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BigorU [14]
3 years ago
7

6x+2y=-2 3x-2y=-5 What is the result of adding these 2 equations? PLEASE HELP

Mathematics
1 answer:
DochEvi [55]3 years ago
7 0

Answer: x=-7/9, y=4/3. (-7/9, 4/3).

Step-by-step explanation: 6x+2y=-2

3x-2y=-5

-----------------

9x=-7

x=-7/9

6(-7/9)+2y=-2

-42/9+2y=-2

2y=-2-(-42/9)

2y=-2+42/9

2y=-18/9+42/9

2y=24/9

y=(24/9)/2

y=(24/9)(1/2)

y=24/18

y=4/3

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The polynomial $f(x)$ has degree 3. If $f(-1) = 15$, $f(0)= 0$, $f(1) = -5$, and $f(2) = 12$, then what are the $x$-intercepts o
Alex_Xolod [135]

Your calculator's cubic regression function can tell you the equation is

... f(x) = 2x³ + 5x² -12x = x(x +4)(2x-3)

The x-intercepts are -4, 0, +1.5.

_____

If you want to solve this "by hand", you can first of all recognize that since there is an x-intercept at 0, the cubic will only have three coefficients. That is, you can write the equation as

... f(x) = ax³ + bx² + cx

Substituting the given points (except (0, 0)) gives three linear equations in a, b, c.

... -a +b -c = 15 . . . . . for x=-1

... a + b + c = -5 . . . . for x=1

... 8a +4b +2c = 12 . . for x=2

adding the first two equations gives 2b=10, or b=5. Now, you can reduce the system to

... a + c = -10

... 4a +c = -4

Subtracting the first of these equations from the second gives 3a=6, or a=2. That tells you c=-12 (from a+c=10).

Then your equation is

... f(x) = x(2x² +5x -12)

Factoring by any of the usual techniques, or graphing, or using the quadratic formula will tell you the zeros (x-intercepts) are as above.

_____

Since the input values are sequential, you can also develop the function from differences of the output values. Those are 15, 0, -5, 12. First differences are -15, -5, +17. Second differences are +10, +22. The third difference is 12. Using the first of these differences in appropriate places in the interpolating polynomial formula, we have

... f(x) = 15 + (x+1)(-15 + (x)/2·(10 + (x-1)/3·(12))) = 2x³ +5x² -12x . . . . as above

4 0
3 years ago
25 mi/hr is equal to m/min
svp [43]
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8 0
3 years ago
Can someone help me please?
11111nata11111 [884]
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3 0
3 years ago
I need help. this is the problem x^2/3+10=7x^1/3
Arlecino [84]

\it x^{\frac{2}{3}}+10=7x^{\frac{1}{3}} \Leftrightarrow  (x^{\frac{1}{3}})^2-7x^{\frac{1}
  {3}} +10=0 
\\\;\\
We\ note\ x^{\frac{1}{3}} =t \ \ and \ the \ equation \ will \ be:
\\\;\\
t^2-7t+10 = 0 \Leftrightarrow t^2 -2t-5t+10=0 \Leftrightarrow t(t-2) -5(t-2)=0

\it \Leftrightarrow (t-2)(t-5)=0 
\\\;\\
t-2=0 \Rightarrow t=2 \Rightarrow x^{\frac{1}{3}}=2 \Rightarrow (x^{\frac{1}{3}})^3 =2^3 \Rightarrow x = 8
\\\;\\
t-5=0 \Rightarrow t=5 \Rightarrow x^{\frac{1}{3}}= 5  \Rightarrow (x^{\frac{1}{3}})^3 = 5^3 \Rightarrow x = 125

S = {8; 125}


7 0
3 years ago
Which of the following best shows the commutative property of addition?
Aleksandr-060686 [28]

Answer:

I think it's number 2 but I'm not sure

6 0
3 years ago
Read 2 more answers
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