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marysya [2.9K]
1 year ago
5

What is the independent variable in the equation y=-5x+2?

Mathematics
1 answer:
Pepsi [2]1 year ago
3 0

Answer: <em>x</em>

Step-by-step explanation:

    For this equation, the variable <em>y</em> is dependent on the value of<em> x</em>. The value of <em>y</em> changes depending on the value of<em> x</em>. This means that <em>x</em> is the independent variable for this equation.

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ki77a [65]

Answer:

16

Step-by-step explanation:

4 0
2 years ago
Find the vaues of x and y
Leni [432]

x=76

y=4

I think that answer should be right. Since it is isosceles the base angles will be the same and the top angle is an angle bisector which means that the angle split on top has even splits. Since no other side lengths- or any lengths at that matter- are give, it is safe to assume that y=4.

4 0
2 years ago
Consider the functions z = 4 e^x ln y, x = ln (u cos v), and y = u sin v.
Katen [24]

Answer:

remember the chain rule:

h(x) = f(g(x))

h'(x) = f'(g(x))*g'(x)

or:

dh/dx = (df/dg)*(dg/dx)

we know that:

z = 4*e^x*ln(y)

where:

y = u*sin(v)

x = ln(u*cos(v))

We want to find:

dz/du

because y and x are functions of u, we can write this as:

dz/du = (dz/dx)*(dx/du) + (dz/dy)*(dy/du)

where:

(dz/dx)  = 4*e^x*ln(y)

(dz/dy) = 4*e^x*(1/y)

(dx/du) = 1/(u*cos(v))*cos(v) = 1/u

(dy/du) = sin(v)

Replacing all of these we get:

dz/du = (4*e^x*ln(y))*( 1/u) + 4*e^x*(1/y)*sin(v)

          = 4*e^x*( ln(y)/u + sin(v)/y)

replacing x and y we get:

dz/du = 4*e^(ln (u cos v))*( ln(u sin v)/u + sin(v)/(u*sin(v))

dz/du = 4*(u*cos(v))*(ln(u*sin(v))/u + 1/u)

Now let's do the same for dz/dv

dz/dv = (dz/dx)*(dx/dv) + (dz/dy)*(dy/dv)

where:

(dz/dx)  = 4*e^x*ln(y)

(dz/dy) = 4*e^x*(1/y)

(dx/dv) = 1/(cos(v))*-sin(v) = -tan(v)

(dy/dv) = u*cos(v)

then:

dz/dv = 4*e^x*[ -ln(y)*tan(v) + u*cos(v)/y]

replacing the values of x and y we get:

dz/dv = 4*e^(ln(u*cos(v)))*[ -ln(u*sin(v))*tan(v) + u*cos(v)/(u*sin(v))]

dz/dv = 4*(u*cos(v))*[ -ln(u*sin(v))*tan(v) + 1/tan(v)]

5 0
2 years ago
Use back-substitution to solve the system. (Enter your answers as a comma-separated list. If there is no solution, enter NO SOLU
nexus9112 [7]

Answer:

(x,y,z) = (5/2, 5/2, 0)

If z = t,

(x,y,z) = ((5-3t)/2, (5-t)/2, t)

Step-by-step explanation:

-x + y - z = 0

2y + z = 5

(1/5)z = 0

From eqn 3, z = 0

2y + z = 5

Substitute for z in eqn 2

2y + 0 = 5

y = 5/2

substituting for y and z in eq 1

-x + (5/2) - 0 = 0

x = (5/2)

(x,y,z) = (5/2, 5/2, 0)

In terms of t, if z = t,

eqn 2 becomes 2y + t = 5

2y = 5 - t

y = (5 - t)/2

Eqn1 becomes

-x + (5-t)/2 - t = 0

-x + (5/2) - (t/2) - t = 0

-x + (5/2) - (3t/2) = 0

x = (5-3t)/2

(x,y,z) = ((5-3t)/2, (5-t)/2, t)

5 0
3 years ago
HELP PLEASE!
Tems11 [23]

f(x) + n - shift the graph n units up

f(x) - n - shift the graph n units down

f(x + n) - shift the graph n units left

f(x - n) - shift the graph n units right

-----------------------------------------------------------------

g(x) = 4x² - 16

Shift 9 units right and 1 unit down. Therfore:

g(x - 9) - 1 = 4(x - 9)² - 16 - 1 = 4(x - 9)² - 17

<h3>Answer: A. h(x) = 4(x - 9)² - 17</h3>
3 0
3 years ago
Read 2 more answers
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