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just olya [345]
3 years ago
15

Please help to finish my math

Mathematics
1 answer:
ELEN [110]3 years ago
5 0
Second line= 3, 3*2=6
third line= v, you have to distribute, 2 times v is 2v and 2 times 3 is 6
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7р + 8р - 12 = 59 what would this be
MArishka [77]

Answer:

4.7333....................

Step-by-step explanation:

First combine like-terms. The equation now is 15p-12=59. Next add 12 to both sides. The equation now is 15p=71. Last divide both sides by 15. Now the equation is 4.73333....

3 0
3 years ago
(m-4)x(m+3) Find the product
Alex

Answer:

m^2-m-12

Step-by-step explanation:

8 0
4 years ago
Look at the sequence : 5, -10, 20, -40, 80
Sphinxa [80]

Answer:

5 (5x1) , -10 -(5x2), 20(10x2), -40 -(20x2), 80(40x2)

so the next number should be previous number in the sequence multiplied by 2 and a negative sign.

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7 0
4 years ago
Read 2 more answers
What is the area of a triangle base 25 feet and height 8 feet
Mariulka [41]
How you get your answer is by multiplying 25*8= 125 
8 0
3 years ago
Read 2 more answers
Use lagrange multipliers to find the point on the plane x − 2y + 3z = 6 that is closest to the point (0, 2, 5).
horsena [70]
Lagrangian:

L(x,y,z,\lambda)=x^2+(y-2)^2+(z-5)^2+\lambda(x-2y+3z-6)

where the function we want to minimize is actually \sqrt{x^2+(y-2)^2+(z-5)^2}, but it's easy to see that \sqrt{f(\mathbf x)} and f(\mathbf x) have critical points at the same vector \mathbf x.

Derivatives of the Lagrangian set equal to zero:

L_x=2x+\lambda=0\implies x=-\dfrac\lambda2
L_y=2(y-2)-2\lambda=0\implies y=2+\lambda
L_z=2(z-5)+3\lambda=0\implies z=5-\dfrac{3\lambda}2
L_\lambda=x-2y+3z-6=0

Substituting the first three equations into the fourth gives

-\dfrac\lambda2-2(2+\lambda)+3\left(5-\dfrac{3\lambda}2\right)=6
11-7\lambda=6\implies \lambda=\dfrac57

Solving for x,y,z, we get a single critical point at \left(-\dfrac5{14},\dfrac{19}7,\dfrac{55}{14}\right), which in turn gives the least distance between the plane and (0, 2, 5) of \dfrac5{\sqrt{14}}.
7 0
3 years ago
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