Answer: For a given quadratic y = ax2 + bx + c, the vertex (h, k)
Step-by-step explanation:
Answer:
24.56 (parallelogram)
Step-by-step explanation:
4 sides of ABCD: AB, BC, AD, DC
perimeter = AB + BC + AD + DC
distance of two points: √(x-x')²+(y-y')²
AB = √3²+4² = √25 = 5
BC = √7²+2² = √53 = 7.28
AD = √7²+2² = √53 = 7.28
DC = √3²+4² = √25 = 5
P = 5*2 + 7.28*2 = 24.56
Answer: 5 Hope this helped-
The question involves the concept & equations associated with projectile motion.
Given:
y₁ = 1130 ft
v₁ = +46 ft/s (note positive sign indicates upwards direction)
t = 6.0 s
g = acceleration due to gravity (assumed constant for simplicity) = -32.2 ft/s²
Of the possible equations of motion, the one we'll find useful is:
y₂ = y₁ + v₁t + 1/2gt²
We can just plug and chug to define the equation of motion:
<u><em>y = (1130 ft) + (46 ft/s)t + 1/2(-32.2 ft/s²)t²</em></u>
<em>(note: if you were to calculate y using t = 6.0 s, you'd find that y = 826.4 ft, instead of 830 ft exactly because of some rounding of g and/or the initial velocity)</em>
Answer:
5–3 = 2
8–5 = 3
12–8 = 4
17–12 = 5
x-17 should be 6, hence x = 6+17 = 23.
Step-by-step explanation:
