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3 years ago

PLEASE HELP ME ANYONE PLEASE HELP I HAVE 15 MI NUTES

Mathematics

1 answer:

OlgaM077 [116]3 years ago

4 0

The answer is A, guess I am a bit late to tell you that though. let me explain why the answer is A. A is the only point that would not be in the blue section. All the other points listed would be in the blue section. If you would like further explanation just ask :)<span />

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A city survey of two neighborhoods asked residents whether they would

masha68 [24]

Answer: 25

Step-by-step explanation:

look at the question

4 0

2 years ago

How to find the maximum or minimum value of a quadratic function?

kipiarov [429]

Answer:

Vertex form

Step-by-step explanation:

You convert to vertex form a(x - b)^2 + c . The coordinates of the maxm/minm will be (b, c).

For example find minimum value of x^2 + 5x - 6:-

x^2 + 5x - 6

= (x + 2.5)^2 - 6.25 - 6

= (x + 2.5)^2 - 12.25

The coordinates of minimumm will be (-2.5, -12.25) The values of the minimum of the function is -12.25

6 0

3 years ago

Add the following rational number =<br> 3/11 and -5/12

tatiyna

Answer:

-19/132

Step-by-step explanation:

To add fractions, find the lowest common multiple of the two denominators and multiply accordingly to make both denominators the same. In this case, the lowest common multiple of 11 and 12 is 132, so we need to multiply the first number by 12 and the second by 11. So we get $\frac{36}{132} + (-\frac{55}{132} )$ = $\frac{36-55}{132}$ = $\frac{-19}{32}$ $-\frac{19}{132}$

4 0

2 years ago

URGENT PLZ HELP ME!!!

hichkok12 [17]

Answer:

You would click at (0,-7)

Step-by-step explanation:

Definition of the minimum point:

"The minimum value of a function is the place where the graph has a vertex at its lowest point. In the real world, you can use the minimum value of a quadratic function to determine minimum cost or area."

Although this is not a quadratic, it still has a minimum point.

The minimum point here would be at it's lowest point

The minimum/lowest point is (0,-7)

4 0

2 years ago

Find the volume of the largest rectangular box in the first octant with three faces in the coordinate planes and one vertex in t

makkiz [27]

Answer:

Step-by-step explanation:

$x + 2y + 3z = 12$

$z=\frac{12-x-2y}{3}$

Volume = $V=f(x,y) = xy(\frac{12-x-2y}{3} )$

find partial derivatives using product rule

$f_x =\frac{y}{3} (12-2x-2y)\\f_y = \frac{x}{3} (12-x-4y)$

i.e.

Using maximum for partial derivatives, we equate first partial derivative to 0.

y=0 or x+y =6

x=0 or x+4y =12

Simplify to get y =2, x = 4

thus critical points are (4,2) (6,0) (0,3)

Of these D the II derivative test gives

D<0 only for (4,2)

Hence maximum volume is when x=4, y=2, z= 4/3

Max volume is = 4(2)(4/3) = 32/3

6 0

2 years ago