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xeze [42]
3 years ago
10

Which graph has a rate of change of zero? ​

Mathematics
2 answers:
inn [45]3 years ago
6 0

Answer:

  • The second (bottom) graph

Step-by-step explanation:

<u>The first line is:</u>

  • x = 3

It has undefined rate of change

<u>The second line is:</u>

  • y = 3

It has rate of change of zero (no change in y-value for any x-value)

salantis [7]3 years ago
5 0

Answer:

  • The first graph is : <u>x=3</u>

Here, the rate of change is not defined.

  • The second graph : <u>y=3</u>

Here, the rate of change is zero .i.e., y is independent of x.

Therefore, 2nd graph has a rate of change of zero.

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A, B &amp; C form the vertices of a triangle.
kirza4 [7]
<h3>Answer:  20.3</h3>

Work Shown:

tan(angle) = opposite/adjacent

tan(B) = AC/AB

tan(67) = x/8.6

8.6*tan(67) = x

x = 8.6*tan(67)

x = 20.2603303460843  ... make sure your calculator is in degree mode

x = 20.3  .... rounding to 3 significant figures

see diagram below

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3 years ago
We did not find results for: A measure of​ malnutrition, called the​ pelidisi, varies directly as the cube root of a​ person's w
asambeis [7]
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now, if that value is less than 100, then the fellow is "undernourished", otherwise, is overfed.
3 0
4 years ago
Are all functions linear or exponential? Or does it go like, “all linear or exponentials are functions”?
Alex_Xolod [135]
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3 0
3 years ago
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Assoli18 [71]

Answer:

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Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
What is the standard form of the equation of a line for which the length of the normal segment to the origin is 8 and the normal
jeka57 [31]
Slope of line = tan(120) = -tan(60) = - &radic;3
Distance from origin = 8

Let equation be Ax+By+C=0
then -A/B=-&radic;3, or
B=A/&radic;3.
Equation becomes
Ax+(A/&radic;3)y+C=0

Knowing that line is 8 units from origin, apply distance formula
8=abs((Ax+(A/&radic;3)y+C)/sqrt(A^2+(A/&radic;3)^2))
Substitute coordinates of origin (x,y)=(0,0)  =>
8=abs(C/sqrt(A^2+A^2/3))
Let A=1 (or any other arbitrary finite value)
solve for positive solution of C
8=C/&radic;(4/3) => C=8*2/&radic;3 = (16/3)&radic;3

Therefore one solution is
x+(1/&radic;3)+(16/3)&radic;3=0
or equivalently
&radic;3 x + y + 16 = 0

Check:
slope = -1/&radic;3  .....ok
distance from origin
= (&radic;3 * 0 + 0 + 16)/(sqrt(&radic;3)^2+1^2)
=16/2
=8  ok.

Similarly C=-16 will satisfy the given conditions.

Answer  The required equations are
&radic;3 x + y = &pm; 16 
in standard form.

You can conveniently convert to point-slope form if you wish.




4 0
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