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ioda
4 years ago
5

What are the solutions to the system of equations? y=2x2−5x−7 y=2x+2

Mathematics
1 answer:
IceJOKER [234]4 years ago
5 0

Answer:

Step-by-step explanation:

y=2x²-5x-7

y=2x+2

2x²-5x-7=2x+2

2x²-7x-9=0

2x²-9x+2x-9=0

x(2x-9)+1(2x-9)=0

(2 x-9)(x+1)=0

x=9/2 ,-1

y=11 ,0

solutions are (9/2,11),(-1,0)

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Answer:

0.7

Step-by-step explanation:

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3 years ago
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1. S(–4, –4), P(4, –2), A(6, 6) and Z(–2, 4) a) Apply the distance formula for each side to determine whether SPAZ is equilatera
Aleksandr [31]

Answer:

a) SPAZ is equilateral.

b) Diagonals SA and PZ are perpendicular to each other.

c) Diagonals SA and PZ bisect each other.

Step-by-step explanation:

At first we form the triangle with the help of a graphing tool and whose result is attached below. It seems to be a paralellogram.

a) If figure is equilateral, then SP = PA = AZ = ZS:

SP = \sqrt{[4-(-4)]^{2}+[(-2)-(-4)]^{2}}

SP \approx 8.246

PA = \sqrt{(6-4)^{2}+[6-(-2)]^{2}}

PA \approx  8.246

AZ =\sqrt{(-2-6)^{2}+(4-6)^{2}}

AZ \approx 8.246

ZS = \sqrt{[-4-(-2)]^{2}+(-4-4)^{2}}

ZS \approx 8.246

Therefore, SPAZ is equilateral.

b) We use the slope formula to determine the inclination of diagonals SA and PZ:

m_{SA} = \frac{6-(-4)}{6-(-4)}

m_{SA} = 1

m_{PZ} = \frac{4-(-2)}{-2-4}

m_{PZ} = -1

Since m_{SA}\cdot m_{PZ} = -1, diagonals SA and PZ are perpendicular to each other.

c) The diagonals bisect each other if and only if both have the same midpoint. Now we proceed to determine the midpoints of each diagonal:

M_{SA} = \frac{1}{2}\cdot S(x,y) + \frac{1}{2}\cdot A(x,y)

M_{SA} = \frac{1}{2}\cdot (-4,-4)+\frac{1}{2}\cdot (6,6)

M_{SA} = (-2,-2)+(3,3)

M_{SA} = (1,1)

M_{PZ} = \frac{1}{2}\cdot P(x,y) + \frac{1}{2}\cdot Z(x,y)

M_{PZ} = \frac{1}{2}\cdot (4,-2)+\frac{1}{2}\cdot (-2,4)

M_{PZ} = (2,-1)+(-1,2)

M_{PZ} = (1,1)

Then, the diagonals SA and PZ bisect each other.

8 0
3 years ago
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Stels [109]

Answer:

See below.

Step-by-step explanation:

First, we can see that \lim_{x \to 2}  (f(x))= -1.

Thus, for the question, we can just plug -1 in:

\lim_{x \to 2} (\frac{x}{f(x)+1})=\frac{(2)}{-1+1}  =und.

Saying undefined (or unbounded) will be correct.

However, note that as x approaches 2, the values of y decrease in order to get to -1. In other words, f(x) will always be greater or equal to -1 (you can also see this from the graph). This means that as x approaches 2, f(x) will approach -.99 then -.999 then -.9999 until it reaches -1 and then go back up. What is important is that because of this, we can determine that:

\lim_{x \to 2} (\frac{x}{f(x)+1})=\frac{(2)}{-1+1}  = +\infty

This is because for the denominator, the +1 will always be greater than the f(x). This makes this increase towards positive infinity. Note that limits want the values of the function as it approaches it, not at it.

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Answer:

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3 years ago
Andre and Noah started tracking their savings at the same time. Andre started with $15 and deposits $5 per week. Noah started wi
Paha777 [63]

Answer:

1. he started with 15 and 5 per week. w is week

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