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Ghella [55]
2 years ago
10

FREE PONITSssssss 2+2

Mathematics
2 answers:
lubasha [3.4K]2 years ago
4 0

4

dgfsdfgsfgdfgdfgdfgdfgdfgdfg

GaryK [48]2 years ago
3 0

Answer:

4

Step-by-step explanation:

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KM bisects ∠LKJ. If m∠LKM = 4x + 12 and m∠MKJ = 6x - 6, find the value of x. *
riadik2000 [5.3K]

Given that,

KM bisects ∠LKJ.

m∠LKM = 4x + 12 and m∠MKJ = 6x - 6

To find,

The value of x.

Solution,

As KM bisects ∠LKJ. It would mean that,

m∠LKM = m∠MKJ

Putting the above values,

4x+12=6x-6

We can solve the above equations as follows :

4x-6x=-12-6

-2x=-18

x=9

Therefore, the value of x is equal to 9.

6 0
2 years ago
Need help on problem 10
Shkiper50 [21]
The Perimeter of Triangle ABC would be 60.
You can solve this by setting up proportions.

3 0
3 years ago
HELP say im getting it wrong <br><br>the perimeter of the polygons is ?​
lara [203]
So the answer would be16*4=64
8 0
2 years ago
(⚠️⚠️HELP!⚠️⚠️)Mary bought 18 books. Mary bought 6 times as many books as Carlos. Let b be the number of books that Carlos bough
kvasek [131]

Answer:

18=6b, b=3

Because Mary has 6 times as much as Carlos does, we put the x with 6. 18 is the number that 6x equals. 6*3=18

8 0
3 years ago
Read 2 more answers
Find the curl of ~V<br> ~V<br> = sin(x) cos(y) tan(z) i + x^2y^2z^2 j + x^4y^4z^4 k
ch4aika [34]

Given

\vec v =  f(x,y,z)\,\vec\imath+g(x,y,z)\,\vec\jmath+h(x,y,z)\,\vec k \\\\ \vec v = \sin(x)\cos(y)\tan(z)\,\vec\imath + x^2y^2z^2\,\vec\jmath+x^4y^4z^4\,\vec k

the curl of \vec v is

\displaystyle \nabla\times\vec v = \left(\frac{\partial h}{\partial y}-\frac{\partial g}{\partial z}\right)\,\vec\imath - \left(\frac{\partial h}{\partial x}-\frac{\partial f}{\partial z}\right)\,\vec\jmath + \left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right)\,\vec k

\nabla\times\vec v = \left(4x^4y^3z^4-2x^2y^2z\right)\,\vec\imath \\\\ - \left(4x^3y^4z^4-\sin(x)\cos(y)\sec^2(z)\right)\,\vec\jmath \\\\ + \left(2xy^2z^2+\sin(x)\sin(y)\tan(z)\right)\,\vec k

\nabla\times\vec v = \left(4x^4y^3z^4-2x^2y^2z\right)\,\vec\imath \\\\ + \left(\sin(x)\cos(y)\sec^2(z)-4x^3y^4z^4\right)\,\vec\jmath \\\\ + \left(2xy^2z^2+\sin(x)\sin(y)\tan(z)\right)\,\vec k

7 0
3 years ago
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