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balandron [24]
3 years ago
11

The prime factorization of 236

Mathematics
2 answers:
nika2105 [10]3 years ago
7 0

Answer:

22 x 591

Step-by-step explanation:

mafiozo [28]3 years ago
3 0

Answer:

The prime factors are: 2 x 2 x 59

or Written in exponential form: 2^2 x 59^1

Step-by-step explanation:

236 ÷ 2 = 118 - No remainder! 2 is one of the factors

118 ÷ 2 = 59 - No remainder! 2 is one of the factors

59 ÷ 59 = 1 - No remainder! 59 is one of the factors

together we have the factors 2 x 2 x 59 = 236

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3 years ago
-5/6x = -10/3<br> solve for x
ololo11 [35]

Answer:

x = 4

Step-by-step explanation:

First you need to multiply both sides of the equation by -6/5

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In parallelogram JKLM, m&lt;L exceeds m&lt;M by 30 degrees. What is the measure of M&lt;J?
Damm [24]

 

Answer:

2)105°

Step-by-step explanation:

In this parallelogram, J should be congruent to L (J=L). We can solve this problem if we find out the value of L.

The sum of the adjacent angle of the parallelogram will be equal to 180 degrees, so  the equation is

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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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