1 = 1 x 10 exponent 0
27.5 = 2.75 x 10 exponent 1
The answer by the sequence:
Width of the garden = w
Length of the garden = 2w+3
Width of the entire plot = w + 0.5w = 1.5w
Length of the entire plot = (2w + 3) 0.5w = 2.5w + 3
Area of the garden = w (2w + 3)
Area of the entire plot = 1.5w (2.5w + 3)
Area of the garden = 2w² + 3w
Area of the entire plot = 3.75w² + 4.5w
Area of border = area of the entire plot - area of the garden
= 3.75w² + 4.5w - (2w² + 3w)
Area of the border = 3.75w² + 4.5w - 2w² - 3w
Area of the border = 1.75w² + 1.5w
Given that PQ and RS are drawn with KL as tranversal intersecting PQ at M and RS at point N. Angle QMN is congruent to angle LNS because they are alternate to each other. The theorem that Kari can use to show that the meansure of QML is supplementary to the measure of angle SNK is Alternate Exterior Angles Theorem.
This is because angle KNR is equal to QML by alternate exterior angles theorem so is angle MLP and SNK
Hi there! The answer is A) x = 3 or x = 6
Divide by 2.
Factor.
Use the rule AB = 0 gives A = 0 or B = 0
The answer is A) x = 3 or x = 6
Answer:
-7(9) < -56
Step-by-step explanation:
-7(9) is -63 which is less than -56 so anything above that is less than 56-
