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3 years ago

Help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics

2 answers:

MA_775_DIABLO [31]3 years ago

7 0

Answer:

x = -10

Step-by-step explanation:

3(-2x - 3) + 4x - 3 = 8

~Simplify

-6x - 9 + 4x - 3 = 8

-2x - 12 = 8

~Add 12 to both sides

-2x = 20

~Divide -2 to both sides

x = -10

Best of Luck!

Tju [1.3M]3 years ago

4 0

Answer:

-10 because you do parenthesis first

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3 years ago

The x intercepts of the function f(x) = 2x(x-5)^2(x+4)^3 are…

Leokris [45]

Answer:

$\boxed{\sf x- intercepts = 0 , 5 \ and \ -4}$

Step-by-step explanation:

A function is given to us and we need to find the x Intercepts of the graph of the given function . The function is ,

$\sf \implies f(x) = 2x( x - 5 ) ^2(x+4)^3$

For finding the x intercept , equate the given function with 0, we have ;

$\sf \implies 2x ( x - 5 )^2(x+4)^3= 0$

Equate each factor with 0 ,

$\sf \implies 2x = 0$

Divide both sides by 2 ,

$\sf \implies\bf x = 0$

Again ,

$\sf \implies ( x - 5)^2=0$

Taking squareroot on both sides,

$\sf \implies x - 5 = 0$

Add 5 to both sides,

$\sf \implies \bf x = 5$

Similarly ,

$\sf \implies \bf x = -4$

Hence the x Intercepts are -4 , 0 and 5 .

{ See attachment also for graph } .

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3 years ago

A piece of cardboard is 13 inches by 26 inches. A square is to be cut from each corner and the sides folded up to make an open-t

Vanyuwa [196]

Answer:

Hence the maximum possible volume will be the 778.53 c.c

Step-by-step explanation:

Given:

A rectangle with 13 x 26 dimensions

And corners are cut to form side squares.

To Find:

Maximum possible volume for box

Solution :

Consider a rectangle of 13 x 26 dimension with and side of square at corner be x.

(Refer the attachment)

Now,

Formulating the volume equation for the box

So corner square sides we are going to fold up which makes height of the box

and remaining part will be length and breadth

As shown in fig,

Length=26-x

breadth=13-x

And height will be x

$V(x)=x*(26-x)*(13-x)$

To get maximum volume differentiate the above equation,

$V(x)=x*(26*13-26*x-13*x+x^2)$

$V(x)=x^3-39x^2+338x\\$

$V'(x)=3x^2-78x+338$

$V''(x)=6x-78$

Now ,Solve the Quadratic Equation to get x values,

$3x^2-78x+338$ =0

x=[-b±(b^2-4ac)^1/2]/2a

x=[78±Sqrt[(78)^2-4*338*3)]/2*3

x=[78±Sqrt(3028)]/6

x=[78±55.027]/6

x=78+55.027/6 or x=78-55.027/6

x=22.17 or x=3.8288

Use these values in 6x-78 to know which value posses the max and min value for the function.

So when x=22.17

6x-78=6*22.17-78

=55.02>0 i.e function will have minimum value .

When x=3.8288

6*3.8288-78

=-55.0272<0 i.e. Function will have maximum value

Now, the function will defines the maximum volume

$V(x)=x^3-39x^2+338x$

$V(x)=3.8288^3-39*(3.82883)^2+338*3.8288$

$V(x)=56.13-571.73+1294.13$

$V(x)=778.53 C.C$

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3 years ago