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ziro4ka [17]
2 years ago
11

) Tara earns twice as much per hour as Kayte.

Mathematics
2 answers:
tekilochka [14]2 years ago
8 0

Answer:

Austin's hourly wage is $8.

Step-by-step explanation:

We can write this problem as a system of linear equations.

We define T: Tara's hourly wage, A: Austin's hourly wage and K: Kayte's hourly wage.

Then, if Tara earns twice as much per hour as Kayte, we have:

T=2K

If Kayte earns $3 more per hour than Austin, we have:

K=A+3

And if they earn $41 per hour as a group, we know:

T+K+A=41

We can use all equations to replace in the third one, as:

T+K+A=41\\\\(2K)+K+A=41\\\\3K+A=41\\\\3(A+3)+A=41\\\\3A+9+A=41\\\\4A=41-9=32\\\\A=32/4=8

Austin's hourly wage is $8.

Cerrena [4.2K]2 years ago
5 0

Answer:

Austin's hourly wage is $8.

Step-by-step explanation:

This question can be solved using a system of equations.

I am going to say that:

Tara's hourly wage is x.

Kayte's hourly wage is y.

Austin's hourly wage is z.

Tara earns twice as much per hour as Kayte.

This means that x = 2y

Kayte earns $3 more per hour than Austin.

This means that y = z + 3

As a group, they earn $41 per hour.

This means that x + y + z = 41

What is Austin's hourly wage?​

This is z.

x + y + z = 41

y = z + 3 and x = 2y, so x = 2(z + 3) = 2z + 6

x + y + z = 41

2z + 6 + z + 3 + z = 41

4z + 9 = 41

4z = 32

z = \frac{32}{4}

z = 8

Austin's hourly wage is $8.

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The correct question is:

Consider the initial value problem

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Step-by-step explanation:

Given the differential equation

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a) We need to find the value of the constant C and r, such that y = Ct^r is a solution to the differential equation together with the initial condition y(-1) = 1.

Since Ct^r is a solution to the initial value problem, it means that y = Ct^r satisfies the said problem. That is

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Suppose F(x, y) is continuous on some region, R = {(x, y) : x_0 − δ < x < x_0 + δ, y_0 −ę < y < y_0 + ę} containing the point (x_0, y_0). Then there exists a number δ1 (possibly smaller than δ) so that a solution y = f(x) to (1) is defined for x_0 − δ1 < x < x_0 + δ1.

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