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alekssr [168]
3 years ago
11

I) If v = u + at find u when u=5, a=3 and t=4​

Mathematics
1 answer:
dusya [7]3 years ago
4 0

Step-by-step explanation:

5+ (3×4)=v

5+ (12)=v

v=17

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Robert wants to use all the ingredients listed in the table at the right to make trail mix. How many ½ -lb packages can he make?
lara31 [8.8K]

Answer:

16 half packages.

Step-by-step explanation:

You basically need to add them all of the weight of each ingredients together, and then you multiply by 2.

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3 years ago
I need HELPPPPPP PLEASEEEEEE!!!!!!!!~!!!!!!
Sedaia [141]

Answer:

Starting from the left.

4 × 3

5 × 5

2 × 5

4 × 5

4 × 4

1.5 × 2

Step-by-step explanation:

Divide all dimensions by 8.

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3 years ago
72.4 divided by 4 please help!!!!
wariber [46]

Answer:

18.1

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8 0
3 years ago
Read 2 more answers
Which of the following CANNOT be given the lengths of the sides of a triangle?
Katena32 [7]

The sides of a triangle must satisfy the triangle inequality, which states the sum of the lengths of any two sides is strictly greater than the length of the remaining side.  

We really only have to check if the sum of the two smaller sides exceeds the largest side.

A. 5+6>7, ok

B. 6+6>10, ok

C. 7+7=14 Not ok, this is a degenerate triangle not a real triangle

D. 4+6>8 ok

Answer: C

6 0
3 years ago
Read 2 more answers
What is the effect in the time required to solve a prob- lem when you double the size of the input from n to 2n, assuming that t
Inga [223]

Answer:

f(2n)-f(n)=log2

b.lg(lg2+lgn)-lglgn

c. f(2n)/f(n)=2

d.2nlg2+nlgn

e.f(2n)/(n)=4

f.f(2n)/f(n)=8

g. f(2n)/f(n)=2

Step-by-step explanation:

What is the effect in the time required to solve a prob- lem when you double the size of the input from n to 2n, assuming that the number of milliseconds the algorithm uses to solve the problem with input size n is each of these function? [Express your answer in the simplest form pos- sible, either as a ratio or a difference. Your answer may be a function of n or a constant.]

from a

f(n)=logn

f(2n)=lg(2n)

f(2n)-f(n)=log2n-logn

lo(2*n)=lg2+lgn-lgn

f(2n)-f(n)=lg2+lgn-lgn

f(2n)-f(n)=log2

2.f(n)=lglgn

F(2n)=lglg2n

f(2n)-f(n)=lglg2n-lglgn

lg2n=lg2+lgn

lg(lg2+lgn)-lglgn

3.f(n)=100n

f(2n)=100(2n)

f(2n)/f(n)=200n/100n

f(2n)/f(n)=2

the time will double

4.f(n)=nlgn

f(2n)=2nlg2n

f(2n)-f(n)=2nlg2n-nlgn

f(2n)-f(n)=2n(lg2+lgn)-nlgn

2nLg2+2nlgn-nlgn

2nlg2+nlgn

5.we shall look for the ratio

f(n)=n^2

f(2n)=2n^2

f(2n)/(n)=2n^2/n^2

f(2n)/(n)=4n^2/n^2

f(2n)/(n)=4

the time will be times 4 the initial tiote tat ratio are used because it will be easier to calculate and compare

6.n^3

f(n)=n^3

f(2n)=(2n)^3

f(2n)/f(n)=(2n)^3/n^3

f(2n)/f(n)=8

the ratio will be times 8 the initial

7.2n

f(n)=2n

f(2n)=2(2n)

f(2n)/f(n)=2(2n)/2n

f(2n)/f(n)=2

5 0
3 years ago
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