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Marina86 [1]
2 years ago
10

Express 5x²-15x+1 in the form p(x+q)²+r, where p,q,r ate constants

Mathematics
1 answer:
AysviL [449]2 years ago
7 0

Answer:

5\bigg(x-\frac{3}{2}\bigg)^2-\frac{41}{4}

Step-by-step explanation:

  • 5x^2-15x +1

  • = 5\bigg(x^2-3x +\frac{1}{5}\bigg)

  • =5\bigg[x^2-2.\frac{3}{2}x +\bigg(\frac{3}{2}\bigg)^2-\bigg(\frac{3}{2}\bigg)^2+\frac{1}{5}\bigg]

  • =5\bigg[\bigg(x-\frac{3}{2}\bigg)^2+\frac{1}{5}-\frac{9}{4}\bigg]

  • =5\bigg[\bigg(x-\frac{3}{2}\bigg)^2+\frac{4-45}{20}\bigg]

  • =5\bigg(x-\frac{3}{2}\bigg)^2-5.\frac{41}{20}

  • =5\bigg(x-\frac{3}{2}\bigg)^2-5.\frac{41}{20}

  • =5\bigg(x-\frac{3}{2}\bigg)^2-\frac{41}{4}
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Describe a situation that the expression -15÷(-15) can represent
FrozenT [24]
-15 so that is negative. so that is negative 15÷15 so 15 ÷15 is 1 so it would be -15÷-15=-1
7 0
3 years ago
A Square was altered so that one side is increased by 9 inches in the other side is decreased by 2 inches.The area of the result
LekaFEV [45]

Let s represent the length of any one side of the original square.  The longer side of the resulting rectangle is s + 9 and the shorter side s - 2.

The area of this rectangle is (s+9)(s-2) = 60 in^2.

This is a quadratic equation and can be solved using various methods.  Let's rewrite this equation in standard form:  s^2 + 7s - 18 = 60, or:

s^2 + 7s - 78 = 0.  This factors as follows:  (s+13)(s-6)=0, so that s = -13 and s= 6.  Discard s = -13, since the side length cannot be negative.  Then s = 6, and the area of the original square was 36 in^2.

4 0
3 years ago
3. (05.06 MC)
AveGali [126]

Answer:

Step-by-step explanation:

y > 3x + 10

y < -3x - 1

<h3> Part A:</h3>

You'd graph the given systems of linear inequalities the same way as you graph the linear equations.  

To graph y > 3x + 10, plot the y-intercept, (0, 10), then use the slope, m = 3 (rise 3, run 1), to plot other points on the graph.  Use a dashed line (because of the ">" symbol).  

Follow the same steps for the other linear inequality. Plot the y-intercept, (0, -1), then use the slope, m = -3 (down 3, run 1) to plot other points. Use a dashed line (because of the "<" symbol).  

Pick a test point on either side of the boundary line and plug it into the original problem.  This will help determine which side of the boundary line is the solution.  Plug in a test point that is not on the boundary line.

Use the point of origin, (0, 0) as the test point. Plug in these values into the given systems of linear inequalities to see whether it will provide a true statement.  

y > 3x + 10

0 > 3(0) + 10

0 > 0 + 10

0 > 10 (False statement).  

y < -3x - 1

0 < -3(0) - 1

0 < 0 - 1

0 < -1 (false statement).  

Since the point of origin provided a false statement to the given systems of linear inequalities, you must shade the half-plane region where it doesn't contain the test point.  

<h3>Part B: </h3>

You'll do the same process as what I've done for the test point. Plug in the values of (8, 10) into the given systems of linear inequalities. If it provides a false statement, then it means that it is not a solution to the system.  

y > 3x + 10

10 > 3(8) + 10

10 > 24 + 10

10 > 34 (False statement).  

y < -3x - 1  

10 < -3(8) - 1

10 < -24 - 1

10 < -25 (false statement).  

Therefore, (8, 10) is not a solution to the system.    

8 0
3 years ago
The system shown has the unique solution (2, y, z). Solve the system and select the values that complete the solution. y = 0 y =
madreJ [45]
So we are given a system:
3x-2y+3z=0\\&#10;-3x - 5y - 5z= -21
Substitute x = 2 we get the system:
-2y+3z=-6\\&#10;- 5y - 5z= -15
Multiply the first equation by -5 and the second by 2 we get the system:
10y-15z=30\\&#10;- 10y - 10z= -30
Adding the two equations we get :
-25z=0\text{ then}z=0.
We find the value of y by using any of the other equations like this:
-2y=-6\\y=3.
Final solution:
z=0,y=3
8 0
3 years ago
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If the simple interest on ​$4000 for 3 years is ​$720​, then what is the interest​ rate?
solong [7]

Answer:

0.06% annually

Step-by-step explanation:

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\frac{240}{4000} = 0.06

4 0
2 years ago
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