Answer:
Perimeter = 28 yards
Area = 49 square yards.
Step-by-step explanation:
Distance formula:
Let as consider the vertices of the floor are A(-2,-3), B(-2,4), C(5,4) and D(5,-3).
Using distance formula, we get
Similarly,
It is conclude that,
.
From the given points it is clear that all sides lie on either vertical or horizontal lines.
Since all sides are equal, and adjacent sides are perpendicular, therefore, the base is a square with edge 7 yards.
Perimeter of square floor is
Area of square floor is
Therefore, the perimeter is 28 yards and the area is 49 square yards.
The table to complete the proof is as follows
Equation statement
1. m∠ABD = 60°, m∠DBC=40° Given
2. m∠ABD + m∠DBC = m∠ABC Angle Addition Postulate
3. 60° + 40° = m∠ABC Substitution Property of Equality
4. 100° = m∠ABC Simplifying
5. ∠ABC is an obtuse angle. greater than 90 degrees
6. △ABC is an obtuse triangle. Definition of obtuse triangle
<h3>What is obtuse angles?</h3>
When an angel is greater than 90 degrees the angle is said to be an obtuse angle.
For the question solved here m∠ABC is greater than 90 degrees hence an obtuse angle.
Read more on angles here: brainly.com/question/25716982
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Answer:
1 and 1/3
Step-by-step explanation:
3a² -4a +1 = 0
3a²- 3a - a+ 1= 0
3a(a-1) - (a-1)= 0
(a-1)(3a -1)= 0
a-1= 0 ⇒ a= 1
3a - 1= 0 ⇒ 3a= 1 ⇒ a= 1/3
The answer is 23 1/3. First, you need to add 280 to both sides of the equations in order cancel out 280 and get 12x by itself. But remember whatever you do to one side of an equation, you must to do the other. So now you have 12x=280. Now divide both sides of the equation by 12 in order to get x completely isolated. 280 divided by twelve is 23 1/3.

Let's find out the gradient (Slope " m ") of line q ;



Now, since we already know the gradient let's find of the equation of line by using its Slope and one of the points using point slope form of line :


Now, plug in the value of gradient ~

here we can clearly observe that, the Area under the curve can easily be represented as :

Since, all the values of y that lies in the shaded region is smaller than the actual value of y for the corresponding values of x in the equation of line q