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4 years ago

What is the difference between “ and “ and “ or “ ?

Mathematics

1 answer:

Nata [24]4 years ago

6 0

Answer:

And is an inclusive conjunction, meaning more can follow and exist at the same time, while or is an exclusive conjunction, meaning only one choice can exist at one time.

Step-by-step explanation:

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Simplify.

Nitella [24]

Answer:

I hope this is the whole question

Step-by-step explanation:

Well I can only see one way and it is this...

$\sqrt{27a}$ = $\sqrt{3.3.3.a}$

= $3\sqrt{3a}$

I got rid of one perfect-square. I dont think you can do much more.

6 0

2 years ago

Solve.<br> 4(x+2)^2=16<br><br><br> HELP PLSSSS

serious [3.7K]

Answer:

x=0; x=-2

Step-by-step explanation:

so first you do (x+2)^2 which is x^2+2x+4

4x^2 +8x +16 = 16

Subtract 16 from both sides

4x^2 +8x = 0

factor out 4x:

4x(x+2) = 0

4 0

2 years ago

$120 is shared among 3 friends Ava, Ben, and Carlos. If Ava receives $20 less than

kenny6666 [7]

Answer:$20

Step-by-step explanation:

If Carlos had $20

Ben would have $60

And Ava would have $40

8 0

3 years ago

The dry cleaning fee for 3 pairs of pants is $18. what is the constant of proportionality

zloy xaker [14]

Every pair of pants would cost 18/3 dollars

6 0

3 years ago

Read 2 more answers

Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software,

vichka [17]

Answer:

A(-1,0) is a local maximum point.

B(-1,0) is a saddle point

C(3,0) is a saddle point

D(3,2) is a local minimum point.

Step-by-step explanation:

The given function is

$f(x,y)=x^3+y^3-3x^2-3y^2-9x$

The first partial derivative with respect to x is

$f_x=3x^2-6x-9$

The first partial derivative with respect to y is

$f_y=3y^2-6y$

We now set each equation to zero to obtain the system of equations;

$3x^2-6x-9=0$

$3y^2-6y=0$

Solving the two equations simultaneously, gives;

$x=-1,x=3$ and $y=0,y=2$

The critical points are

A(-1,0), B(-1,2),C(3,0),and D(3,2).

Now, we need to calculate the discriminant,

$D=f_{xx}(x,y)f_{yy}(x,y)-(f_{xy}(x,y))^2$

But, we would have to calculate the second partial derivatives first.

$f_{xx}=6x-6$

$f_{yy}=6y-6$

$f_{xy}=0$

$\Rightarrow D=(6x-6)(6y-6)-0^2$

$\Rightarrow D=(6x-6)(6y-6)$

At A(-1,0),

$D=(6(-1)-6)(6(0)-6)=72\:>\:0$ and $f_{xx}=6(-1)-6=-18\:$

Hence A(-1,0) is a local maximum point.

See graph

At B(-1,2);

$D=(6(-1)-6)(6(2)-6)=-72\:$

Hence, B(-1,0) is neither a local maximum or a local minimum point.

This is a saddle point.

At C(3,0)

$D=(6(3)-6)(6(0)-6)=-72\:$

Hence, C(3,0) is neither a local minimum or maximum point. It is a saddle point.

At D(3,2),

$D=(6(3)-6)(6(2)-6)=72\:>\:0$ and $f_{xx}=6(3)-6=12\:>\:0$

Hence D(3,2) is a local minimum point.

See graph in attachment.

3 0

3 years ago