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soldi70 [24.7K]
3 years ago
5

Need help on this question

Mathematics
1 answer:
UNO [17]3 years ago
7 0
List of multiples of 2/3
\frac{2}{3}, \frac{4}{6}, \frac{6}{9}, \frac{8}{12}

List of multiples of 3/4
\frac{3}{4}, \frac{6}{8}, \frac{9}{12}, \frac{12}{16}

They have a common denominator at 12.

Compare 8/12  and 9/12.
9>8 ,  therefore 9/12 > 8/12  and  3/4 > 2/3
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Pavel [41]

Answer: T (The first option).


Step-by-step explanation:

1. By definition, coplanar points are three or more points that are in the same plane.

2. Based on this information, as you can see in the figure attached, the points A, C, D, E, S and U are coplanars because they are in the same plane, which is: ACDE.

3. Therefore, you can conclude that the point T is not coplanar with points A, C and D because this is not in the plane ACDE.


3 0
3 years ago
Subtract the following polynomials.<br><br> (5x2+6)−(6x−5x2)<br><br> im bad at math ;-;
Eddi Din [679]

Answer:

First off if you want this answered make sure it's clear when your multiplying or using X

5 0
3 years ago
Read 2 more answers
#3 need help with this math problem please.
kupik [55]

Answer:

undefined

Step-by-step explanation:

f(x) = 2/ (-5x-10)

Let x = -2

f(-2) = 2/ (-5(-2)-10)

         =2/(10-10)

         =2/0

         = undefined

5 0
3 years ago
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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