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Contact [7]
3 years ago
14

Solve the following system of equations. In other words, which  values for x and y make both equations true? 

Mathematics
1 answer:
Angelina_Jolie [31]3 years ago
3 0
I'm assuming the two equations are as follows:
y = 5x
7x + 2y = -17

You can substitute 5x for y in the second equation since they are equivalent. Remember that whatever you do to one side you must do to the other as well.
7x + 2y = -17
7x + 2(5x) = -17
7x + 10x = -17
17x = -17
17x /17 = -17 /17
x = -1

Now we sub in the value we have for x to find y:
y = 5x
y = 5(-1)
y = -5

We now check that both answers are right by substituting them both into the second equation:
7x + 2y = -17
7(-1) + 2(-5) = -17
-7 + (-10) = -17
-17 = -17

And it looks like it works! So x = -1, and y = -5. Hope this helps! :)
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You spend 2 hours and 15 minutes commuting to work each day. You work five days per week. How much time do you spend commuting t
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11 hours and 15 minutes because doing 2.25, which is equal to 2 hours and 15 minutes, times 5 you end up getting 11.25, and converting that to time you get 11 hours and 15 minutes.

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3 years ago
A car travels at a speed of 55 miles per hour. How far will it travel in 5 hours?
kiruha [24]

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55*5=275

Step-by-step explanation:

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A leaking faucet drips into a bucket. The faucet drips at a constant rate. The level of water in the bucket rises
Sveta_85 [38]

Answer:

The per hour rate is 0.5 inches per hour

Step-by-step explanation:

Here, we want to calculate the rate at which the level of the water increased

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7 0
3 years ago
I am having trouble with this relative minimum of this equation.<br>​
Norma-Jean [14]

Answer:

So the approximate relative minimum is (0.4,-58.5).

Step-by-step explanation:

Ok this is a calculus approach.  You have to let me know if you want this done another way.

Here are some rules I'm going to use:

(f+g)'=f'+g'       (Sum rule)

(cf)'=c(f)'          (Constant multiple rule)

(x^n)'=nx^{n-1} (Power rule)

(c)'=0               (Constant rule)

(x)'=1                (Slope of y=x is 1)

y=4x^3+13x^2-12x-56

y'=(4x^3+13x^2-12x-56)'

y'=(4x^3)'+(13x^2)'-(12x)'-(56)'

y'=4(x^3)'+13(x^2)'-12(x)'-0

y'=4(3x^2)+13(2x^1)-12(1)

y'=12x^2+26x-12

Now we set y' equal to 0 and solve for the critical numbers.

12x^2+26x-12=0

Divide both sides by 2:

6x^2+13x-6=0

Compaer 6x^2+13x-6=0 to ax^2+bx+c=0 to determine the values for a=6,b=13,c=-6.

a=6

b=13

c=-6

We are going to use the quadratic formula to solve for our critical numbers, x.

x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}

x=\frac{-13 \pm \sqrt{13^2-4(6)(-6)}}{2(6)}

x=\frac{-13 \pm \sqrt{169+144}}{12}

x=\frac{-13 \pm \sqrt{313}}{12}

Let's separate the choices:

x=\frac{-13+\sqrt{313}}{12} \text{ or } \frac{-13-\sqrt{313}}{12}

Let's approximate both of these:

x=0.3909838 \text{ or } -2.5576505.

This is a cubic function with leading coefficient 4 and 4 is positive so we know the left and right behavior of the function. The left hand side goes to negative infinity while the right hand side goes to positive infinity. So the maximum is going to occur at the earlier x while the minimum will occur at the later x.

The relative maximum is at approximately -2.5576505.

So the relative minimum is at approximate 0.3909838.

We could also verify this with more calculus of course.

Let's find the second derivative.

f(x)=4x^3+13x^2-12x-56

f'(x)=12x^2+26x-12

f''(x)=24x+26

So if f''(a) is positive then we have a minimum at x=a.

If f''(a) is negative then we have a maximum at x=a.

Rounding to nearest tenths here:  x=-2.6 and x=.4

Let's see what f'' gives us at both of these x's.

24(-2.6)+25

-37.5  

So we have a maximum at x=-2.6.

24(.4)+25

9.6+25

34.6

So we have a minimum at x=.4.

Now let's find the corresponding y-value for our relative minimum point since that would complete your question.

We are going to use the equation that relates x and y.

I'm going to use 0.3909838 instead of .4 just so we can be closer to the correct y value.

y=4(0.3909838)^3+13(0.3909838)^2-12(0.3909838)-56

I'm shoving this into a calculator:

y=-58.4654411

So the approximate relative minimum is (0.4,-58.5).

If you graph y=4x^3+13x^2-12x-56 you should see the graph taking a dip at this point.

3 0
3 years ago
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