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Vanyuwa [196]
2 years ago
10

Graph the points K(-2, -1), L(-1, 2), M(2, 4), and N(1,1) in desmos geometry. What is the distance from point M to point N?

Mathematics
1 answer:
Rudik [331]2 years ago
5 0

Quadrilateral KLMN has vertices K(-2,-2), L(1,1), M(0,4), and N(-3,5)

It is first translated by (x + 2,y - 1)

then

K(-2,-2) ---> K'(-2 + 2,-2 - 1) = (0,-3)

L(1,1), ---> L'(1 + 2,1 - 1) = (3,0)

M(0,4), ---> M' (0 + 2,4 - 1) = (2,3)

and

N(-3,5) ---> N' (-3 + 2,5 - 1) = (-1,4)

and then translated by (x - 3,y + 4)

K'(0,-3) ---> K''(0 - 3,-3 + 4) = (-3,1)

L'(3,0) ---> L''(3 - 3,0 + 4)= (0,4)

M'(2,3)  ---> M'' (2 - 3, 3 + 4)= (-1,7)

and

N'(-1,4) ---> N'' (-1 - 3,4 + 4) = (-4,8)

the coordinates of K" L" M" N" after both translations are

K''= (-3,1)

L''= (0,4)

M''= (-1,7)

and

N'' = (-4,8)

hope im right if not im sorry but it going to be right cuz im good at this stuff for you

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How much money has to be invested at 5.9% interest compounded continuously to have $15,000 after 12 years?
scoray [572]

The amount of $7389.43 has to be invested at 5.9% interested continuously to have $15,000 after 12 years.

Step-by-step explanation:

The given is,

                Future value, F  = $15,000

                           Interest, i = 5.9%

              ( compounded continuously )

                            Period, t = 12 years

Step:1

           Formula to calculate the present with compounded continuously,

                                       F=Pe^{(i)(t)}...............(1)

           Substitute the values in equation (1) to find the P value,

                                  15000=Pe^{(0.059)(12)}          ( ∵ i = \frac{5.9}{100}=0.059 )

                                  15000=Pe^{0.708}

                                  15000=P(2.0299)             ( ∵ e^{o.708} =2.0299 )

            We change the P (Present value) into the left side,

                                        P=\frac{15000}{2.0299}

                                            =7389.427

                                            ≅ 7389.43

                                         P = $ 7389.43

Result:

           The amount of $7389.43 has to be invested at 5.9% interested continuously to have $15,000 after 12 years.  

                       

8 0
3 years ago
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