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nordsb [41]
3 years ago
10

7. In A.ABC. JB and KA are medians, JK = 10x - 12, AB = 9x + 18, JM = 21, KM = 23. AJ = 60. and

Mathematics
1 answer:
Juliette [100K]3 years ago
3 0

Answer:

A) JK and AB =

[tex] \boxed{( \theta) = \f{12}{( \theta)} } So, =》 \dfrac{10}{5} = \dfrac{1}{ \( \theta) } =》 \( \theta) = \dfrac{5}{2}[/tex]

B) AABC

[tex] \boxed{( \theta) = {1}{( \theta)} } So, 23}{-18=\\Ans } \boxed

C) AABM:

[tex] e \fbox{: } The value of k for which \sin(jm) = \cos(x) is \dfrac{\p}{2} ans[/tex]

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Marrrta [24]

Answer:

4. -\tan 32°

Step-by-step explanation:

Given:

The tangent of the angle is given as:

\tan (-212\°)

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-212° when measured from counter clockwise direction will be equal to:

-212\°(Clockwise)=360-212=148\°(CCW)

Therefore, \tan (-212\°) = \tan (148\°)

Now, we have the identity for tan as:

\tan(\pi-\theta)=-\tan \theta

Here, \theta=148\°

Therefore,

-\tan (148\°)=\tan(180-148)\\-\tan(148\°)=\tan(32\°)\\\textrm{Multiplying both sides by -1, we get:}\\-1(-\tan(148\°)=-1\times \tan(32\°)\\\\\tan(148\°)=-\tan(32\°)

Hence, \tan (-212\°) = \tan (148\°) = -\tan 32°

5 0
3 years ago
6xy (1/2x ^2 - 1/2xy + 1/2y ^2 )
nevsk [136]
Is this correct on what you typed above -> \bf 6xy\left( \cfrac{1}{2x^2}-\cfrac{1}{2xy}+\cfrac{1}{2y^2} \right) ?
6 0
3 years ago
Please come answer this, I need help asap
Vadim26 [7]

Answer:

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Step-by-step explanation:

To evaluate this expression, all we have to do is substitute the given values of the variables into the expression. That is, replace every m in the expression with a 1 and every p in the expression with a 5.

Doing so, we get:

p^{2} +m

=5^{2} +1 (Substitute given values into the expression)

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Hope this helps!

8 0
3 years ago
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Answer:

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Step-by-step explanation:

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3 0
3 years ago
In the diagram, polygon ABCD is flipped over a line of reflection to form a polygon with its vertices at A′, B′, C′, and D′. Poi
babunello [35]

Answer: The line of reflection must be x=5.


Step-by-step explanation:

Given: In the diagram, polygon ABCD is flipped over a line of reflection to form a polygon with its vertices at A′, B′, C′, and D′. Points A′, B′, and D′ are shown, but the line of reflection and point C′ are not .

coordinates of C from the graph = (4,2)

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Therefore, the line of reflection must be x=5.

3 0
3 years ago
Read 2 more answers
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