Answer:
12a^3+30a^2
Step-by-step explanation:
6a^2(2a+5)
12a^3+30a^2
Answer:
Dr hmm fruncfyhn
Step-by-step explanation:
fudjcjdh FC hey DC KFC TV HD hmm vuejcwsm t Ufa hmm c CT DC x GB chv
Answer:
19ft
Step-by-step explanation:
Given the height of a ball above the ground after x seconds given by the quadratic function y = -16x2 + 32x + 3, we can find the maximum height reached by the ball since we are not told what to look for.
The velocity of the ball is zero at maximum height and it is expressed as:
V(x) = dy/dx
V(x) = -32x+32
Since v(x) = 0
0 = -32x+32
32x = 32
x = 32/32
x = 1s
Get the height y
Recall that y = -16x² + 32x + 3.
Substitute x = 1
y = -16(1)²+32(1)+3
y = -16+32+3
y = -16+35
y = 19ft
Hence the maximum height reached by the ball is 19ft
T=time
d=st
r=train speed
p=plane speed
<span>160=rt
</span><span>720=pt
speed of plan is 20kmph less than 5 times speed of train
p=5r-20
we notice that 160 times 4.5=720
times first equaton by 4.5
720=4.5rt
720=pt
set equal
4.5rt=pt
divide both sides by t
4.5r=p
sub </span><span>p=5r-20 for p
4.5r=5r-20
minus 4.5r from both sides
0=0.5r-20
add 20 to both sides
20=0.5r
times 2 both sides
40=r
sub back</span><span>
p=5r-20
p=5(40)-20
p=200-20
p=180
plane is 180kmph
train is 40kmph
</span>
Answer:
The answer is ΔJMK ≈ ΔMLK ≈ ΔJLM ⇒ answer (A)
Step-by-step explanation:
* Lets start with the equal angles i the three triangles
- In ΔJMK
∵ m∠JKM = 90°
∴ m∠KJM + m∠KMJ = 90 ⇒ (1)
- IN ΔMLK
∵ m∠MKL = 90°
∴ m∠KML + m∠KLM = 90° ⇒ (2)
∵ m∠KMJ + m∠KML = 90° ⇒ (3)
- From (1) , (2) , (3)
∴ m∠KJM = m∠KML
∴ m∠KMJ = m∠KLM
* Now lets check the condition of similarity in the 3 triangles
- At first ΔJMK and ΔMLK
- In triangles JMK , MLK
∵ m∠KJM = m∠KML
∵ m∠KMJ = m∠KLM
∵ m∠JKM = m∠MKL
∴ ΔJMK ≈ ΔMLK ⇒ (4)
- At second ΔJMK and ΔJLM
∵ m∠KJM = m∠MJL
∵ m∠KMJ = m∠MLJ
∵ m∠JKM = m∠JML
∴ ΔJMK ≈ ΔJLM ⇒ (5)
* If two triangles are similar to one triangle, then they are
similar to each other
- From (4) and(5)
∴ ΔJMK ≈ ΔMLK ≈ ΔJLM
