Evaluate \dfrac32y-3+\dfrac53z 2 3 y−3+ 3 5 zstart fraction, 3, divided by, 2, end fraction, y, minus, 3, plus, start fraction, 5, divided by, 3, end fraction, z when y=6y=6y, equals, 6 and z=3z=3z, equals, 3.

11.

Step-by-step explanation:

We are asked to evaluate the expression , when y=6 and z=3.

To evaluate our expression at given values we will just substitute y=6 and z=3 in our given expression.

According order of operations (PEMDAS) we will multiply or divide our expression from left to right.

Therefore, the value of our given expression at y=6 and z=3 is 11.

answer: bearing in mind that a rectangle has a perimeter of length + length + width + width.

step-by-step explanation:

p = perimeter

l = length

x=width

p = 2l + 2x

p = 2(l+x)     < ==== perimeter of the base of the box.

it would open into a square depending on the type of envelope i’m not sure about the second part