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2 years ago

The school cafeteria offers three choices for the main course (peanut

Mathematics

1 answer:

Elena-2011 [213]2 years ago

6 0

Answer: 18 different combinations

Step-by-step explanation: When finding out combinations, you just multiply all the numbers. In this case, it would be 3x2x3 which equals 18. Please mark brainliest <3

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Let a is in {−3, −1, 0}. Evaluate −a for each element of the set.

12345 [234]

Ok so a∈{-3,-1,0}. then just {-3*-1,-1*-1,0*-1} which comes to {3,1,0}

7 0

3 years ago

A patrolman spends 25% every day completing paperwork. The patrolman’s shift each day is 8 hours. How much of his time does he s

Mrrafil [7]

$8 \times 25\% \\ = 8 \times \dfrac{25}{100} \\ = 2$

Answer : He spends 2 hours doing paperwork each day.

Hope this helps. - M

4 0

3 years ago

Find a point on the ellipsoid x2+y2+4z2=36x2+y2+4z2=36 where the tangent plane is perpendicular to the line with parametric equa

Anvisha [2.4K]

Answer:

A point on the ellipsoid is (-4,2,2) or (4,-2,-2)

Step-by-step explanation:

Given equation of ellipsoid f(x,y,z) : $x^2+y^2+4z^2=36$

Parametric equations:

x=-4t-1

y=2t+1

z=8t+3

Finding the gradient of function

$\nabla f(x,y,z)=\\\nabla f(x,y,z)=$

So, The directions vectors=(-4,2,8)

Now the line is perpendicular to plane when direction vector is parallel to the normal vector of line

$\nablaf(x,y,z)=(2x,2y,8z)=\lambda(-4,2,8)$

So, $2x=-4\lambda$

$\Rightarrow x=-2\lambda$

$2y=2\lambda\\\Rightarrow y=\lambda\\8z=8\lambda\\\Rightarrow z=\lambda$

Substitute the value of x , y and z in the ellipsoid equation

$(2\lambda)^2+(\lambda)^2+4(\lambda)^2=36\\9(\lambda)^2=36\\\lambda^2=4\\\lambda=\pm 2$

With $\lambda = 2$

x=-2(2)=-4

y=2

z=2

With $\lambda =- 2$

x=-2(-2)=4

y=-2

z=-2

Hence a point on the ellipsoid is (-4,2,2) or (4,-2,-2)

7 0

3 years ago

Pls helpppp ASAP I’m so confused

Sidana [21]

Answer:

2268 in3

Step-by-step explanation:

7 0

2 years ago

Please solve this sheet.

lyudmila [28]

Hello,

A: roots: -1,-3
a point (-2,1)
Vertex=((-2,1)

y=k*(x+1)(x+3) using roots
but k*(-2+1)(-2+3)=1==>k*(-1)*1=1==>k=-1

eq: y=-(x+1)(x+3)
==>y=-(x²+3x+x+3)
==>y=-x²-4x-3
y=k(x+2)²+1 if x=-1,y=0 ==>k*1+1=0==>k=-1
==>y=-(x+2)²+1

Answer :A--> R,K

B)
y=k(x+4)²-2 and k=-1/2
y=-1/2(x+4)²-2
y=-1/2x²-4x-10

answer B--> I,≈W if it is written -1/2*x² (square has been forgotten)

C:
y=2x²-16x+30
y=2(x-4)²-2
answer : C-->S,J

D:
y=-(x+3)(x+1)
y=-x²-4x-3
=-(x+2)²+1
answer D--> V,L

E:
Here there is a problem: or the graph is wrong, or 2 equations are missing!

y=1(x+1)(x-3) using roots
y=x²-2x-3 ≈ T si it were -2x and not +2x.

y=(x-1)²-4 ≈H is it were -1 in place of +1 [H:y=(x+1)²-4]

6 0

3 years ago