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PIT_PIT [208]
3 years ago
12

What is the value of exppresion if m=3 and n=2 (2m)3 +3(6-n)n?

Mathematics
1 answer:
shtirl [24]3 years ago
7 0

Answer:

102

Step-by-step explanation:

2(3)³ + 3(6-2)2

2 (27) + 3(4)2

54 + 48

102

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Find the distance between the two points rounding to the nearest tenth (if necessary).
lara31 [8.8K]
The formula is called the “midpoint formula”. It looks like this
m=(x1+x2)/2 , (y1+y2)/2 where m means midpoint. In your case it looks like this.

6 0
3 years ago
The curve is the graph of function f. Use the graph to find the value of the function for the following values of x: -2,-.5,0,2.
VikaD [51]

Answer:

1, 0.75, 1.25, 3, 0

Step-by-step explanation:

To find the value of the function, we find the y coordinate of the point with the given x value.

7 0
2 years ago
Find the distance between a point (– 2, 3 – 4) and its image on the plane x+y+z=3 measured across a line (x + 2)/3 = (2y + 3)/4
dimaraw [331]

Answer:

Distance of the point from its image = 8.56 units

Step-by-step explanation:

Given,

Co-ordinates of point is (-2, 3,-4)

Let's say

x_1\ =\ -2

y_1\ =\ 3

z_1\ =\ -4

Distance is measure across the line

\dfrac{x+2}{3}\ =\ \dfrac{2y+3}{4}\ =\ \dfrac{3z+4}{5}

So, we can write

\dfrac{x-x_1+2}{3}\ =\ \dfrac{2(y-y_1)+3}{4}\ =\ \dfrac{3(z-z_1)+4}{5}\ =\ k

=>\ \dfrac{x-(-2)+2}{3}\ =\ \dfrac{2(y-3)+3}{4}\ =\ \dfrac{3(z-(-4))+4}{5}\ =\ k

=>\ \dfrac{x+4}{3}\ =\ \dfrac{2y-3}{4}\ =\ \dfrac{3z+16}{5}\ =\ k

=>\ x\ =\ 3k-4,\ y\ =\ \dfrac{4k+3}{2},\ z\ =\ \dfrac{5k-16}{3}

Since, the equation of plane is given by

x+y+z=3

The point which intersect the point will satisfy the equation of plane.

So, we can write

3k-4+\dfrac{4k+3}{2}+\dfrac{5k-16}{3}\ =\ 3

=>6(3k-4)+3(4k+3)+2(5k-16)\ =\ 18

=>18k-24+12k+9+10k-32\ =\ 18

=>\ k\ =\dfrac{13}{8}

So,

x\ =\ 3k-4

   =\ 3\times \dfrac{13}{8}-4

   =\ \dfrac{7}{4}

y\ =\ \dfrac{4k+3}{2}

   =\ \dfrac{4\times \dfrac{13}{8}+3}{2}

   =\ \dfrac{19}{4}

z\ =\ \dfrac{5k-16}{3}

  =\ \dfrac{5\times \dfrac{13}{8}-16}{3}

   =\ \dfrac{-21}{8}

Now, the distance of point from the plane is given by,

d\ =\ \sqrt{(x-x_1)^2+(y-y_1)^2+(z-z_1)^2}

   =\ \sqrt{(-2-\dfrac{7}{4})^2+(3-\dfrac{19}{4})^2+(-4+\dfrac{21}{8})^2}

   =\ \sqrt{(\dfrac{-15}{4})^2+(\dfrac{-7}{4})^2+(\dfrac{9}{8})^2}

   =\ \sqrt{\dfrac{225}{16}+\dfrac{49}{16}+\dfrac{81}{64}}

   =\ \sqrt{\dfrac{1177}{64}}

   =\ 4.28

So, the distance of the point from its image can be given by,

D = 2d = 2 x 4.28

            = 8.56 unit

So, the distance of a point from it's image is 8.56 units.

4 0
3 years ago
3x - 4/5 = 7/12 help me with this question
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Answer:

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5 0
3 years ago
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Janelle ate 82% of the pie. What fraction of the pie remained? help meee
pav-90 [236]

Answer:

There is 18% left of the pie.

Step-by-step explanation:

There is only 100% of a pie. If she at 82% then there is 18% left. 100-82=18

4 0
3 years ago
Read 2 more answers
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