I am going to guess you want to find a line which passes through the points (0,4) and (2,10)...
y₂-y₁ / x₂-x₁
10 - 4 / 2 - 0
6 / 2
3....slope
(0,4) .....b
y = 3x + 4
Answer:
31.9secs
6,183.3m
Step-by-step explanation:
Given the equation that models the height expressed as;
h(t ) = -4.9t²+313t+269
At the the max g=height, the velocity is zero
dh/dt = 0
dh/dt = -9,8t+313
0 = -9.8t + 313
9.8t = 313
t = 313/9.8
t = 31.94secs
Hence it takes the rocket 31.9secs to reach the max height
Get the max height
Recall that h(t ) = -4.9t²+313t+269
h(31.9) = -4.9(31.9)²+313(31.9)+269
h(31.9) = -4,070.44+9,984.7+269
h(31.9) = 6,183.3m
Hence the maximum height reached is 6,183.3m
Given:
The given expression is:
To find:
The value of the given expression at 
.
Solution:
We have,
Substituting , we get
![\dfrac{3\sin (-45)-4\sin [4(-45)]}{\sin [5(-45)]}](https://tex.z-dn.net/?f=%5Cdfrac%7B3%5Csin%20%28-45%29-4%5Csin%20%5B4%28-45%29%5D%7D%7B%5Csin%20%5B5%28-45%29%5D%7D)
On substituting 
, we get,
Therefore, the value of the given expression at is 
.
Answer:
8x⁶y⁵ - 12x⁴y⁷
Step-by-step explanation:
→ 4xy² × 2x² - 3y² × x³y³
→ 4x⁴y⁵ × 2x² - 3y²
→ 8x⁶y⁵ - 12x⁴y⁷
HOPE THIS HELPS YOU
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