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xeze [42]
3 years ago
14

Find the inverse of the function. y = 2x2 –4

Mathematics
2 answers:
Orlov [11]3 years ago
6 0

Answer:

y = ± √(x+4)/2

Step-by-step explanation:

VashaNatasha [74]3 years ago
4 0

Answer:

(2÷2)+4

step by step explanation

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Can somebody write this is quadratic form please
galina1969 [7]

To write an equation in standard form, move each term to the left side of the equation and simplify.

ax2+bx+c = 0ax2+bx+c = 0

Move all the expressions to the left side of the equation.

−17−8x2−4x+3x2 = 0-17-8x2-4x+3x2 = 0

Add −8x2-8x2 and 3x23x2.

−17−5x2−4x = 0-17-5x2-4x = 0

Reorder the polynomial.

5x2+4x+17 = 0

Step-by-step explanation:

I hope this helped-

7 0
3 years ago
Which equation is equivalent to 63-3(2-10x)=150
kogti [31]

Answer:

D: 57 + 30x = 150

Step-by-step explanation:

63-3(2-10x)=150

63 - 6 + 30x = 150

57 + 30x = 150

4 0
3 years ago
Read 2 more answers
Graphs of what functions are shown below?
shtirl [24]

Answer:

-3/5

Step-by-step explanation:

6 0
3 years ago
I need help with 8-11....
MaRussiya [10]
8. Quadrants
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8 0
3 years ago
Use the fundamental theorem of calculus to find the area of the region between the graph of the function x^5 + 8x^4 + 2x^2 + 5x
BaLLatris [955]

Answer:

The area of the region is 25,351 units^2.

Step-by-step explanation:

The Fundamental Theorem of Calculus:<em> if </em>f<em> is a continuous function on </em>[a,b]<em>, then</em>

                                   \int_{a}^{b} f(x)dx = F(b) - F(a) = F(x) |  {_a^b}

where F is an antiderivative of f.

A function F is an antiderivative of the function f if

                                                    F^{'}(x)=f(x)

The theorem relates differential and integral calculus, and tells us how we can find the area under a curve using antidifferentiation.

To find the area of the region between the graph of the function x^5 + 8x^4 + 2x^2 + 5x + 15 and the x-axis on the interval [-6, 6] you must:

Apply the Fundamental Theorem of Calculus

\int _{-6}^6(x^5+8x^4+2x^2+5x+15)dx

\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx\\\\\int _{-6}^6x^5dx+\int _{-6}^68x^4dx+\int _{-6}^62x^2dx+\int _{-6}^65xdx+\int _{-6}^615dx

\int _{-6}^6x^5dx=0\\\\\int _{-6}^68x^4dx=\frac{124416}{5}\\\\\int _{-6}^62x^2dx=288\\\\\int _{-6}^65xdx=0\\\\\int _{-6}^615dx=180\\\\0+\frac{124416}{5}+288+0+18\\\\\frac{126756}{5}\approx 25351.2

3 0
3 years ago
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