-8-5b+7+5b
add +5b to -5b to get 0.
-8+7 = -1
Answer is -1.
Slope of AB=(2-3)/(1+1)=-1/2; AC=(-1-3)/(-3+1)=-4/-2=2; BC=(-1-2)/(-3-1)=-3/-4=3/4.
The product of two sides that are perpendicular is -1. Slopes of AB and AC are perpendicular so angle A is a right angle. ABC is a right-angled triangle.
Answer:
Sum of the sequence (Sn) = 33,859
Step-by-step explanation:
Given:
Sequence = 685+678+671+664+...+6
Find:
Sum of the sequence (Sn)
Computation:
a = 685
d = 678 - 985 = -7
an = 6
an = a+(n-1)d
6 = 685+(n-1)(-7)
-679 = (n-1)(-7)
97 = n-1
n = 98
So,
Sum of the sequence (Sn) = (n/2)[a+an]
Sum of the sequence (Sn) = (98/2)[685+6]
Sum of the sequence (Sn) = (49)(691)
Sum of the sequence (Sn) = 33,859
Answer:
66yx , 30x2y, hope its right :)
Step-by-step explanation:
Answer:
titutex=cos\alp,\alp∈[0:;π]
\displaystyle Then\; |x+\sqrt{1-x^2}|=\sqrt{2}(2x^2-1)\Leftright |cos\alp +sin\alp |=\sqrt{2}(2cos^2\alp -1)Then∣x+
1−x
2
∣=
2
(2x
2
−1)\Leftright∣cos\alp+sin\alp∣=
2
(2cos
2
\alp−1)
\displaystyle |\N {\sqrt{2}}cos(\alp-\frac{\pi}{4})|=\N {\sqrt{2}}cos(2\alp )\Right \alp\in[0\: ;\: \frac{\pi}{4}]\cup [\frac{3\pi}{4}\: ;\: \pi]∣N
2
cos(\alp−
4
π
)∣=N
2
cos(2\alp)\Right\alp∈[0;
4
π
]∪[
4
3π
;π]
1) \displaystyle \alp \in [0\: ;\: \frac{\pi}{4}]\alp∈[0;
4
π
]
\displaystyle cos(\alp -\frac{\pi}{4})=cos(2\alp )\dotscos(\alp−
4
π
)=cos(2\alp)…
2. \displaystyle \alp\in [\frac{3\pi}{4}\: ;\: \pi]\alp∈[
4
3π
;π]
\displaystyle -cos(\alp -\frac{\pi}{4})=cos(2\alp )\dots−cos(\alp−
4
π
)=cos(2\alp)…
1
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