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2 years ago

How is -5x + 2y = 9 is not constant of variation

2 answers:

7 0

A constant of variation is only in the form ay=bx where b/a=constant so given any point y/x=constant let us check here by substituting x=1 and x=2
when x=1---> 2y=14,y=7 here y/x=7
when x=2--->2y=19,y=9.5 here y/x=4.75
so this equation is not constant of variation
let us check an equation of the form 3y=7x
when x=1--->y=7/3--->y/x=7/3
when x=2--->y=14/3---->y/x=7/3
so 3y=7x is a constant of variation

sukhopar [10]2 years ago

4 0

Yes because there is no constant

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3 years ago

Jose is selling peanuts at the carnival. The original bag he purchased weighs 6 3/4 lbs. If the bags he plans on selling will we

Aleks04 [339]

Answer:

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Step-by-step explanation:

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2 years ago

Find the distance between the points A(2, 3) and B(4, 1).

kherson [118]

Answer:

sqrt(8) or 2.828

Step-by-step explanation:

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6 0

2 years ago

Read 2 more answers

Solve with cramer's rule x+2y+3z=11, 2x+y+2z=10, 3x+2y+z=9

nalin [4]

Answer:

x = 2 , y = 0 , z = 3

Step-by-step explanation:

Cramer's rule is a rule through which we can find the solution of linear equation.

we have the three linear equations as

x+2y+3z=11

2x+y+2z=10

3x+2y+z=9

AX=B

A: coefficient matrix

X= unknown vectors(x,y,z)

D = values of the linear equation (11 , 10 , 9)

now we find the determinant of the given linear equation

determinant of the matrix will be

A = $\left[\begin{array}{ccc}1&2&3\\2&1&2\\3&2&1\end{array}\right]$ = 1(1-4) - 2(2-6) + 3(4 - 3)

= 1(-3) - 2(-4) + 3(1)

= -3+8+3 = 8

also D $\neq 0$

so the determinant is Non zero we can apply Cramer's rule

we will be replacing the first column of the coefficient matrix A with the values of D

by replacing the first column we will get the value of the variable 'x'

Dx = $\left[\begin{array}{ccc}11&2&3\\10&1&2\\9&2&1\end{array}\right]$ = 11(1-4) -2(10-18) + 3(20-9) = -33+16+33 = 16

x = $\frac{Dx}{D}$ = $\frac{16}{8}$ = 2

similarly

Dy = $\left[\begin{array}{ccc}1&11&3\\2&10&2\\3&9&1\end{array}\right]$ = 1(10-18) -11(2-6) + 3(18 -30) = -8 +44 -36 = 0

y = $\frac{Dy}{D}$ = 0

Dz= $\left[\begin{array}{ccc}1&2&11\\2&1&10\\3&2&9\end{array}\right]$ = 1(9 - 20) -2(18 - 30) + 11(4 -3) = -11 +24 +11 = 24

z = $\frac{Dz}{D}$ = $\frac{24}{8} = 3$

so we have the solution as

x = 2 , y = 0 , z = 3

Therefore the solution for the given linear equations is (2,0,3).

3 0

2 years ago

an initial deposit of $200 is made into an account that pays 4.2% annual interest. What is a model to represent the balance in t

Sholpan [36]

You would use the equation y = A(1+r)^t

A = Original amount = 200
r = interest rate = 0.042 (when converted from percent to decimal)
t = time = x

Plug them in and you get the model: y = 200(1+0.042)^x

or just simplify the parentheses to (1.042)

5 0

3 years ago