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Leno4ka [110]
2 years ago
6

Construct Arguments For rectangles that have a constant perimeter, the length

Mathematics
2 answers:
Karo-lina-s [1.5K]2 years ago
7 0

Answer:

lol

Step-by-step explanation:

Anika [276]2 years ago
3 0
Add the length and the width times height
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Solve the following System of Three Equations:<br> x−3y+z=−15<br> 2x+y−z=−2<br> x+y+2z=1
SashulF [63]

Answer:

x = -3 , y = 4 , z = 0

Step-by-step explanation:

Solve the following system:

{x - 3 y + z = -15

2 x + y - z = -2

x + y + 2 z = 1

Hint: | Choose an equation and a variable to solve for.

In the first equation, look to solve for z:

{x - 3 y + z = -15

2 x + y - z = -2

x + y + 2 z = 1

Hint: | Solve for z.

Subtract x - 3 y from both sides:

{z = 3 y + (-x - 15)

2 x + y - z = -2

x + y + 2 z = 1

Hint: | Perform a substitution.

Substitute z = -15 - x + 3 y into the second and third equations:

{z = -15 - x + 3 y

15 + 3 x - 2 y = -2

x + y + 2 (-15 - x + 3 y) = 1

Hint: | Expand the left hand side of the equation x + y + 2 (-15 - x + 3 y) = 1.

x + y + 2 (-15 - x + 3 y) = x + y + (-30 - 2 x + 6 y) = -30 - x + 7 y:

{z = -15 - x + 3 y

15 + 3 x - 2 y = -2

-30 - x + 7 y = 1

Hint: | Choose an equation and a variable to solve for.

In the second equation, look to solve for x:

{z = -15 - x + 3 y

15 + 3 x - 2 y = -2

-30 - x + 7 y = 1

Hint: | Isolate terms with x to the left hand side.

Subtract 15 - 2 y from both sides:

{z = -15 - x + 3 y

3 x = 2 y - 17

-30 - x + 7 y = 1

Hint: | Solve for x.

Divide both sides by 3:

{z = -15 - x + 3 y

x = (2 y)/3 - 17/3

-30 - x + 7 y = 1

Hint: | Perform a substitution.

Substitute x = (2 y)/3 - 17/3 into the third equation:

{z = -15 - x + 3 y

x = (2 y)/3 - 17/3

(19 y)/3 - 73/3 = 1

Hint: | Choose an equation and a variable to solve for.

In the third equation, look to solve for y:

{z = -15 - x + 3 y

x = (2 y)/3 - 17/3

(19 y)/3 - 73/3 = 1

Hint: | Isolate terms with y to the left hand side.

Add 73/3 to both sides:

{z = -15 - x + 3 y

x = (2 y)/3 - 17/3

(19 y)/3 = 76/3

Hint: | Solve for y.

Multiply both sides by 3/19:

{z = -15 - x + 3 y

x = (2 y)/3 - 17/3

y = 4

Hint: | Perform a back substitution.

Substitute y = 4 into the first and second equations:

{z = -x - 3

x = -3

y = 4

Hint: | Perform a back substitution.

Substitute x = -3 into the first equation:

{z = 0

x = -3

y = 4

Hint: | Sort results.

Collect results in alphabetical order:

Answer:  {x = -3 , y = 4 , z = 0

4 0
3 years ago
YALL HELP!!!!!!!!!!!!!!
QveST [7]

Answer:

The second choice

Step-by-step explanation:

In most cases, the dependent variable is the y value in the graph.

The dependent variable is the thing that is being measured or changing.

In this example it can't be the time because time is always moving at a constant rate and there is nothing that can be done to change it.

8 0
3 years ago
What are the restrictions for a? 2a^2+a-15/5a^2+16a+3
koban [17]

ANSWER


The restrictions are

a\ne -3,a\ne -\frac{1}{5}


EXPLANATION


We were given the rational function,


\frac{2a^2+a-15}{5a^2+16a+3}


The function is defined for all values of a, except



5a^2+16a+3=0


This has become a quadratic trinomial, so we need to split the middle term.


We do that by multiplying the coefficient of x^2 which is 5 by the constant term which is 3. This gives us 15.


The factors of 15 that adds up to 16 are 1 and 15.


We use these factors to split the middle term.




5a^2+15a+a+3=0


We now factor to get,


5a(a+3)+1(a+3)=0


We factor further to get,


(a+3)(5a+1)=0



This implies that,


(a+3)=0,(5a+1)=0


This gives


a=-3,a=-\frac{1}{5}


These are the restrictions.





8 0
3 years ago
5 wars of corn for $1.55. What was the cost per ear?
Delvig [45]

Answer:

$0.31

Step-by-step explanation:

1.55/5=0.31

So if each ear of corn cost $0.31 than 5 times .31 should equal $1.55

5x.31=1.55

Yes!

7 0
3 years ago
I need this to be answered by 11:59 pm pls help
Gnesinka [82]

Answer:

Step-by-step explanation:

1) 8

2) ?

3) g=7

4) v=6

5)x=-4

6) p=8

now #2 i havent learnd yet but all the others. EASYYYYY i love math just choose me if you are having troubles in algebra at all i got you bro <3

5 0
3 years ago
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