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il63 [147K]
3 years ago
6

Find a set of scalar parametric equations for the line formed by the two intersecting planes. 3x+3y+2z+2=0 2x−3y+2z−2=0

Mathematics
1 answer:
vivado [14]3 years ago
8 0

The normal vectors to the two planes are (3, 3, 2) and (2, -3, 2). The cross product of these will be the direction vector of the line of intersection, (12, -2, -15).

Using x=0, we can find a point on this line by solving the simultaneous equations that remain:

... 3y +2z = -2

... -3y +2z = 2

Adding these, we get

... 4z = 0

... z = 0

so the point we're looking for is (x, y, z) = (0, -2/3, 0). This gives rise to the parametric equations ...

  • x = 12t
  • y = -2/3 -2t
  • z = -15t

By letting t=2/3, we can find a point on the line that has integer coefficients. That will be (x, y, z) = (8, -2, -10).

Then our parametric equations can be written as

  • x = 8 +12t
  • y = -2 -2t
  • z = -10 -15t
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After a baby was born, he began to gain weight at a rate of 1.5 pounds per month. The
myrzilka [38]

Answer:

W(t) = 10 + 1.5*t

Step-by-step explanation:

Given:

The weight gain of the baby is 1.5 pounds per month.

After 4 months, baby's weight = 16 pounds

Let us the say the weight of the baby when it was born be x.

Then:

x + 1.5*(4) = 16

x = 10

Then weight of the baby as a function of number of months(t) will be=

initial weight + incremental weight per month*no of months(t).

W(t) = 10 + 1.5*t

5 0
2 years ago
I don't know how to do this it is hard
Solnce55 [7]
(-4) means that it would be +4
3 0
3 years ago
A force of 12 lb is required to hold a spring stretched 8 in. beyond its natural length. How much work W is done in stretching i
sattari [20]

Answer:

Work done in stretching the spring = 7.56 lb-ft.

Step-by-step explanation:

Normal length of the spring = 8 in or \frac{8}{12} ft

= \frac{2}{3} ft

If the spring has been stretched to 11 inch then the stretched length of the spring is = 11 in

= \frac{11}{12} ft

Force applied to stretch the spring = 12 lb

By Hook's law,

F = kx [where k is the spring constant and x = length by which the spring is stretched]

12 = k(\frac{2}{3})

k = 18

Work done (W) to stretch the spring by \frac{11}{12} ft will be

W = \int\limits^\frac{11}{12} _0 {kx} \, dx

    = \int\limits^\frac{11}{12} _0 {(18x)} \, dx

    = 18[\frac{x^{2}}{2}]^{\frac{11}{12}}_0

    = 9(\frac{11}{12})²

    = 7.56 lb-ft

6 0
3 years ago
- 7h + 2(-4h +5) > -4h +1 +10. Solve for h​
navik [9.2K]

Hello.

-7h + 2(-4h + 5) > -4h + 1 + 10 ; solve for h

Our first step is to get rid of all the parenthesis. We can do this by simplifying everything out.

-7h - 8h + 10 > -4h + 11

Add like terms.

-15h + 10 > -4h + 11

Now, we must isolate our variables. We do this by moving all variables to one side and moving our other numbers to the other.

Subtract 10 from both sides. Then, add 4h to both sides.

-15h + 4h > 11 - 10

Simplify.

-h > 1

Divide both sides by -1.

However, keep in mind that when dividing with a negative, you must change the sign. Therefore, > turns into <

-h ÷ -1 < 1 ÷ -1

h < -1

6 0
3 years ago
A pair of dice is rolled. Find the probability for P(not 2 or not 12)
irakobra [83]

Answer:

There’s only one way to get 12 when 2 dice are tossed, both have to equal 6. There are 6 ways tossing a single die can come out (1,2,3,4,5,6), so if you toss dice, the second die could have any one of six values with each of the numbers that could result from the first toss (e.g., 1 from die 1 and 1,2,3,4,5, or 6 from die 2). So, considering there are 6 ways to fill each of two slots, there are 6 x 6 = 36 possible outcomes of tossing two dice. Only one of them equals 12, so p(12 given 2 dice tossed) = 1/36 = 0.02777777777778.

3 0
3 years ago
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