Hmmm the object, is at rest, when dropped, so it has a velocity of 0 ft/s
the only force acting on the object, is gravity, using feet will then be -32ft/s²,
was wondering myself on -32 or 32.. but anyhow... we'll settle for the negative value, since it seems to be just a bit of convention issues
so, we'll do the integral to get v(t) then
when will it reach the ground level? let's set s(t) = 0
part B) check the picture below
Answer:
d = 14/5
Step-by-step explanation:
The point (-4,2) means that;
At x = -4, y = 2
Now general form of a linear equation is;
Ax + By + C = 0
We are given;
4y = 3x + 6
Rearranging to the form of a linear equation gives;
3x - 4y + 6 = 0
Thus, A = 3, B = -4 and C = 6
Thus, at point (-4,2), distance between them is;
d = (3(-4) - 4(2) + 6)/√(3² + (-4)²)
d = -14/5
We will take the absolute value.
Thus; d = 14/5
Answer:
y
Step-by-step explanation:
Answer:
Step-by-step explanation:
The attached photo shows the diagram of quadrilateral QRST with more illustrations.
Line RT divides the quadrilateral into 2 congruent triangles QRT and SRT. The sum of the angles in each triangle is 180 degrees(98 + 50 + 32)
The area of the quadrilateral = 2 × area of triangle QRT = 2 × area of triangle SRT
Using sine rule,
q/SinQ = t/SinT = r/SinR
24/sin98 = QT/sin50
QT = r = sin50 × 24.24 = 18.57
Also
24/sin98 = QR/sin32
QR = t = sin32 × 24.24 = 12.84
Let us find area of triangle QRT
Area of a triangle
= 1/2 abSinC = 1/2 rtSinQ
Area of triangle QRT
= 1/2 × 18.57 × 12.84Sin98
= 118.06
Therefore, area of quadrilateral QRST = 2 × 118.06 = 236.12
Answer:
2
Step-by-step explanation:
Given the data : 29, 2, 28, 30, 26, 31
Outlier ;
Lower :Q1 - (1.5 * IQR)
Upper : Q3 + (1.5 * IQR)
Q1 = Lower quartile ; Q3 = upper quartile ; IQR = Interquartile range
Using calculator :
Q1 = 26
Q3 = 30
IQR = (Q3 - Q1) = 30 - 26 = 4
Lower : 26 - (1.5 * 4) = 20
Upper : 30 + (1.5 * 4) = 36
Hence, the number in the given data which falls outside the range is 2
