Evaluate (7^2))^-2/7^-4

PLEASE HELP ASAP!!! CORRECT ANSWERS ONLY PLEASE!!! Graph g(x) = 3x^2 - 12x - 3

Paul [167]

Answer: Vertex = (2, -15) 2nd point = (0, -3)

Step-by-step explanation:

g(x) = 3x² - 12x - 3

= 3(x² - 4x - 1)

a=1 b=-4 c=-1

Find the x-value of the vertex by using the formula for the axis of symmetry: $x = \dfrac{-b}{2a}$

$x = \dfrac{-(-4)}{2(1)}$

$= \dfrac{4}{2}$

= 2

Find the y-value of the vertex by plugging the x-value (above) into the given equation: g(x) = 3x² - 12x - 3

g(2) = 3(2)² - 12(2) - 3

= 12 - 24 - 3

= -15

So, the vertex is (2, -15) ← PLOT THIS COORDINATE

Now, choose a different x-value. Plug it into the equation and solve for y. I chose x = 0

g(0) = 3(0)² - 12(0) - 3

= 0 - 0 - 3

= -3

So, an additional point is (0, -3) ← PLOT THIS COORDINATE

5 0

2 years ago

CAn someone help me with thiss???

sashaice [31]

I used to have these in my middle school, i think searching up the lesson and lesson name will give you the entire answer key for that specific sheet

6 0

2 years ago

Just numbers 43-45 In the picture please!!

steposvetlana [31]

43) 50% of the time, Store A has 550 or more customers per day.
44) Distribution A appears to be skewed to the left with no outliers and a center at 500 customers. Distribution B appears to be Normal with no outliers and a center at 400 customers.
45) Store A has more customers.

7 0

2 years ago

Expand can mean "To stretch or unfold." How can this definition help you remember this definition

mihalych1998 [28]

There are several examples that can help one remember the definition of expand. To begin with, a balloon can expand when air or water enters it, depending on the type. Cookies often expand when baking due to the leavening agent used in the recipe. When we stretch our bodies, our muscles often expand.

7 0

3 years ago

Use Lagrange multipliers to find the maximum and minimum values of (i) f(x,y)-81x^2+y^2 subject to the constraint 4x^2+y^2=9. (i

sp2606 [1]

i. The Lagrangian is

$L(x,y,\lambda)=81x^2+y^2+\lambda(4x^2+y^2-9)$

with critical points whenever

$L_x=162x+8\lambda x=0\implies2x(81+4\lambda)=0\implies x=0\text{ or }\lambda=-\dfrac{81}4$

$L_y=2y+2\lambda y=0\implies2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

$L_\lambda=4x^2+y^2-9=0$

If $x=0$ , then $L_\lambda=0\implies y=\pm3$ .
If $y=0$ , then $L_\lambda=0\implies x=\pm\dfrac32$ .
Either value of $\lambda$ found above requires that either $x=0$ or $y=0$ , so we get the same critical points as in the previous two cases.

We have $f(0,-3)=9$ , $f(0,3)=9$ , $f\left(-\dfrac32,0\right)=\dfrac{729}4=182.25$ , and $f\left(\dfrac32,0\right)=\dfrac{729}4$ , so $f$ has a minimum value of 9 and a maximum value of 182.25.

ii. The Lagrangian is

$L(x,y,z,\lambda)=y^2-10z+\lambda(x^2+y^2+z^2-36)$

with critical points whenever

$L_x=2\lambda x=0\implies x=0$ (because we assume $\lambda\neq0$ )

$L_y=2y+2\lambda y=0\implies 2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

$L_z=-10+2\lambda z=0\implies z=\dfrac5\lambda$

$L_\lambda=x^2+y^2+z^2-36=0$

If $x=y=0$ , then $L_\lambda=0\implies z=\pm6$ .
If $\lambda=-1$ , then $z=-5$ , and with $x=0$ we have $L_\lambda=0\implies y=\pm\sqrt{11}$ .

We have $f(0,0,-6)=60$ , $f(0,0,6)=-60$ , $f(0,-\sqrt{11},-5)=61$ , and $f(0,\sqrt{11},-5)=61$ . So $f$ has a maximum value of 61 and a minimum value of -60.

5 0

3 years ago