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Margaret [11]
2 years ago
7

A soccer player gets 2 points for a goal and 1 point for an assist. If the combined number of goals and assists that a player ha

s is 24, and the total number of points that the player has is 42, which system of equations can be used to determine the number of goals and assists the player has? Assume g represents the number of goals and a represents the number of assists. G a = 24. G 2 a = 42. G a = 24. 2 g a = 42. G a = 42. G 2 a = 24. G a = 42. 2 g a = 24.
Mathematics
1 answer:
irakobra [83]2 years ago
3 0

The player has a total of 18 goals scored and 6 assists made.

<h3>Equation</h3>

Equation is an expression used to show the relationship between two or more numbers and variables.

Let x represent the number of goals scored and y represent the number of assists. Since the combined goals and assist is 24, hence:

x + y = 24     (1)

Also:

2x + y = 42    (2)

From equation 1 and 2:

x = 18, y = 6

The player has a total of 18 goals scored and 6 assists made.

Find out more on Equation at: brainly.com/question/2972832

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y = -2.5x + 5

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7 is the number we inserted for x, so it is the x value. -12.5 is the number we got out of the equation, so it is the y value.

The points will now look like this: (x, y) but for the example they would look like (7, -12.5)

Plot the numbers, x being on the horizontal axis, y being on the vertical axis.

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the 2nd one, 3rd one, and 5th one

Step-by-step explanation:

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As part of a screening process, computer chips must be operated in an oven at 145 °C. Ten minutes after starting, the temperatur
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Answer:

Step-by-step explanation:

I solved this using initial conditions and calculus, so I hope that's what you are doing in math.  It's actually NOT calculus, just a concept that is taught in calculus.

The initial condition formula we need is

y=Ce^{kt}

Filling in our formula with the 2 conditions we are given:

65=Ce^{10k}   and   85=Ce^{15k}

With those 2 equations, we have 2 unknowns, the C (initial value) and the k (the constant). We know that the initial value (or starting temp) for both conditions is the same, so we solve for C in one equation, sub it into the other equation and solve for k.  If

65=Ce^{10k} then

\frac{65}{e^{10k}}=C which, by exponential rules is the same as

C=65e^{-10k}

Since that value of C is the same as the value of C in the other equation, we sub it in:

85=65e^{-10k}(e^{15k})

Divide both sides by 65 and use the rules of exponents again to get

\frac{85}{65}=e^{-10k+15k} which simplifies down to

\frac{85}{65}=e^{5k}

Take the natural log of both sides to get

ln(\frac{85}{65})=5k

Do the log thing on your calculator to get

.2682639866 = 5k and divide both sides by 5 to find k:

k = .0536527973

Now that we have k, we sub THAT value in to one of the original equations to find C:

65=Ce^{10(.0536527973)}

which simplifies down to

65=Ce^{.536527973}

Raise e to that power on your calculator to get

65 = C(1.710059171) and divide to solve for C:

C = 38.01038064

Now sub in k and C to the final problem when t = 23:

y=38.01038064e^{(.0536527973)(23)} which simplifies a bit to

y=38.01038064e^{1.234014338}

Raise e to that power on your calculator to get

y = 38.01038064(3.434991111) and

finally, the temp at 23 minutes is

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Answer:

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Step-by-step explanation:

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