Which quadratic function in standard form can be represented by the graph that has a vertex at (6, 3) and passes through the poi
1 answer:
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Answer:
line1：
line2:
Step-by-step explanation:
Here is all the work and answers for these questions they are all correct i checked them all and ive done this assignmnet before
Answer:
segment YZ ≈ 19.4 in
angle X ≈ 85.3°
angle Z ≈ 26.7°
Explanation:
1) Given two side lenghts and one angle you can use sine law:
2) Using the sides with length 43 in and 40in, and the corresponding opposite angles, Z and 68°, that leads to:
From which you can clear sinZ and get:
sinZ = 43 × sin(68) / 40 = 0.9967
⇒ Z = arcsine(0.9967) ≈ 85.36°
3) The third angle can be determined using 85.36° + 68° + X = 180°
⇒ X = 180° - 85.36° - 68° = 26.64°.
4) Finally, you can apply the law of sine to obtain the last missing length:
From which: x = 40 × sin(26.64°) / sin(68°) = 19.34 in
The answer, then is:
segment YZ ≈ 19.4 in
angle X ≈ 85.3°
angle Z ≈ 26.7°
segment YZ ≈ 19.4 in
angle X ≈ 85.3°
angle Z ≈ 26.7°
Explanation:
1) Given two side lenghts and one angle you can use sine law:
2) Using the sides with length 43 in and 40in, and the corresponding opposite angles, Z and 68°, that leads to:
From which you can clear sinZ and get:
sinZ = 43 × sin(68) / 40 = 0.9967
⇒ Z = arcsine(0.9967) ≈ 85.36°
3) The third angle can be determined using 85.36° + 68° + X = 180°
⇒ X = 180° - 85.36° - 68° = 26.64°.
4) Finally, you can apply the law of sine to obtain the last missing length:
From which: x = 40 × sin(26.64°) / sin(68°) = 19.34 in
The answer, then is:
segment YZ ≈ 19.4 in
angle X ≈ 85.3°
angle Z ≈ 26.7°