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3 years ago

HELP!!!!!!!!!!!!QUICK!!!!!!!!!!!Plzzzzzzzzzzzz,!!!!!!!!!!!!!

Mathematics

1 answer:

Novay_Z [31]3 years ago

6 0

Answer:

the thirs one

Step-by-step explanation:

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Consider the function f(x)=-3x^2 +7x -k. [3 Marks] a) For what values of k will the function have no zeros? b) For what values o

Anna35 [415]

Answer: a) k >4.08

b) k = 4.08

c) k<4.08

Step-by-step explanation:

Since we have given that

$f(x)=-3x^2+7x-k$

a) For what values of k will the function have no zeros?

It mean it has no real zeroes i.e. Discriminant < 0

As we know that

$D=b^2-4ac$

Here, a =-3

b = 7

c = -k

So, it becomes,

$D$

b) For what values of k will the function have one zero?

It means it has one real root i.e equal roots.

So, in this case, D = 0

So, it becomes,

$D=b^2-4ac=0\\\\D=7^2-4\times -3\times -k=0\\\\49-12k=0\\\\49=12k\\\\k=\dfrac{49}{12}\\\\k=4.08$

c) For what values of k will the function have two zeros?

It means it has two real roots.

In this case, D>0

So, it becomes,

$D=7^2-4\times -3\times -k>0\\\\49-12k>0\\\\-12k>-49\\\\12k$

Hence, a) k >4.08

b) k = 4.08

c) k<4.08

3 0

3 years ago

Given the following equation where A = Area of a rectangle and w = width of the rectangle, what value of 'w' would maximize the

Papessa [141]

Answer:

the second option : w should be 25 units

Step-by-step explanation:

the area of the rectangle is length×width = L×W

the perimeter of a rectangle = 2L + 2W

now, we know that the perimeter is 100 units.

and we have to find the best length of W, that will then define L (to keep the 100 units of perimeter) and maximizes the area of the rectangle.

in other words, what is the maximum area of a rectangle with perimeter of 100 (and what are the corresponding side lengths)?

now, w = 625 is impossible. that side alone would be bigger than the whole perimeter.

W = 0 would render the whole rectangle to a flat line with L = 50 because of

100 = 2L + 2W = 2L + 0 = 2L

L = 50

and A = L×W = 50×0 = 0

an area of 0 is for sure not the largest possible area.

w = 50 would cause L = 0

100 = 2L + 2W = 2L + 2×50 = 2L + 100

0 = 2L

L = 0

and with L = 0 the same thing happens as with W = 0 : a flat line with 0 area.

so, the only remaining useful answer is W = 25

100 = 2L + 2W = 2L + 2×25 = 2L + 50

50 = 2L

L = 25

A = L×W = 25×25 = 625 units²

and indeed, the maximum area for a given perimeter is achieved by arranging the sides to create a square.

8 0

3 years ago