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Zinaida [17]
3 years ago
10

List two pairs of an alternate interior angle and write them as a congress statements​

Mathematics
1 answer:
Sliva [168]3 years ago
6 0

Answer:

Step-by-step explanation:

1.  Angle 4 is congruent to angle 6

2.  Angle 3 is congruent to angle 5

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PLS ANSWER!!!! i need help plsssssss
ohaa [14]

Answer:

Sorry dude, I don't know, but maybe someone else will come and answer!!!

Step-by-step explanation:

4 0
3 years ago
Findℒ{f(t)}by first using a trigonometric identity. (Write your answer as a function of s.)f(t) = 12 cost −π6
allsm [11]

Answer:

L(f(t)) = \dfrac{6}{S^2+1} [\sqrt{3} \ S +1 ]

Step-by-step explanation:

Given that:

f(t)  = 12 cos (t- \dfrac{\pi}{6})

recall that:

cos (A-B) = cos AcosB + sin A sin B

∴

f(t) = 12 [cos\  t \  cos \dfrac{\pi}{6}+ sin \ t  \ sin \dfrac{\pi}{6}]

f(t) = 12 [cos \  t \ \dfrac{3}{2}+ sin  \ t  \ sin \dfrac{1}{2}]

f(t) = 6 \sqrt{3} \ cos \ (t) + 6 \ sin \ (t)

L(f(t)) = L ( 6 \sqrt{3} \ cos \ (t) + 6 \ sin \ (t) ]

L(f(t)) = 6 \sqrt{3} \ L [cos \ (t) ] + 6\ L [ sin \ (t) ]

L(f(t)) = 6 \sqrt{3}  \dfrac{S}{S^2 + 1^2}+ 6 \dfrac{1}{S^2 +1^2}

L(f(t)) = \dfrac{6 \sqrt{3} +6 }{S^2+1}

L(f(t)) = \dfrac{6( \sqrt{3} \ S +1 }{S^2+1}

L(f(t)) = \dfrac{6}{S^2+1} [\sqrt{3} \ S +1 ]

7 0
3 years ago
If FG=10 and HJ = 10. Which additional facts guarantee that FGHJ is a parallelogram? check all that apply.
Kamila [148]
Yes, if FJ || GH, and also if FG || HJ, since a parallelogram has two pairs of parallel sides.

now, we know FG = HJ = 10, but if it's a parallelogram, then the other two sides must also have the same length, so chances are FJ = GH = 15.
5 0
3 years ago
4) -4y= -x + 4<br> 6) y + 3x = 3
wel

Answer:

x=16/13

y= -9/13

Step-by-step explanation:

A:         -4y=-x+4

B:          y+3x=3

4B:       4y+12x=12

A+4B:   12x= -x+16

----->    13x=16

               x=16/13

-4y= -16/13+4

4y=16/13-4

y=4/13-1= -9/13

4 0
3 years ago
Find the mean mode median,range<br> 5.4, 3, 4.2, 9, 2.1, 3, 4.2
RideAnS [48]
Range = 6.9
mode = 4.2
median= 3
7 0
3 years ago
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