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Ivan
3 years ago
13

Convert to the nearest tenth.

Mathematics
2 answers:
Finger [1]3 years ago
8 0
1. 27.3 2.4.8 3.11.4 4.10.9
Rasek [7]3 years ago
4 0
1 metre = 1.0936 yards
A. 27.34
B. 4.75
C. 11.43
D. 10.94
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Step-by-step explanation:

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Then we have (x-5)(x-1)=0 if x=5 or x=1.

Intersections on x are points (1,0) and (5,0), middle is (3,0).

Intersection with y is when put x=0 in equation, so you will get y=0-6*0+5, y=5. The point is (0,5).

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If vector u has its initial point at (-7, 3) and its terminal point at (5, -6), u =
attashe74 [19]

First of all, let <span>θθ</span> be some angle in <span><span>(0,π)</span><span>(0,π)</span></span>. Then

<span><span><span>θθ</span> is acute <span>⟺⟺</span> <span><span>θ<<span>π2</span></span><span>θ<<span>π2</span></span></span> <span>⟺⟺</span> <span><span>cosθ>0</span><span>cos⁡θ>0</span></span>.</span><span><span>θθ</span> is right <span>⟺⟺</span> <span><span>θ=<span>π2</span></span><span>θ=<span>π2</span></span></span> <span>⟺⟺</span> <span><span>cosθ=0</span><span>cos⁡θ=0</span></span>.</span><span><span>θθ</span> is obtuse <span>⟺⟺</span> <span><span>θ><span>π2</span></span><span>θ><span>π2</span></span></span> <span>⟺⟺</span> <span><span>cosθ<0</span><span>cos⁡θ<0</span></span>.</span></span>

Now, to see if (say) angle <span>AA</span> of the triangle <span><span>ABC</span><span>ABC</span></span> is acute/right/obtuse, we need to check whether <span><span>cos∠BAC</span><span>cos⁡∠BAC</span></span> is positive/zero/negative. But what is <span><span>cos∠BAC</span><span>cos⁡∠BAC</span></span>? It is the angle made by the vectors <span><span><span>AB</span><span>−→−</span></span><span><span>AB</span>→</span></span> and <span><span><span>AC</span><span>−→−</span></span><span><span>AC</span>→</span></span>. (When you are computing the angle at a particular vertex <span>vv</span>, you should make sure that both the vectors corresponding to the two adjacent sides have that vertex <span>vv</span> as the initial point.) We will first compute these two vectors:

<span><span><span><span>AB</span><span>−→−</span></span>=(0,0,0)−(1,2,0)=(−1,−2,0)</span><span><span><span>AB</span>→</span>=(0,0,0)−(1,2,0)=(−1,−2,0)</span></span><span><span><span><span>AC</span><span>−→−</span></span>=(−2,1,0)−(1,2,0)=(−3,−1,0)</span><span><span><span>AC</span>→</span>=(−2,1,0)−(1,2,0)=(−3,−1,0)</span></span>Therefore, the angle between these vectors is given by:<span><span><span>cos∠BAC=<span><span><span><span>AB</span><span>−→−</span></span>⋅<span><span>AC</span><span>−→−</span></span></span><span>|<span><span>AB</span><span>−→−</span></span>||<span><span>AC</span><span>−→−</span></span>|</span></span>=…</span>(1)</span><span>(1)<span>cos⁡∠BAC=<span><span><span><span>AB</span>→</span>⋅<span><span>AC</span>→</span></span><span>|<span><span>AB</span>→</span>||<span><span>AC</span>→</span>|</span></span>=…</span></span></span>Can you take it from here? From the sign of this value, you should be able to decide if angle <span>AA</span> is acute/right/obtuse.

Now, do the same procedure for the remaining two angles <span>BB</span> and <span>CC</span> as well. That should help you solve the problem.

A shortcut. Since you are not interested in the actual values of the angles, but you need only whether they are acute, obtuse or right, it is enough to compute only the sign of the numerator (the dot product between the vectors) in formula (1). The denominator is always positive.

6 0
3 years ago
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