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The length of a rectangle is 3 units greater than its width. Which expression would correctly represent the perimeter of the rectangle?

A. 2W+2[W+3]

B. W+W+3

C. W[W+3]

D. 2W+2[3W]

Answer Explanation:

Let width of the rectangle be W. We are told that length, L is 3 units greater than width. Then,

L=W+3

And perimeter of the rectangle is given by the relation, P=2(L+W]=2L+2W

In place of L substitute W+3. Thus,

P=2[W+3]+2W

Therefore, the Correct Answer is A.

More Questions on TEAS 7 Math

  • Q #1: Which of the following is the correct simplification of the expression below? 12 ÷ 3 * 4 - 1 + 23

    A. 6

    B. 21

    C. 38

    D. 23

    Answer Explanation

    Use the order of operations to solve the given equation. The acronym PEDMAS will help us to solve the given problem.

    P- Parenthesis

    E-Exponents

    D-Divide

    M-Multiply

    A-Add

    S-Subtract

    From the given equation start with solving :-

    12 ÷3 * 4 - 1 + 23

    Next, carry out division 12÷3=4

    4*4-1+23

    Multiply is 4*4=16

    16-1+23

    Add or subtract can be conducted in any order. From left to right, we subtract followed by addition

    Subtract 16-1=15

    15+23

    The last step is to add 15+23

    15+23=38

    Thus, the answer to the given equation is 38.

  • Q #2: A sport utility vehicle travels 386.4 miles on 14 gallons of gas. Which of the following represents the constant proportionality of miles per gallon?

    A. 372.4

    B. 13.8

    C. 27.6

    D. 5,409.6

    Answer Explanation

    The constant of proportionality is the distance traveled divided by gallons of gas consumed

    Constant of proportionality = 386.4 miles/14 gallons = 27.6 miles per gallon.

  • Q #3: Which of the following is the total number of whole boxes that measure 2 ft * 2 ft * 2 ft that can be stored in a room that measures 9 ft * 9 ft * 9 ft, if the size of the boxes cannot not be altered?

    A. 125

    B. 64

    C. 18

    D. 92

    Answer Explanation

    Number of boxes in the room is equal to the volume of the room divided by the volume of one box

    Volume of the room=9 ft*9 ft*9 ft= 729 ft3

    Volume of one box = 2 ft* 2 ft * 2 ft= 8 ft3

    Number of boxes= 729 ft3/ 8 ft3= 91.125 boxes

    Thus, the room can store 91.125 boxes, which is about 92 boxes