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nlexa [21]
3 years ago
9

Follow these instructions to make your own triangle?

Mathematics
1 answer:
ExtremeBDS [4]3 years ago
6 0
Make a diagonal line like this / connected to it make the same / but mirrored or backwards underneath make a connecting like this _
You might be interested in
Please Help With This
pav-90 [236]

Answer:

I think; 1400 m

I hope I helped you^_^

4 0
3 years ago
The position of an object at time t is given by s(t) = 3 - 4t. Find the instantaneous velocity at t = 8 by finding the derivativ
zysi [14]
For this case, we have that the equation of the position is given by:
 s (t) = 3 - 4t

 To find the velocity, we must derive the equation from the position.
 We have then:
 s' (t) = - 4

 Then, we evaluate the derivative for time t = 8.
 We have then:
 s' (8) = - 4
 Answer:
 
the instantaneous velocity at t = 8 is:
 
s' (8) = - 4
7 0
3 years ago
Read 2 more answers
Let f(x)=3(12)x+1. Which is equal to f(1)?
aalyn [17]

Answer:

just substitute the value of x as 1

f(x)=3(12)x+1

f(1)=3(12)(1)+1

f(1)=3(12)(1)+1

f(1)=36+1

f(1)=37

5 0
3 years ago
The x-coordinate of the vertex of the equation y = 2x2 − 4x + 12 The x-coordinate of the vertex of the equation y = 4x2 + 8x + 3
dolphi86 [110]

So since the vertex falls onto the axis of symmetry, we can just solve for that to get the x-coordinate of both equations. The equation for the axis of symmetry is x=\frac{-b}{2a}, with b = x coefficient and a = x^2 coefficient. Our equations can be solved as such:

y = 2x^2 − 4x + 12: x=\frac{4}{2*2}=\frac{4}{4}=1

y = 4x^2 + 8x + 3: x=\frac{-8}{2*4}=\frac{-8}{8}=-1

In short, the vertex x-coordinate's of y = 2x^2 − 4x + 12 is 1 while the vertex's x-coordinate of y = 4x^2 + 8x + 3 is -1.

6 0
3 years ago
explain how it is possible that all proportional relationships are linear functions but not all linear functions are proportiona
OLEGan [10]

Proportional and linear functions are almost identical in form. The only difference is the addition of the “b” constant to the linear function. Indeed, a proportional relationship is just a linear relationship where b = 0, or to put it another way, where the line passes through the origin (0,0).

You’re welcome

6 0
3 years ago
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