1) The length of one of the equal legs of an isisceles triangle is 8cm less than 4 time the length of the base. If the perimeter is 29 cm, find the length of one of the equal legs.

a) 4 cm
B) 5 cm
C) 11 cm
d) 12 cm

2) The perimeter of a rectangle is to be no greater than 300 in., and the length must be 125 in. Find the maximum width of the rectangle.
A) 20 in
B) 25 in
C) 50 in
D) 175 in

1. Solve these two eq
x + 2y = 29
y = 4x - 8

2.

2. 2x + 2y <or= 300
x = 125
250 + 2y <or= 300
2y <or= 50
y <or= ?

would the answer be 25?

Yeah

To solve the first question, we have two equations:

x + 2y = 29 (equation 1)
y = 4x - 8 (equation 2)

To find the length of one of the equal legs, we need to solve these equations simultaneously.

1. Rearrange equation 1 to solve for x:
x = 29 - 2y

2. Substitute the value of x from equation 1 into equation 2:
y = 4(29 - 2y) - 8

3. Distribute and simplify:
y = 116 - 8y - 8

4. Combine like terms:
9y = 108

5. Divide both sides by 9:
y = 12

6. Now substitute the value of y into equation 1 to find x:
x + 2(12) = 29
x + 24 = 29
x = 5

Therefore, the length of one of the equal legs is 5cm. So the correct answer is option B) 5cm.

Now let's solve the second question:

We are given:
Perimeter of the rectangle (P) ≤ 300
Length (L) = 125

We need to find the maximum width (W) of the rectangle.

1. The formula for the perimeter of a rectangle is:
P = 2L + 2W

2. Substitute the given values into the formula:
300 = 2(125) + 2W

3. Simplify and solve for W:
300 = 250 + 2W
50 = 2W
W = 25

Therefore, the maximum width of the rectangle is 25 inches. So the correct answer is option B) 25 in.