So anyone doing a zoom because I’m bored and I’ll join

The volume of a square prism is 144x^3 +216x^2 +81x

Svetach [21]

Answer:

Equation for the perimeter of prism's square face: 16x + 12

Step-by-step explanation:

Volume of Square prism = Length * Width * Height

= 144 x^3 + 216 x^2 +81 x

taking 9x common = 9x( 16 x^2 + 24 x + 9)

= 9x ( (4x)^2 + 2(4x)(3) + (3)^2 )

= 9x ( 4x+3)^2

so the length is 9x, width is 4x+3 and height is 4x+3

Now, Perimeter of prism's square face = 2* Width + 2 * Height

= 2* (4x+3) + 2* (4x+3)

= 8x +6 + 8x + 6

= 16x +12

6 0

3 years ago

HELP PLEASE !!!

Lana71 [14]

972-4=968 and 62-4=58
The highest common number between 968 and 58 is 2 so 968 in fractions is 484/2 and 58 is 29/2

If you multiply these two together you will get 14036/2

5 0

3 years ago

the net of the rectangular prism is shown below. use a ruler to measure the dimensions of the net to the nearest half centimeter

Arte-miy333 [17]

Answer:

the answer is D. 40.5 cm

8 0

3 years ago

This 1 seems really complicated

Fofino [41]

The solution to this system set is: "x = 4" , "y = 0" ; or write as: [4, 0] .
________________________________________________________
Given:
________________________________________________________
y = - 4x + 16 ;

4y − x + 4 = 0 ;
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"Solve the system using substitution" .
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First, let us simplify the second equation given, to get rid of the "0" ;

→ 4y − x + 4 = 0 ;

Subtract "4" from each side of the equation ;

→ 4y − x + 4 − 4 = 0 − 4 ;

→ 4y − x = -4 ;
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So, we can now rewrite the two (2) equations in the given system:
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y = - 4x + 16 ; ===> Refer to this as "Equation 1" ;

4y − x = -4 ; ===> Refer to this as "Equation 2" ;
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Solve for "x" and "y" ; using "substitution" :
________________________________________________________
We are given, as "Equation 1" ;

→ " y = - 4x + 16 " ;
_______________________________________________________
→ Plug in this value for [all of] the value[s] for "y" into {"Equation 2"} ;

to solve for "x" ; as follows:
_______________________________________________________
Note: "Equation 2" :

→ " 4y − x = - 4 " ;
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Substitute the value for "y" {i.e., the value provided for "y"; in "Equation 1}" ;
for into the this [rewritten version of] "Equation 2" ;
→ and "rewrite the equation" ;

→ as follows:
_________________________________________________

→ " 4 (-4x + 16) − x = -4 " ;
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Note the "distributive property" of multiplication :
_________________________________________________

a(b + c) = ab + ac ; AND:

a(b − c) = ab <span>− ac .
_________________________________________________
As such:

We have:
</span>
→ " 4 (-4x + 16) − x = - 4 " ;
_________________________________________________
AND:

→ "4 (-4x + 16) " = (4* -4x) + (4 *16) = " -16x + 64 " ;
_________________________________________________
Now, we can write the entire equation:

→ " -16x + 64 − x = - 4 " ;

Note: " - 16x − x = -16x − 1x = -17x " ;

→ " -17x + 64 = - 4 " ; Solve for "x" ;

Subtract "64" from EACH SIDE of the equation:

→ " -17x + 64 − 64 = - 4 − 64 " ;

to get:

→ " -17x = -68 " ;

Divide EACH side of the equation by "-17" ;
to isolate "x" on one side of the equation; & to solve for "x" ;

→ -17x / -17 = -68/ -17 ;

to get:

→ x = 4 ;
______________________________________
Now, Plug this value for "x" ; into "{Equation 1"} ;

which is: " y = -4x + 16" ; to solve for "y".
______________________________________

→ y = -4(4) + 16 ;

= -16 + 16 ;

→ y = 0 .
_________________________________________________________
The solution to this system set is: "x = 4" , "y = 0" ; or write as: [4, 0] .
_________________________________________________________
Now, let us check our answers—as directed in this very question itself ;
_________________________________________________________
→ Given the TWO (2) originally given equations in the system of equation; as they were originally rewitten;

→ Let us check;

→ For EACH of these 2 (TWO) equations; do these two equations hold true {i.e. do EACH SIDE of these equations have equal values on each side} ; when we "plug in" our obtained values of "4" (for "x") ; and "0" for "y" ??? ;

→ Consider the first equation given in our problem, as originally written in the system of equations:

→ " y = - 4x + 16 " ;

→ Substitute: "4" for "x" and "0" for "y" ; When done, are both sides equal?

→ "0 = ? -4(4) + 16 " ?? ;   → "0 = ? -16 + 16 ?? " ; → Yes! ;

{Actually, that is how we obtained our value for "y" initially.}.

→ Now, let us check the other equation given—as originally written in this very question:

→ " 4y − x + 4 = ?? 0 ??? " ;

→ Let us "plug in" our obtained values into the equation;

{that is: "4" for the "x-value" ; & "0" for the "y-value" ;

→ to see if the "other side of the equation" {i.e., the "right-hand side"} holds true {i.e., in the case of this very equation—is equal to "0".}.

→ " 4(0)  − 4 + 4 = ? 0 ?? " ;

→ " 0 − 4 + 4 = ? 0 ?? " ;

→ " - 4 + 4 = ? 0 ?? " ; Yes!
_____________________________________________________
→ As such, from "checking [our] answer (obtained values)" , we can be reasonably certain that our answer [obtained values] :
_____________________________________________________
→ "x = 4" and "y = 0" ; or; write as: [0, 4] ; are correct.
_____________________________________________________
Hope this lenghty explanation is of help! Best wishes!
_____________________________________________________

7 0

3 years ago

State the converse, contrapositive, and inverse of each of these conditional statements a) If it snows tonight, then I will stay

marissa [1.9K]

Step-by-step explanation:

Consider the provided information.

For the condition statement $p \rightarrow q$ or equivalent "If p then q"

The rule for Converse is: Interchange the two statements.
The rule for Inverse is: Negative both statements.
The rule for Contrapositive is: Negative both statements and interchange them.

Part (A) If it snows tonight, then I will stay at home.

Here p is If it snows tonight, and q is I will stay at home.

Converse: If I will stay at home then it snows tonight.

$q \rightarrow p$

Inverse: If it doesn't snows tonight, then I will not stay at home.

$\sim p \rightarrow \sim q$

Contrapositive: If I will not stay at home then it doesn't snows tonight.

$\sim q \rightarrow \sim p$

Part (B) I go to the beach whenever it is a sunny summer day.

Here p is I go to the beach, and q is it is a sunny summer day.

Converse: It is a sunny summer day whenever I go to the beach.

$q \rightarrow p$

Inverse: I don't go to the beach whenever it is not a sunny summer day.

$\sim p \rightarrow \sim q$

Contrapositive: It is not a sunny summer day whenever I don't go to the beach.

$\sim q \rightarrow \sim p$

Part (C) When I stay up late, it is necessary that I sleep until noon.

P is I sleep until noon and q is I stay up late.

Converse: If I sleep until noon, then it is necessary that i stay up late.

$q \rightarrow p$

Inverse: When I don't stay up late, it is necessary that I don't sleep until noon.

$\sim p \rightarrow \sim q$

Contrapositive: If I don't sleep until noon, then it is not necessary that i stay up late.

$\sim q \rightarrow \sim p$

7 0

2 years ago