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makvit [3.9K]
3 years ago
13

1. Use the Pythagorean Theorem to find then find the perimeter and area of each

Mathematics
1 answer:
Reil [10]3 years ago
6 0

haf ki liksa sa na

hs k

hsvst

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find the component equation of the plane which is normal to the vector -2i+5j+k and which contains the point (-10;7;5).​
maksim [4K]

Given:

A plane is normal to the vector = -2i+5j+k

It contains the point (-10,7,5).​

To find:

The component equation of the plane.

Solution:

The equation of plane is

a(x-x_0)+b(y-y_0)+c(z-z_0)=0

Where, (x_0,y_0,z_0) is the point on the plane and \left< a,b,c\right> is normal vector.

Normal vector is -2i+5j+k and plane passes through (-10,7,5). So, the equation of the plane is

-2(x-(-10))+5(y-7)+1(z-5)=0

-2(x+10)+5y-35+z-5=0

-2x-20+5y-35+z-5=0

-2x+5y+z-60=0

-2x+5y+z=60

Therefore, the equation of the plane is -2x+5y+z=60.

4 0
3 years ago
Find all the factors of 45
scoundrel [369]

Answer:

So, the factors of 45 are 1, 3, 5, 9, 15 and 45.

Common Factors.

Prime Numbers.

Highest Common Factor.

Step-by-step explanation:

45 is a composite number. 45 = 1 x 45, 3 x 15, or 5 x 9. Factors of 45: 1, 3, 5, 9, 15, 45. Prime factorization: 45 = 3 x 3 x 5, which can also be written 45 = 3² x 5.

8 0
3 years ago
The plane x+y+2z=8 intersects the paraboloid z=x2+y2 in an ellipse. Find the points on this ellipse that are nearest to and fart
DiKsa [7]

Answer:

The minimum distance of   √((195-19√33)/8)  occurs at  ((-1+√33)/4; (-1+√33)/4; (17-√33)/4)  and the maximum distance of  √((195+19√33)/8)  occurs at (-(1+√33)/4; - (1+√33)/4; (17+√33)/4)

Step-by-step explanation:

Here, the two constraints are

g (x, y, z) = x + y + 2z − 8  

and  

h (x, y, z) = x ² + y² − z.

Any critical  point that we find during the Lagrange multiplier process will satisfy both of these constraints, so we  actually don’t need to find an explicit equation for the ellipse that is their intersection.

Suppose that (x, y, z) is any point that satisfies both of the constraints (and hence is on the ellipse.)

Then the distance from (x, y, z) to the origin is given by

√((x − 0)² + (y − 0)² + (z − 0)² ).

This expression (and its partial derivatives) would be cumbersome to work with, so we will find the the extrema  of the square of the distance. Thus, our objective function is

f(x, y, z) = x ² + y ² + z ²

and

∇f = (2x, 2y, 2z )

λ∇g = (λ, λ, 2λ)

µ∇h = (2µx, 2µy, −µ)

Thus the system we need to solve for (x, y, z) is

                           2x = λ + 2µx                         (1)

                           2y = λ + 2µy                       (2)

                           2z = 2λ − µ                          (3)

                           x + y + 2z = 8                      (4)

                           x ² + y ² − z = 0                     (5)

Subtracting (2) from (1) and factoring gives

                     2 (x − y) = 2µ (x − y)

so µ = 1  whenever x ≠ y. Substituting µ = 1 into (1) gives us λ = 0 and substituting µ = 1 and λ = 0  into (3) gives us  2z = −1  and thus z = − 1 /2 . Subtituting z = − 1 /2  into (4) and (5) gives us

                            x + y − 9 = 0

                         x ² + y ² +  1 /2  = 0

however, x ² + y ² +  1 /2  = 0  has no solution. Thus we must have x = y.

Since we now know x = y, (4) and (5) become

2x + 2z = 8

2x  ² − z = 0

so

z = 4 − x

z = 2x²

Combining these together gives us  2x²  = 4 − x , so

2x²  + x − 4 = 0 which has solutions

x =  (-1+√33)/4

and

x = -(1+√33)/4.

Further substitution yeilds the critical points  

((-1+√33)/4; (-1+√33)/4; (17-√33)/4)   and

(-(1+√33)/4; - (1+√33)/4; (17+√33)/4).

Substituting these into our  objective function gives us

f((-1+√33)/4; (-1+√33)/4; (17-√33)/4) = (195-19√33)/8

f(-(1+√33)/4; - (1+√33)/4; (17+√33)/4) = (195+19√33)/8

Thus minimum distance of   √((195-19√33)/8)  occurs at  ((-1+√33)/4; (-1+√33)/4; (17-√33)/4)  and the maximum distance of  √((195+19√33)/8)  occurs at (-(1+√33)/4; - (1+√33)/4; (17+√33)/4)

4 0
3 years ago
The number of students in the period 7 study hall at jin's school is 4 times the number of students in jin's home room. how many
k0ka [10]
1/4 the students in study hall (per 7) = 16

16*4 = 64 students in study hall.

<span>Hope I helped! (Pick my answer as brainliest!)</span>
5 0
4 years ago
Optimization problem: what are the dimensions of the lightest open-top right circularcylindrical can that will hold a volume of
vivado [14]
So Volume of cylinder is pi*r^2*h = 1,000 

Then lightest one means you have the smallest surface area. Which is one base and then the area of the surface. SA = pi*r^2 + 2pi*r*h 

So now you have 2 equations, so: 

h = 1,000/(pi*r^2) 
So then SA = pi*r^2 + 2pi*r*(1,000/(pi*r^2) = pi*r^2 + 2,000/r 

Derivative of SA is then 2pi*r -2,000/r^2. Set to 0 

2pi*r-2,000/r^2 =0 --> 2pi*r^3 = 2,000 --> r^3 = 1,000/pi --> r = 10/pi^(1/3) 

Now go back to the volume function: pi*r^2*h =1,000 --> 1,000/(pi*100/pi^(2/3)) = h 
<span>h = 10 / pi^(1/3)</span>
3 0
3 years ago
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