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Alja [10]
3 years ago
8

What is 2+2 I don't know the answer to that

Mathematics
2 answers:
Leya [2.2K]3 years ago
5 0
It's 4 2+2=4
this the answer
Debora [2.8K]3 years ago
3 0
It's 4.

If you use a number line, with the numbers 0-10 on it, it'll be easier to see. Put your finger on the number two, and move your finger to the right 2 numbers. You should end up on the number 4.

Hope that helps!
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HELP PLEASE
irinina [24]

Answer:

A. 3x+3

Step-by-step explanation:

The perimeter is found by adding all the side lengths:

(x)+(x-3)+(x+6)

=1x+1x+1x-3+6

=3x+3

(Remember, add x-terms to x-terms, and numbers to numbers)

5 0
3 years ago
Write down the size of angle ABC give a reason for your answer
solong [7]

Answer:

∠ ABC = 90°

Step-by-step explanation:

∠ ABC is the angle in a semi circle and is right, that is

∠ ABC = 90°

6 0
3 years ago
PLEASE HELP FOR 30 POINTS....GEOMETRY/TRIG
Advocard [28]

Answer:

10/cot 62 degrees

Step-by-step explanation:

5 0
3 years ago
A.x=48<br><br> B.x=24<br><br> C.x=12<br><br> D.x= -18
Roman55 [17]
Answer: “B” is the answer.


Explanation: I did it in my head so I am correct.
6 0
3 years ago
Determine the equation of the line that is perpendicular to the lines r(t)=(-2+3t,2t,3t)
Mnenie [13.5K]
<span>Vector Equation
(Line)</span>(x,y) = (x,y) + t(a,b);tERParametric Formx = x + t(a), y = y + t(b); tERr = (-4,-2) + t((-3,5);tERFind the vector equation of the line passing through A(-4,-2) & parallel to m = (-3,5)<span>Point: (2,5)
Create a direction vector: AB = (-1 - 2, 4 - 5) 
= (-3,-1) or (3,1)when -1 (or any scalar multiple) is divided out.
r = (2,5) + t(-3,-1);tER</span>Find the vector equation of the line passing through A(2,5) & B(-1,4)<span>x = 4 - 3t
y = -2 + 5t
;tER</span>Write the parametric equations of the line passing through the line passing through the point A(4,-2) & with a direction vector of m =(-3,5)<span>Create Vector Equation first:
AB = (2,8)
Point: (4,-3)
r = (4,-3) + (2,8); tER
x = 4 + 2t 
y = -3 + 8t
;tER</span>Write the parametric equations of the line through A(4,-3) & B(6,5)<span>Make parametric equations: 
x = 5 + 4t
y = -2 + 3t ; tER
For x sub in -3
-3 = 5 + 4t 
(-8 - 5)/4 = t
-2 = t
For y sub in -8 
-8 = -2 + 3t
(-8 + 2)/3 = t
-2 = t
Parameter 't' is consistent so pt(-3,-8) is on the line.</span>Given the equation r = (5,-2) + t(4,3);tER, is (-3,-8) on the line?<span>Make parametric equations: 
x = 5 + 4t
y = -2 + 3t ; tER
For x sub in 1
-1 = 5 + 4t 
(-1 - 5)/4 = t
-1 = t
For y sub in -7 
-7 = -2 + 3t
(-7 + 2)/3 = t
-5/3 = t
Parameter 't' is inconsistent so pt(1,-7) is not on the line.</span>Given the equation r = (5,-2) + t(4,3);tER, is (1,-7) on the line?<span>Use parametric equations when generating points: 
x = 5 + 4t
y = -2 + 3t ;tER
X-int:
sub in y = 0
0 = -2 + 3t
solve for t
2/3 = t (this is the parameter that will generate the x-int) 
Sub t = 2/3 into x = 5 + 4t 
x = 5 + 4(2/3)
x = 5 + (8/3)
x = 15/3 + (8/3) 
x = 23/3
The x-int is (23/3, 0)</span>What is the x-int of the line r = (5,-2) + t(4,3); tER?Note: if they define the same line: 1) Are their direction vectors scalar multiples? 2) Check the point of one equation in the other equation (LS = RS if point is subbed in)What are the two requirements for 2 lines to define the same line?
3 0
3 years ago
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