the length of a rectangle is twice its width. if the perimeter of the rectangle is 54m find its area

Mathematics

1 answer:

mixer [17]3 years ago

7 0

Answer:

You can denote the width of the rectangle as x,

than the length of the rectangle will be 2x

and the perimeter of the rectangle will be two length plus two width and it should be equal to 54 in

and the equation will look 2(2x)+2x=54

solving the equation 6x=54, x=9in

So the width of the rectangle is 9in and the length is 9*2=18in

and the area of the rectangle is 9*18=162 sq in.

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X + 3y = 14 -5x + 6y = -28 Solve systems by substitution

mote1985 [20]

Rearrange the first system to get x = -3y + 14. Substitute that into the other system.

$-5x+6y=-28\\\\-5(-3y+14)+6y=-28\\\\15y-70+6y=-28\\\\15y-70+6y=-28\\\\21y-70=-28\\\\21y=42\\\\y=2$

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4 0

3 years ago

For how many three digit numbers (100 to 999) is the sum of the digits even? (For example, 343 has an even sum of digits: 3 + 4

n200080 [17]

Answer:

there are 425 numbers of three digits whose the sum of the digits is even.

Step-by-step explanation:

Option 1:

It is necessary to accomplish one of the next conditions:

(1) all of the digits are even.

(2) exactly 2 of the digits are odd

Then, for the first condition:

4x5x5 = 100 -> this is because the first digit can not be 0.

for the second condition:

(3C1)*5*5*5-1*5*5 =3*125-25=350

where 3C1 is a combination. The 3C1 determines which of the three digits is even, and it is necessary to subtract the numbers under 100 with the other 2 digits odd.

So, there are 450 numbers of three digits whose the sum of the digits is even.

Option 2:

there are 900 three digit number and 5*5*5 =125 has 3 odd numbers and 5*5*5+4*5*5+4*5*5=325 has 1 odd number, so there are 450 numbers of three digits whose the sum of the digitis is odd, so 900-450 whose the sum of the digits is even.

4 0

3 years ago

Which of these relations on the set {0, 1, 2, 3} are equivalence relations? If not, please give reasons why. (In other words, if

Delicious77 [7]

Answer:

(1)Equivalence Relation

(2)Not Transitive, (0,3) is missing

(3)Equivalence Relation

(4)Not symmetric and Not Transitive, (2,1) is not in the set

Step-by-step explanation:

A set is said to be an equivalence relation if it satisfies the following conditions:

Reflexivity: If $\forall x \in A, x \rightarrow x$
Symmetry: $\forall x,y \in A, $if x \rightarrow y,$ then y \rightarrow x$
Transitivity: $\forall x,y,z \in A, $if x \rightarrow y,$ and y \rightarrow z, $ then x \rightarrow z$