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dangina [55]
2 years ago
13

HELLLPPPPPP PLLEEEEEAAAASSSEEEE!!!!!!!!!!!!

Mathematics
1 answer:
seraphim [82]2 years ago
5 0
It would be Kellen's pumpkin because :
Jermaine = 42.009
Kim = 42.19
Kellen = 42.9
And after the 9 in 42.9 there is a 0, so: 42.90
And by comparing that to the other people's pumpkins you can see that Kellen's was the heaviest
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determine whether the lines intersect and if so find the point of intersection and the cosine of the angle of intersection x=4t+
Fittoniya [83]

Answer:

As the lines are not intersecting nor parallel, they must be skew.

Step-by-step explanation:

Question is incomplete, we consider the nearest match available online.

Parametric equations of two lines are:

L₁ : x=4t+2 , y = 3 , z =-t+1

L₂: x=2s+2 , y= 2s+5 , z = s+1

If lines are parallel then parametric coordinates must be equal scalar multiple of each other which s not true here.

4t+2=2s+2 ---(1)\\ \\3=2s+5---(2) \\\\-t+1=s+1---(3)

If lines are intersecting then parametric coordinates must be equal for some value of t and s.

From (3)\\-t=s\\From (2)\\\\s=\frac{3-5}{2}\\\\s=-1\\\implies t=1\\\\To\,\,check\,\, the \,\,values \,\,of \,\,s\,\, and \,\,t \,\,for\,\,intersection\,\,put \,\,them\,\, in\,\, (1)\\\\4(1)+2= 2(-1)+2\\ 6\ne0

Hence the lines are not intersecting nor parallel, they must be skew.

6 0
3 years ago
Simplify: 1.3 + (–6) + (–4.25)
Dmitrij [34]

Answer:

-8.95

Step-by-step explanation:

1.3 - 6 - 4.25

-4.7 - 4.25

-8.95

Hope this helps!

Have a nice day!

If you find my answer helpful

<em>Pls consider marking my answer as </em><em>Brainliest</em><em>! It would mean a lot!</em>

6 0
2 years ago
Find the value of x.<br> When a leg is 6<br> units and the<br> hypotenuse is 12<br> units.
DIA [1.3K]

Answer:

6√3

Step-by-step explanation:

6^2+x^2=12^2

36+x^2=144

-36          -36

x^2=108

x=6√3

3 0
3 years ago
It 5/2 not -5/2 and sorry for telling u the
babunello [35]
What does this mean buddy?
4 0
3 years ago
Read 2 more answers
Hepp I am being timeddd​
Lorico [155]
The answer is Megan



following me on ig for tutoring
7 0
3 years ago
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