Please some-one help! Given: m∠JNM =4x+6m ∠KNL = 7x−21 What is the m∠JNM ?

Mathematics

1 answer:

Nostrana [21]3 years ago

6 0

Answer:

x = 5

Step-by-step explanation:

If it's a number by itself then that's right cuz JNM and KNL is equal, so u do 4x + 6 = 7x - 21. When u do that, u get x = 5.

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$\bf \textit{let's start by grouping}
\\\\\\
x^2+4x+y^2-2y=20\implies (x^2+4x)+(y^2-2y)=20
\\\\\\
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now, recall $\bf \begin{array}{cccccllllll}
{{ a}}^2& + &2{{ a}}{{ b}}&+&{{ b}}^2\\
\downarrow && &&\downarrow \\
{{ a}}&& &&{{ b}}\\
&\to &({{ a}} + {{ b}})^2&\leftarrow 
\end{array}\qquad 
% perfect square trinomial, negative middle term
\begin{array}{cccccllllll}
{{ a}}^2& - &2{{ a}}{{ b}}&+&{{ b}}^2\\
\downarrow && &&\downarrow \\
{{ a}}&& &&{{ b}}\\
&\to &({{ a}} - {{ b}})^2&\leftarrow 
\end{array}$

so.. the middle term of a perfect square trinomial, is really 2 * the other 2 guys

so hmm we can say 2 * x * ? = 4x, thus 2x * ? = 4x, thus cross-multiplying
? = 2

now, for the other, 2 * y * ? = 2y, thus 2y* ? = 2y thus ? = 1

so our guys are 2 and 1

now, bear in mind, all we're doing, is borrowing from our very good friend Mr Zero, 0, so, if we add whatever, we also have to subtract whatever

thus $\bf (x^2+4x+2^2)+(y^2-2y+1^2)-1^2-2^2=20
\\\\\\(x^2+4x+4)+(y^2-2y+1)-5=20\\\\\\ \boxed{(x+2)^2+(y-1)^2=25}\\\\
-----------------------------\\\\
(x-{{ h}})^2+(y-{{ k}})^2={{ r}}^2
\qquad center\ ({{ h}},{{ k}})\qquad
radius={{ r}}\\\\
-----------------------------\\\\
(x-(-2))^2+(y-1)^2=5^2\qquad center\ (-2,1)\qquad radius=5$

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