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zhenek [66]
3 years ago
10

What equation is the inverse of y=2x^2-8?

Mathematics
1 answer:
Fynjy0 [20]3 years ago
5 0

Step-by-step explanation:

To get the inverse, swap the x- and y-variables, then solve for y. We should have the equation: x = 2y^2 - 8

Solving for y:    

y = 2x^{2} - 8    because  y + 8 = 2x^{2} or  x =\sqrt\frac{y +  8}{2}

So on interchanging the variable, we get the equation:

x = 2y^2 - 8

x + 8 = 2y^2

y = \sqrt\frac{x + 8}{2}

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On a math test, you obtained a score of 66 points. This was 88% ofthe total number of points on the test. How many total points
Lyrx [107]

Answer:

75

Step-by-step explanation:

Let x be the total points

66/x  = 0.88 (88%)

x = 75

3 0
3 years ago
Jane and Sue decide to make friendship bracelets for all their classmates. Jane brings 4.5 feet of string, and together they hav
Olenka [21]
5.5 ft

10 - 4.5 = 5.5
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Please help with this practice quiz!!
Nady [450]

Answer:

slope is 9

Step-by-step explanation:

use the formula y=mx+b wherein m is the slope and b is the y-intercept

to solve for m, use formula (y2-y1)/(x2-x1):

\frac{127-1}{5-(-9)} = \frac{126}{14} (you add 5 and 9 together because negative + negative is positive)

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5 0
3 years ago
7+3√5/3+√5 - 7-3√5/3-√5 = a + b√5
daser333 [38]
<h3>Given:-</h3>

\\ \sf \implies\frac{7 + 3 \sqrt{5} }{3 +  \sqrt{5} }  -  \frac{7  - 3 \sqrt{5} }{3  -   \sqrt{5} }  = a +  \sqrt{5} b \\

<h3>To Find:-</h3>

  • The value of a and b

<h3>Solution:-</h3>

\\ \sf \implies\frac{7 + 3 \sqrt{5} }{3 +  \sqrt{5} }  -  \frac{7  - 3 \sqrt{5} }{3  -   \sqrt{5} }  = a +  \sqrt{5} b \\

\\ \sf \implies\frac{( \: 7 + 3 \sqrt{5} \:  \: (  3  -   \sqrt{5}) \:  \:  - 7  -  3 \sqrt{5} \:  \: (  3   +    \sqrt{5}) \: }{(3 +  \sqrt{5})  \:  \:(3 +  \sqrt{5}) }   = a +  \sqrt{5}  \:  b\\

\\ \sf \implies\frac{( \: 21 - 7 \sqrt{5} \:   +  9    \sqrt{5} - 15) \:  \:  - ( \: 21  + 7 \sqrt{5} \:    -  9    \sqrt{5}  +  15)\: }{(3 +  \sqrt{5})  \:  \:(3 +  \sqrt{5}) }   = a +  \sqrt{5}  \:  b\\

\\ \sf \implies\frac{( \: 6 + 2 \sqrt{5} ) \:  \:  - ( \: 6 - 2 \sqrt{5} )\: }{(3 +  \sqrt{5})  \:  \:(3 +  \sqrt{5}) }   = a +  \sqrt{5}  \:  b\\

\\ \sf \implies\frac{\: 6 + 2 \sqrt{5}  \:  \:  -  \: \: 6 - 2 \sqrt{5} \: }{(3 +  \sqrt{5})  \:  \:(3 +  \sqrt{5}) }   = a +  \sqrt{5}  \:  b\\

\\ \sf \implies\frac{\: 4 \sqrt{5}  \:  \:   }{3  {}^{2}   -  {\sqrt{5} }^{2}  }   = a +  \sqrt{5}  \:  b\\

\\ \sf \implies\frac{\: 4 \sqrt{5}  \:  \:   }{ \:  \:  \:  \: 9 - 5 \:  \:  \:   }   = a +  \sqrt{5}  \:  b\\

\\ \sf \implies\frac{\: 4 \sqrt{5}  \:  \:   }{ \:  \:  \:  \: 4 \:  \:  \:   }   = a +  \sqrt{5}  \:  b\\

\\ \sf \implies\frac{\: \cancel{4 } \sqrt{5}  \:  \:   }{ \:  \:  \:  \:  \cancel{4 }\:  \:  \:   }   = a +  \sqrt{5}  \:  b\\

\\ \sf \implies \: \sqrt{5}  = a +  \sqrt{5}  \:  b\\

we can also write it as ;

\\ \sf \implies \: 0 + \sqrt{5}  = a +  \sqrt{5}  \:  b\\

★<u> </u><u>Henceforth, the value of a and b are</u> :

→ a = 0

→ b = 1

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Solve the system using elimination.
Alchen [17]
\left \{ {{7x+5y=15\ \ |*4} \atop {-6x-4y=-14\ \ |*5}} \right. \\\\ \left \{ {{28x+20y=60} \atop {-30x-20y=-70}} \right. \\+------\\&#10;-2x=-10\ \ \ |Divide\ by\ -2\\&#10;x=5\\\\&#10;5y=15-7x\\&#10;5y=15-7*5\\&#10;5y=15-35\\&#10;5y=-20\ \ \ |:5\\&#10;y=-4\\\\\ Solution\ is\ B(5,-4).
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3 years ago
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