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jeyben [28]
3 years ago
14

Plz help!!I can't find the answer, the one I put in is wrong.

Mathematics
2 answers:
hichkok12 [17]3 years ago
8 0

Answer:

3:23 to 3:24

Step-by-step explanation:

explantion between these to points the slope of the graph hits 3:24 hard

Inessa [10]3 years ago
4 0

Answer:

see explanation

Step-by-step explanation:

The function is increasing when the graph goes upwards from left to right.

The function is decreasing when the graph goes downwards from left to right.

Here the graph is decreasing from 3.23  to 3.25 and is decreasing the fastest from 3.23 to 3.24




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The height h of a projectile is a function of the time t it is in the air. the height in feet for t seconds is given by the func
alexgriva [62]

Domain means the values of independent variable(input) which will give defined output to the function.

Given:

The height h of a projectile is a function of the time t it is in the air. The height in feet for t seconds is given by the function

h(t)=-16t^2 + 96t

Solution:

To get defined output, the height h(t) need to be greater than or equal to zero. We need to set up an inequality and solve it to find the domain values.

To \; find \; domain:\\\\h(t) \geq0\\\\-16t^2+96t \geq  0\\Factoring \; -16t \; in \; the \; left \; side \; of \; the \; inequality\\\\-16t(t-6) \geq  0\\Step \; 1: Find \; Boundary \; Points \; by \; setting \; up \; above \; inequality \; to \; zero.\\\\t(t-6)=0\\Use \; zero \; factor \; property \; to \; solve\\\\t=0 \; (or) \; t = 6\\\\Step \; 2: \; List \; the \; possible  \; solution \; interval \; using \; boundary \; points\\(- \infty,0], \; [0, 6], \& [6, \infty)

Step \; 3:Pick \; test \; point \; from \; each \; interval \; to \; check \; whether \\\; makes \; the \; inequality \; TRUE \; or \; FALSE\\\\When \; t = -1\\-16(-1)(-1-6) \geq  0\\-112 \geq  0 \; FALSE\\(-\infty, 0] \; is \; not \; solution\\Also \; Logically \; time \; t \; cannot \; be \; negative\\\\When \; t = 1\\-16(1)(1-6) \geq  0\\80 \geq  0 \; TRUE\\ \; [0, 6] \; is \; a \; solution\\\\When \; t = 7\\-16(7)(7-6) \geq  0\\-112 \geq  0 \; FALSE\\ \; [6, -\infty) \; is \; not \; solution

Conclusion:

The domain of the function is the time in between 0 to 6 seconds

0 \leq  t \leq  6

The height will be positive in the above interval.

7 0
3 years ago
Read 2 more answers
Hello,
balu736 [363]

Answer:

yesss it makes sense!!!!!!!!!!!

6 0
2 years ago
Which expression is equivalent to -35 3/5
Alex787 [66]
You can look at -32 as being -2^5 so if you write (-2^5)^3/5 the 5's cancel out making it -2^3 which equals -8
5 0
3 years ago
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NO LINKS!! Please help me with this problem​
Anastaziya [24]

Answer:

  y²/324 -x²/36 = 1

Step-by-step explanation:

Where (0, ±b) are the ends of the transverse axis and y = ±(b/a)x describes the asymptotes, the equation of the hyperbola can be written as ...

  y²/b² -x²/a² = 1

<h3>Application</h3>

Here, we have transverse axis endpoints of (0, ±18) and asymptotes of y = ±3x, so we can conclude ...

  b = 18

  b/a = 3   ⇒   a = 18/3 = 6

The equation of the hyperbola in standard form is ...

  y²/324 -x²/36 = 1

7 0
2 years ago
Complete the steps of the derivation of the quadratic formula step 1: (x+b/2a)^2-b^2-4ac/4a^2=0
CaHeK987 [17]
(x+b/2a)^2-(b^2-4ac)/2a=0
Step 2:
Re-write the expression:
(x+b/2a)^2=(b^2-4ac)/4a^2

Step 3:
get the square root of both sides:
x+b/2a=sqrt[(b^2-4ac)/4a^2]

Step 4:
Simplifying we get:
x+b/2a=sqrt[b^2-4ac]/2a

Step 5
Make x the subject:
x=-b/2a+/-sqrt[b^2-4ac]/2a


5 0
3 years ago
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