Answer:
a) s = -16
+ 192t + 96
b) Height of the football after 2 seconds is 448 feet.
ii. The football would not hit the ground after 2 seconds.
Step-by-step explanation:
Given:
s = -16
+
t + 
where: s is the height of the football above the ground, t is the time,
is the initial velocity and
is the initial height.
The height of the building =
= 192 feet and the initial speed of the ball =
= 96 feet per second, then;
s = -16
+ 192t + 96
a) The polynomial is given as:
s = -16
+ 192t + 96
b) The height of the football when t = 2 seconds is;
s = -16(2) + 192(2) + 96
= -32 + 384 + 96
= -32 + 480
s = 448
The height of the football when t = 2 seconds is 448 feet.
ii. The football would not hit the ground after 2 seconds.
Option D:
ΔCAN ≅ ΔWNA by SAS congruence rule.
Solution:
Given data:
m∠CNA = m∠WAN and CN = WA
To prove that ΔCAN ≅ ΔWNA:
In ΔCAN and ΔWNA,
CN = WA (given side)
∠CNA = ∠WAN (given angle)
NA = NA (reflexive side)
Therefore, ΔCAN ≅ ΔWNA by SAS congruence rule.
Hence option D is the correct answer.
Answer:
So each interior angle measurement of the regular triangle is 60 degrees.
Step-by-step explanation:
If the triangle is regular, this means all the side measurements are congruent to each other and that all the angle measurements are congruent to each other.
If the sum of the angles in a triangle is 180 degrees and you know they are each the same then you could either solve x+x+x=180 or know we are just dealings with an equiangular triangle in which all angles have measurement 60 degrees.
If need more convincing, let's actually solve:
x+x+x=180
3x=180
x=180/3
x=60
So each interior angle measurement of the regular triangle is 60 degrees.
Answer:
3. 6
Step-by-step explanation:
X - y = 4 ( Equation 1 )
X + y =8 ( Equation 2 )
equation 1 + equation 2
; 2x = 12
x = 12÷2
x =6
hope it helps ☺️
Answer:
1/3(5.2)h cm³
Step-by-step explanation:
A solid right pyramid has a regular hexagonal base with an area of 5.2 cm2 and a height of h cm. Which expression represents the volume of the pyramid?
One-fifth(5.2)h cm3 StartFraction 1 Over 5 h EndFraction(5.2)h cm3
One-third(5.2)h cm3 StartFraction 1 Over 3 h EndFraction(5.2)h cm3
Volume of the pyramid = 1/3 × area × height
Area = 5.2 cm²
Height = h cm
Volume of the pyramid = 1/3 × 5.2 cm² × h cm
= 1/3(5.2)h cm³