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8_murik_8 [283]
3 years ago
12

Find the equation in standard form of the parabola passing through the giving points.

Mathematics
1 answer:
zaharov [31]3 years ago
4 0

Let y = ax²+bx+c then substitute all of orders. we'll be doing equations system.

a-b+c=0\\c=3\\a+b+c=2

Ok we are at this.

Substitute c = 3 into equation 1 and equation 3.

a-b+3=0\\a-b=-3

a+b+3=2\\a+b=2-3\\a+b=-1

then addition and subtract for a-b = 3 and a+b = -1

a-b=-3\\a+b=-1\\2a=-4\\a=-2

Therefore a = -2, substitute a = -2 in a-b+c=0 or a+b+c=2 but i'm doing a-b+c=0 one.

a-b+c=0\\\\-2-b+3=0\\-b+1=0\\-b=-1\\b=1

Therefore b = 1, substitute both a, b and c into y=ax²+bx+c as we get,

y=-2x^2+x+3 #

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There are three separate, equal-size boxes, and inside each box there are four separate small boxes, and inside each of the smal
katen-ka-za [31]

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There are 51 boxes all together.

Step-by-step explanation:

Since there are three separate, equal-size boxes, and inside each box there are four separate small boxes, and inside each of the small boxes there are three even smaller boxes, to determine how many boxes are there all together, the following calculation must be done:

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6 0
2 years ago
Suppose ten students in a class are to be grouped into teams. (a) If each team has two students, how many ways are there to form
ValentinkaMS [17]

Answer:

(a) There are 113,400 ways

(b) There are 138,600 ways

Step-by-step explanation:

The number of ways to from k groups of n1, n2, ... and nk elements from a group of n elements is calculated using the following equation:

\frac{n!}{n1!*n2!*...*nk!}

Where n is equal to:

n=n1+n2+...+nk

If each team has two students, we can form 5 groups with 2 students each one. Then, k is equal to 5, n is equal to 10 and n1, n2, n3, n4 and n5 are equal to 2. So the number of ways to form teams are:

\frac{10!}{2!*2!*2!*2!*2!}=113,400

For part b, we can form 5 groups with 2 students or 2 groups with 2 students and 2 groups with 3 students. We already know that for the first case there are 113,400 ways to form group, so we need to calculate the number of ways for the second case as:

Replacing k by 4, n by 10, n1 and n2 by 2 and n3 and n4 by 3, we get:

\frac{10!}{2!*2!*3!*3!}=25,200

So, If each team has either two or three students, The number of ways  form teams are:

113,400 + 25,200 = 138,600

6 0
3 years ago
What is the image of the point (3,4) after a rotation of 90 degree counterclockwise about the origin
jeka57 [31]

Answer:

(- 4, 3 )

Step-by-step explanation:

Under a counterclockwise rotation about the origin of 90°

a point (x, y ) → (- y, x ) , then

(3, 4 ) → (- 4, 3 )

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2 years ago
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