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3 years ago

Help!! Picking Brainliest! I need an answer fast...

Mathematics

1 answer:

IRISSAK [1]3 years ago

6 0

3 * 5 = 15 miles during weekdays

6.5*2 = 13 miles on weekends

13 +15 = 28 miles total

28 / 7 = 4 miles average per day

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X^2-4x+3 factorize into its factors and give me answer I will giv u anda burger

Ghella [55]

Answer:

(x - 3)(x - 1)

Step-by-step explanation:

Consider the factors of the constant term (+ 3) which sum to give the coefficient of the x- term.

The factors are - 3 and - 1, since

- 3 × - 1 = + 3 and - 3 - 1 = - 4, hence

x² - 4x + 3 = (x - 3)(x - 1) ← in factored form

6 0

3 years ago

Read 2 more answers

What is the area of the shaded region?

ella [17]

33cm

Because add all of them then subtract then do the parenthasees and you get 33

3 0

3 years ago

Stacy uses a spinner with six equal sections numbered 2, 2, 3, 4, 5, and 6 to play a game. Stacy spins the pointer 120 times and

Alex17521 [72]

Answer:

The solution is given below;

Step-by-step explanation:

The computation is shown below:

There is six equal sections are numbered i.e. 2, 2, 3, 4, 5, and 6.

And, Number of Outcome of 2 on the Spinner is 2

Also, total Number of Sections is 6

As the winning spin is a 2

So, The theoretical probability of winning

= 2 ÷6 × 100

= 33.33%

Now

There is a total no of trials be 120

And, the number of outcome of 2 be 30

So, the experimental probability is 30 ÷ 120 = 1 ÷ 4 × 100 = 25%

As 33.3%>25%, this represent that the theoretical probability of winning would be greater.

6 0

3 years ago

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Pls help: Write 2^100 as a power with the following base. For example: 4^50<br> 1)16 2)32 3)1024

geniusboy [140]

$2^{100} = 16^{25} =32^{20} =1024^{10}$

P/s: $2^{100} = 2^{4.25} =16^{25}$ ......

ok done. Thank to me :>

7 0

2 years ago

Let G be a connnectde graph with n vertices and m edges. supposed also that m = n. prove that G contains exactly one cycle

ElenaW [278]

Answer:

Contradiction

Step-by-step explanation:

Suppose that G has more than one cycle and let C be one of the cycles of G, if we remove one of the edges of C from G, then by our supposition the new graph G' would have a cycle. However, the number of edges of G' is equal to m-1=n-1 and G' has the same vertices of G, which means that n is the number of vertices of G. Therefore, the number of edges of G' is equal to the number of vertices of G' minus 1, which tells us that G' is a tree (it has no cycles), and so we get a contradiction.

7 0

3 years ago