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Neporo4naja [7]
3 years ago
5

For lunch, Kile can eat a sandwich with either ham or a bologna and with or without cheese. Kile also has the choice of drinking

water or juice with his sandwich. The total number of lunches Kile can choose is
Mathematics
1 answer:
Scorpion4ik [409]3 years ago
4 0

Answer:

8

Step-by-step explanation:

Ham with or without cheese-2 choices

Bologna with or without cheese-2 choices

Bologna with cheese with water or juice-2 choices

Bologna without cheese with juice or water-2 choices

Ham with cheese with juice or water -2 choices

Ham without cheese with juice or water -2 choices

2+2+2+2=8

Kile has 8 choices for lunch

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A parabola has a vertex at (3, -2). What is the axis of symmetry? x = _____
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x =3

Step-by-step explanation:

The axis of symmetry is at the vertex

The x coordinate gives the axis of symmetry

x =3

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an online book store has 173 boxes of books in its warehouse each box has 48 books how many books are in the warehouse​
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How do you write 9.085 x 10-9 in standard form?
Levart [38]

Answer: 0.000000009085

Step-by-step explanation: This expression is rewritten like this because it is multiplied by a negative power of ten. Therefore, applying this principle to scientific notation, the answer would have to be less than 0. That would be your answer. Hope this helps!

3 0
2 years ago
Simplify these expressions a) 5x +3x + 2y + 4y b) 6x - 2x + 8y - 6y​
cestrela7 [59]

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3 0
3 years ago
Find the vectors T, N, and B at the given point. r(t) = < t^2, 2/3t^3, t >, (1, 2/3 ,1)
maxonik [38]

Answer with Step-by-step explanation:

We are given that

r(t)=< t^2,\frac{2}{3}t^3,t >

We have to find T,N and B at the given point t > (1,2/3,1)

r'(t)=

\mid r'(t) \mid=\sqrt{(2t)^2+(2t^2)^2+1}=\sqrt{(2t^2+1)^2}=2t^2+1

T(t)=\frac{r'(t)}{\mid r'(t)\mid}=\frac{}{2t^2+1}

Now, substitute t=1

T(1)=\frac{}{2+1}=\frac{1}{3}

T'(t)=\frac{-4t}{(2t^2+1)^2} +\frac{1}{2t^2+1}

T'(1)=-\frac{4}{9}+\frac{1}{3}

T'(1)=\frac{1}{9}=

\mid T'(1)\mid=\sqrt{(\frac{-2}{9})^2+(\frac{4}{9})^2+(\frac{-4}{9})^2}=\sqrt{\frac{36}{81}}=\frac{2}{3}

N(1)=\frac{T'(1)}{\mid T'(1)\mid}

N(1)=\frac{}{\frac{2}{3}}=

N(1)=

B(1)=T(1)\times N(1)

B(1)=\begin{vmatrix}i&j&k\\\frac{2}{3}&\frac{2}{3}&\frac{1}{3}\\\frac{-1}{3}&\frac{2}{3}&\frac{-2}{3}\end{vmatrix}

B(1)=i(\frac{-4}{9}-\frac{2}{9})-j(\frac{-4}{9}+\frac{1}{3})+k(\frac{4}{9}+\frac{2}{9})

B(1)=-\frac{2}{3}i+\frac{1}{3}j+\frac{2}{3}k

B(1)=\frac{1}{3}

5 0
3 years ago
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