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iren [92.7K]
3 years ago
9

WHERE DO I DRAG THE LAST TWO

Mathematics
1 answer:
Anastasy [175]3 years ago
4 0
-3/4 goes between -1 and 0 it is the 3rd line to the left and n would go between positive 0 and 1 on the 3rd line to the right
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Describe a sequence of transformations that can be performed to polygon ABCDE to prove that it is similar to polygon FGHIJ
maxonik [38]

Answer:

answer below

Step-by-step explanation:

ABCDE go through dilation over center (6 , -2) with factor of 1/2 to FGHIJ

AB // FG slope: -2 , √20:√5 = 2: 1

BC // GH // X axis 8:4 = 2:1

CD // HI, slope= 1 , √8:√2 = 2:1

DE // IJ // x axis, 4:2 = 2:1

EA // JF // y axis, 2:1

8 0
2 years ago
Circumference of a circle of radius 13.4cm
Sedbober [7]

Answer:

26.8 pi or approx. 84.19

Step-by-step explanation:

Circumference can be found by multiplying the diameter by pi. The diameter is twice the radius, which in this case is 13.4, so it's 13.4 * 2 = 26.8.

Multiply by pi, and you've got your answer!

5 0
3 years ago
Cosx+1/sin^3x=cscx/1-cosx
ANTONII [103]
<span> I am assuming you want to prove:
csc(x)/[1 - cos(x)] = [1 + cos(x)]/sin^3(x).

 </span>
<span>If we multiply the LHS by [1 + cos(x)]/[1 + cos(x)], we get:
LHS = csc(x)/[1 - cos(x)]
= {csc(x)[1 + cos(x)]/{[1 + cos(x)][1 - cos(x)]}
= {csc(x)[1 + cos(x)]}/[1 - cos^2(x)], via difference of squares
= {csc(x)[1 + cos(x)]}/sin^2(x), since sin^2(x) = 1 - cos^2(x).

 </span>
<span>Then, since csc(x) = 1/sin(x):
LHS = {csc(x)[1 + cos(x)]}/sin^2(x)
= {[1 + cos(x)]/sin(x)}/sin^2(x)
= [1 + cos(x)]/sin^3(x)
= RHS.

 </span>
<span>I hope this helps! </span>
8 0
3 years ago
Am I right about this or..?
Arisa [49]

Answer:

Correct, as 4 is the only number repeating twice

4 0
3 years ago
Determine if the question posed is a statistical question (yes) or not (no). What is 2 plus 2?
dimulka [17.4K]

Answer:yes?

Step-by-step explanation:

it has #s in it lol

5 0
3 years ago
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