3. 90%
4. 24
5. 55
If you need the work or anything just let me know! :)
The solution of the sin (10x) cos (7x) is 1/2 sin(17x ) + sin(3x).
<h3>What are the trigonometric identities?</h3>
We can calculate with the help of one of the trigonometric identities;
sin(A)cos(B) = 1/2 sin(A+B) + sin(A - B)
WE have given
sin (10x) cos (7x)
Here, A = 10x
B= 7x
So, sin(A)cos(B) = 1/2 sin(A+B) + sin(A - B)
sin(10x) cos(7x) = 1/2 sin(10x + 7x ) + Sin (10x - 7x)
sin(10x) cos(7x) = 1/2 sin(17x ) + sin(3x)
sin(10x) cos(7x) = 1/2 sin(17x ) + sin(3x)
Learn more about trigonometric ratios here:
brainly.com/question/22599614
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Answer:
Given : EFGH is a parallelogram.
Prove: EG bisects HF and HF bisects EG.
Since, a parallelogram has two pairs of congruent and parallel sides.
Therefore, In parallelogram EFGH,
EF≅GH and EH≅FG
Also, EF║GH and EH║FG
Since,In triangles EKF and GKH,
EF≅HG
Since, EF║HG
⇒∠FEK ≅ ∠HGK and ∠EFK ≅ ∠ GHK ( When two parallel lines are cut by a transversal alternative interior angles are congruent )
⇒ Δ EKF ≅ Δ GKH ( ASA congruence postulate )
⇒ EK ≅ GK and FK ≅ HK ( CPCTC)
⇒ EG bisects HF and HF bisects EG ( Definition of bisector)
Hence proved.
Answer:
m > ![\frac{10}{11}](https://tex.z-dn.net/?f=%5Cfrac%7B10%7D%7B11%7D)
Step-by-step explanation:
-3m + 3 < 8m - 7
-3m - 8m < -3 - 7
-11m < -10
m > ![\frac{10}{11}](https://tex.z-dn.net/?f=%5Cfrac%7B10%7D%7B11%7D)
Answer:
60
Step-by-step explanation:
The triangles are similar so corresponding angles will be congruent. We also know this because both angles have two lines marking they are congruent.