***I know I've posted this several times but it's like getting skipped and or no one has gone over it so I'm posting it one more time before I go to bed. I need to have these checked by tommorrow PLEASE!***
Out of 31 questions these are ones I struggled with. after the = sign is what I think the answer( except for #4, I don't know what to do at all), is so if I'm wrong can someone correct it with an explanation/formula with it please?
1.) Find the area of the parallelogram. ( base(s)=10cm, side length(s)=6cm, height=? and the angle of elevation=60'
a.)15√3 cm^2
b.)30√3 cm^2
c.)15√2 cm^2
=30√3 cm^2
2.) What is the length of the radius of a circle with the center at O(-3,4) and passes through the point P(2,-5)?
a.)9
b.)12.5
c.)10.3
=9
3.) What is the equation for a circle with a diameter having the following endpoints: N(0,1) and Q(18,1)?
a.)(x - 9)^2 + (y - 1)^2 = 81
b.)(x)^2 + (y)^2 = 72
c.)(x + 2)^2 + (y - 7)^2 = 56
=(x)^2 + (y)^2 = 72
4.) Using coordinate geometry to prove that the diagonals of a square are perpendicular to each other.
Given: Vertices are at A(0,0), B(a,0), C(a,a) and D(0,a) Slope of AC=1; Slope of BD=-1
Type your proof into the text box below.
=
1. correct
2. R=√((-3-2)²+(4-(-5))²)
=10.3
3. If the end-points are (0,1) and (18,1), it means that the circle's diameter passes through theses two points.
Therefore the diameter is(18-0)=18, radius is 9.
The centre is ((0+18)/2, (1+1)/2)=(9,1).
The standard equation of a circle with radius r and passing through (x1,y1) is
(x-x1)²+(y-y1)²=r&su2p;
Therefore the answer is
(x-9)²+(y-1)²=9²
4. The product of the slopes of two lines intersecting at right angles is -1. You can take it from here.
Thanks:)
Glad to be of help!
Find the area of a parallelogram with sides of 6 and 12 and an angle of 60°.
1.) To find the area of a parallelogram, you can use the formula A = base * height. In this case, the base is given as 10cm, but the height is missing. However, the angle of elevation is given as 60 degrees.
You can use the equation height = side length * sin(angle of elevation) to find the height. Plugging in the values, we get height = 6cm * sin(60 degrees) = 6cm * (√3/2) = 3√3 cm.
Now, we can calculate the area using the formula A = base * height: A = 10cm * 3√3 cm = 30√3 cm^2. Therefore, the correct answer is b) 30√3 cm^2.
2.) To find the length of the radius of a circle, you can use the distance formula between two points. The distance formula is given by d = √[(x2 - x1)^2 + (y2 - y1)^2].
In this case, the center of the circle is at O(-3,4) and it passes through the point P(2,-5).
Using the distance formula, the radius is calculated as: r = √[(-3 - 2)^2 + (4 - (-5))^2] = √[(5)^2 + (9)^2] = √[25 + 81] = √106.
Therefore, the correct answer is a) 9.
3.) The equation for a circle with a diameter can be written in the form (x - h)^2 + (y - k)^2 = r^2, where (h,k) represents the center of the circle and r represents the radius.
In this case, the diameter has endpoints N(0,1) and Q(18,1). The x-coordinates of both points are the same, indicating that the center of the circle lies on the x-axis. Therefore, the y-coordinate of the center is the midpoint of the y-coordinates of the endpoints, which is (1 + 1)/2 = 1.
The radius of the circle can be found using the distance formula between the center and one of the endpoints: r = √[(0 - 9)^2 + (1 - 1)^2] = √[81 + 0] = 9.
Now, we can write the equation for the circle as: (x - 9)^2 + (y - 1)^2 = 9^2 = 81.
Therefore, the correct answer is a) (x - 9)^2 + (y - 1)^2 = 81.
4.) To prove that the diagonals of a square are perpendicular to each other using coordinate geometry, we need to show that the slopes of the diagonals are negative reciprocals of each other.
Given that the vertices of the square are at A(0,0), B(a,0), C(a,a), and D(0,a), we can calculate the slopes of the two diagonals.
The slope of AC can be found using the formula: slope = (y2 - y1)/(x2 - x1). In this case, (x1, y1) = (0,0) and (x2, y2) = (a,a). So, the slope of AC is (a - 0)/(a - 0) = 1.
The slope of BD can be found using the formula: slope = (y2 - y1)/(x2 - x1). In this case, (x1, y1) = (a,0) and (x2, y2) = (0,a). So, the slope of BD is (a - 0)/(0 - a) = -1.
Since the slope of AC is 1 and the slope of BD is -1, we can see that they are negative reciprocals of each other.
Therefore, the diagonals of a square are perpendicular to each other.