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Alja [10]
3 years ago
14

Qhich equation is the inverse of y=9x^2-4

Mathematics
1 answer:
Anvisha [2.4K]3 years ago
4 0
To find the inverse of this equation, switch the positions of the variables 'x' and 'y' to create an equation for 'x' in terms of 'y'. x=9 y^{2} -4

Now we solve this equation for 'y'.
x=9 y^{2} -4 add 4 to both sides
x+4=9 y^{2} divide both sides by 9
\frac{x+4}{9} = y^{2} take the square root of both sides
\frac{ \sqrt{x+4} }{3} =y
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If Sean earns $8.50 per hour sand made a total of $80.75 today, How many hours did he work?
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Pepsi [2]

I could not see the first question so i couldn't answer.

2) Draw a horizontal line and cut it in half with a vertical angle. 4 equal quarters will form. There each quarter is 90° (right angle is exactly 90°angle)

3) Just add the two angles.

x =13+19

x= 32° (Answer)

4) We must know the other angle value in order to find x. So considering both the angles are equal,

2x = 44

x= 22° (Answer)

5) Yes, x = 80° because the line cutted is a straight line and the value of a straight line is 180°

6) Supplementary angles total 180°

Complementary angles total 90°

7)Opposite angles is equal when two straight lines intersect. So,

14x - 16 = 12x + 22

14x-12x=22+16

2x= 38

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3 0
2 years ago
Write an expression for the 12th partial sum of the series 3/2+7/3+19/6+... using summation notation
lapo4ka [179]

Answer:

S_{12}=\sum_{i=1}^{12} [\frac{3}{2}+(i-1)\times \frac{5}{6}]

S_{12}=73

Step-by-step explanation:

First\ term\ of\ the\ series(a_1)=\frac{3}{2}\\\\Second\ term\ of\ the\ series(a_2)=\frac{7}{3}\\\\Third\ term\ of\ the\ series(a_3)=\frac{19}{6}\\\\a_2-a_1=\frac{7}{3}-\frac{3}{2}=\frac{5}{6}\\\\a_3-a_2=\frac{19}{6}-\frac{7}{3}=\frac{5}{6}\\\\Hence\ it\ is\ an\ Arithmetic\ Series\ with\ first\ term=\frac{3}{2}\ and\ constant\ difference=\frac{5}{6}

a_1=\frac{3}{2}+0\times \frac{5}{6}\\\\a_2=\frac{3}{2}+1\times \frac{5}{6}\\\\a_3=\frac{3}{2}+2\times \frac{5}{6}\\\\.\\.\\.\\a_n=\frac{3}{2}+(n-1)\times \frac{5}{6}\\\\S_n=a_1+a_2+a_3+......+a_n\\\\S_n=(\frac{3}{2}+0\times \frac{5}{6})+(\frac{3}{2}+1\times \frac{5}{6})+(\frac{3}{2}+2\times \frac{5}{6})+....+(\frac{3}{2}+[n-1]\times \frac{5}{6})\\\\S_n=\sum_{i=1}^n [\frac{3}{2}+(i-1)\times \frac{5}{6}]\\\\S_n=(\frac{3}{2}+\frac{3}{2}+\frac{3}{2}+...n\ times)+\frac{5}{6}(1+2+3+4+...+(n-1))\\\\

S_n=\frac{3}{2}\times n+\frac{5}{6}\times \frac{n(n-1)}{2}\\\\

S_{12}=\sum_{i=1}^{12} [\frac{3}{2}+(i-1)\times \frac{5}{6}]

S_{12}=\frac{3}{2}\times 12+\frac{5}{6}\times \frac{(12)(12-1)}{2}\\\\S_{12}=18+55\\\\S_{12}=73

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3 years ago
At the Rights of Mankind High School, there is a ratio of two boys to every three girls. If there are 122 boys, how many student
Marina86 [1]
122 students total thats how many students are at romhs
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