In parallelogram defg, dh = x + 1, hf = 3y, g h = 3 x − 4 , and he = 5y + 1. find the values of x and y. the diagram is not drawn to scale.

see explanation

Step-by-step explanation:

Using the property of parallelograms

• The diagonals bisect each other, hence

DH = HF and GH = HE

x + 1 = 3y and 3x - 4 = 5y + 1 ⇒ 3x = 5y + 5

Solving the 2 equations simultaneously

x + 1 = 3y → (1)

3x = 5y + 5 → (2)

rearrange (1) expressing x in terms of y

x = 3y - 1 → (3)

substitute x = 3y - 1 in (2)

3(3y - 1) = 5y + 5

9y - 3 = 5y + 5 ( subtract 5y from both sides )

4y - 3 = 5 ( add 3 to both sides )

4y = 8 ( divide both sides by 4 )

y = 2

substitute y = 2 into (3)

x = (3 × 2) - 1 = 6 - 1 = 5

Hence x = 5, y = 2

1 number

if we use a graphing calculator, there is only one such number that satisfies the conclusion of the mean value theorem, that is, a number such that

where the right-hand side is the average rate of change of f on the interval [0.5,0.75], which is the slope of the secant line.

in the attached graph, the black line is the secant line which has its slope as the average rate of change of f on [0.5,0.75], while the orange line is the only tangent line to the graph of f in the interval that satisfies the mvt conclusion.

(5x-1)^2= (5x)^2-2(5x)×1+(1)^2=25x^2-10x+1