. Mason began swimming 3 seconds earlier than Nora.

hram777 [196]

What does 5.3+1 mean

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2 years ago

The Venn diagram shows the results of two events resulting from rolling a number cube.

gulaghasi [49]

Answer: $P(A\cap B)=\frac{1}{3}$

$P(A)=\frac{1}{3}$

$P(B)=1$

$P(A|B)=\frac{1}{3}$

Step-by-step explanation:

From the given figure it can be seen that

Total number on cube= n(S)=6

Intersection of A and B = n(A ∩ B)= 2

Therefore,

$P(A\cap B)=\frac{n(A\cap B)}{n(s)}=\frac{2}{6}=\frac{1}{3}$

Also, The number of elements in A = 2

Therefore,

$P(A)=\frac{n(A)}{n(s)}=\frac{2}{6}=\frac{1}{3}$

Similarly, The number of elements in B= 6

Therefore,

$P(B)=\frac{n(B)}{n(s)}=\frac{6}{6}=1$

The formula to find the conditional probability is given by :-

$P(A|B)=\frac{P(A\cap B)}{P(B)}\\\\\Rightarrow P(A|B)=\frac{\frac{1}{3}}{1}=\frac{1}{3}$

4 0

3 years ago

Read 2 more answers

Grace wants to find the probability of a family of 3 children having 2 boys and 1 girl. Use a tree diagram to list the possibili

pychu [463]

Answer:

Step-by-step explanation:

My approach was to draw out the probabilities, since we have 3 children, and we are looking for 2 boys and 1 girl, the probabilities can be Boy-Boy-Girl, Boy-Girl-Boy, and Girl-Boy-Boy. So a 2/3 chance if you think about it, your answer 2/3 can't be correct. If we assume that boys and girls are born with equal probability, then the probability to have two girls (and one boy) should be the same as the probability to have two boys and one girl. So you would have two cases with probability 2/3, giving an impossible 4/3 probability for both cases. Also, your list "Boy-Boy-Girl, Boy-Girl-Boy, and Girl-Boy-Boy" seems strange. All of those are 2 boys and 1 girl, so based on that list, you should get a 100 percent chance. But what about Boy-Girl-Girl, or Girl-Girl-Girl? You get 2/3 if you assume that adjacencies in the (ordered) list are important, i.e., "2 boys and a girl" means that the girl was not born between the boys.

5 0

1 year ago

4x -y -16 = 0 in slope intercept form

xxMikexx [17]

Y=-4x+16 is slope intercept form. I hope this helped

4 0

2 years ago