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Bond [772]
3 years ago
14

If three points are coplanar, they are also collinear true or false​

Mathematics
2 answers:
Darina [25.2K]3 years ago
5 0
I am sorry i don’t understand can you just type it again ?
kakasveta [241]3 years ago
3 0

Answer: True

Step-by-step explanation: Coplanar points are points that lie in the same plane. Collinear points are points that lie in the same line. It does not mean that coplanar points are immediately collinear.

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ILL GIVE BRAINLIEST TO THE RIGHT ANSWER!!!!!
Paraphin [41]

Answer:

-3<lxl<5

Step-by-step explanation:

take least number than use x in absolute value bars than take highest number sorry if this is wrong I'm just a kid

4 0
3 years ago
Mary can paint a house in 12 hours.it takes them 2hours.without help,how long would it take Nancy to complete the same job?
frez [133]
It Takes Nancy 10 Hours.
6 0
4 years ago
Help pls need help show your solution and then graph​
Stella [2.4K]

Answer:

Step-by-step explanation:

We can use the quadratic formula or factor, This looks hard to factor so we should use the quadratic formula.

ax^2 + bx + c

so

a=4

b=-9

c=2

you can try it by hand or use this:

https://www.calculatorsoup.com/calculators/algebra/quadratic-formula-calculator.php

you get x=2 and x=0.25 these are the x intercepts (where Y is zero and where the graph crosses the x axis)

so mark x=2 and x=0.25 with a dot on a number line and you can draw a straight line between them since that is the part of the graph that is below the x axis (because the equation has <0) it is a positive parabola because the "a" value is positive and the presence of a "b" value means part of the graph will be below the x axis

5 0
2 years ago
What is the range for the following function? y=1/(x+2)+3
LUCKY_DIMON [66]

Answer:

\large \boxed{\text{C) }\{y: y \in \mathbb{R}, y \ne 3\} }

Step-by-step explanation:

The range is the spread of the y-values (minimum to maximum distance travelled).

The graph of your function is a hyperbola shifted two units left and three units up from the origin.

There is a vertical asymptote at x = -2, so y does not exist when x = -2.  However,

\displaystyle \lim_{x \rightarrow -{2}^{+}}f(x) = \lim_{x \rightarrow -{2}^{+}}\left (\dfrac{1}{x+2}+3 \right ) = 0 + 3 = 3\\\\\lim_{x \rightarrow -{2}^{-}}f(x) = \lim_{x \rightarrow -{2}^{-}}\left (\dfrac{1}{x+2}+3 \right ) = 0 + 3 = 3

Since the limit from either side is the same,

\displaystyle \lim_{x \rightarrow -{2}}f(x) = 3

The graph below shows the asymptotes of your function.

Thus. y can take any value  except 3.

In set builder notation, the range is  

\large \boxed{\mathbf{\{y: y \in \mathbb{R}, y \ne 3\} }}

4 0
3 years ago
What is another way to write the expression ​ t⋅(14−5) ​?
grigory [225]

t⋅(14−5)

distribute

14t -5t

Choice A

8 0
3 years ago
Read 2 more answers
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