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dusya [7]
1 year ago
15

The table shows some values of a function of the form y

Mathematics
1 answer:
kherson [118]1 year ago
4 0

The value of c, the constant of the function y = ax² + bx + c, exists -3.

<h3>What is an equation?</h3>

An equation exists as an expression that indicates the relationship between two or more numbers and variables.

Given that: y = ax² + bx + c

At point (4, 21)

21 = a(4²) + 4b + c .......(1)

At point (5, 32)

32 = a(5²) + 5b + c .........(2)

At points (6, 45)

45 = a(6²) + 6b + c .......(3)

Therefore, the value of a = 1, b = 2 and c = -3.

The value of c, the constant of the function y = ax² + bx + c, exists -3.

To learn more about equations refer to:

brainly.com/question/2972832

#SPJ9

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Given:

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1 year ago
Solve 32+6•5-7+12 divided by 4
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Find a unit vector orthogonal to the plane containing the points A= 1,0,0 , B= 3,−1,−3 , and C= 1,3,−2 .
zysi [14]

Answer:

unit normal vector n will be n=(a,b,c) = (4/√171,11/√171,6/√171)

Step-by-step explanation:

There are several ways to solve this problem

1) build 2 vectors AB and BC such that the vectorial product ABxBC is the orthogonal vector to the plane , then find unit vector

2) since the 3 points belongs to the plane solve a linear system of 4 equation with 4 variables

for the second solution , the equation of the plane with normal vector n=(a,b,c) and containing the point (x₀,y₀,z₀) is

a*(x-x₀)+b*(y-y₀)+c*(z-z₀) =0

and

a²+b²+c² = 1 (unit vector)

then choosing A=(x₀,y₀,z₀)=(1,0,0)

a*(x-1)+b*(y-0)+c*(z-0) =0

for B

a*(3-1)+b*(-1-0)+c*(-3-0) =0

1) 2*a - b - 3*c =0

for C

a*(1-1)+b*(3-0)+c*(-2-0) =0

2) 3*b - 2*c=0 → b= 2/3*c

replacing in 1)

2*a -  2/3*c - 3*c =0

2*a-11/3*c=0 → a=11/6*c

thus

a²+b²+c² = 1

(11/6*c)²+(2/3*c)²+c² = 1

(121/36+4/9+1)*c² = 1

171/36*c²=1 → c= 6/√171

therefore

a=11/6*c = 11/6*6/√171= 11/√171

b=2/3*c= 2/3*6/√171= 4/√171

then the unit normal vector n will be

n=(a,b,c) = (4/√171,11/√171,6/√171)

5 0
4 years ago
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