I need help for both of these !
1 answer:
4. 
<-8
First, you can multiply both sides of the inequality by 2 to cancel out the fraction to get:
w+3<-16
Then subtract 3 from both sides to get:
w<-19
Above is your answer
5. h≤
Like we did in the first one, multiply by 5 on both sides to get:
5h≤6h+3
Then subtract 6h from both sides to get
-1h≤3
Divide by -1 into both sides, and when you divide or multiply an inequality by a negative number, you flip the sign, so you would get
h≥-3
Above is your final answer
Hope this helped!
First, you can multiply both sides of the inequality by 2 to cancel out the fraction to get:
w+3<-16
Then subtract 3 from both sides to get:
w<-19
Above is your answer
5. h≤
Like we did in the first one, multiply by 5 on both sides to get:
5h≤6h+3
Then subtract 6h from both sides to get
-1h≤3
Divide by -1 into both sides, and when you divide or multiply an inequality by a negative number, you flip the sign, so you would get
h≥-3
Above is your final answer
Hope this helped!
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<h3>Answer: 5</h3>
===================================================
Explanation:
Everywhere you see an x, replace it with -5. Then use the order of operations PEMDAS to simplify
Keep in mind that for the x^2 term, you are squaring all of x = -5, which includes the negative as well. So that means
x^2 = (-5)^2 = (-5)*(-5) = 25
but sticking a negative out front makes
-x^2 = -(-5)^2 = -25
Answer:
(a) t = ±2
(b) t ∈ {0, 1}
(c) In navigation terms: east by north. The slope is about 0.42 at that point.
Step-by-step explanation:
(a) dy/dx = 0 when dy/dt = 0
dy/dt = 3t^2 -12 = 0 = 3(t -2)(t +2)
The slope is zero at t = ±2.
__
(b) dy/dx = (dy/dt)/(dx/dt) = <em>undefined</em> when dx/dt = 0
dx/dt = 6t^2 -6t = 6(t)(t -1) = 0
The slope is undefined for t ∈ {0, 1}.
__
(c) At t=3, dy/dx = (dy/dt)/(dx/dt) = 3(3-2)(3+2)/(6(3)(3-1)) = 15/36 = 5/12
The general direction of movement is away from the origin along a line with a slope of 5/12, about 22.6° CCW from the +x direction.
_____
The first attachment shows the derivative and its zeros and asymptotes. It also shows some of the detail of the parametric curve near the origin.
The second attachment shows the parametric curve over the domain for which it is defined, along with the point where t=3.
