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1 year ago

PLEASE HELP... Students are making necklaces of six beads for their mothers. There are

Mathematics

1 answer:

nasty-shy [4]1 year ago

3 0

Answer:

Step-by-step explanation:

#1 Number of ways picking 2 from 6 = 6C2 = 15

#2 Arranging them in a circle (consider 1 pair AB)

can be done

1,5 (ABBBBB) or (5,1)

2,4 ( AABBBB) or (4,2)

3,3

1,1 (ABABAB)

all up = 7 ways

#3 = product of #1 and #2 = 15x7 = 105

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Ann took a taxi home from the airport. The taxi fare was

Zinaida [17]

Answer:

She would have traveled 21 miles, because it you multiply 2.1 by 21 you get 44.1 and then Ann gave a $5 tip so that would bring it up to $49.10.

Step-by-step explanation:

6 0

3 years ago

X - y = 6

Dominik [7]

the system has no solution.

Option C is correct.

Step-by-step explanation:

We need to solve the system of equations by substitution

$x - y = 6\,\,\,eq(1)\\x = y + 2\,\,\,eq(2)$

Putting value of x from eq(2) into eq(1)

$(y+2)-y=6\\y+2-y=6\\0+2=6\\0=6-2\\0\neq4$

As as 0≠4is not true, we cannot find the value of y so the system has no solution.

Option C is correct.

Keywords: System of equations

Learn more about system of equations at:

#learnwithBrainly

4 0

3 years ago

The first decision is a choice between

pantera1 [17]

Answer:

min-mid-max-mid-min

-cosine

Step-by-step explanation:

This is the correct answer, further proof in the file attached.

4 0

3 years ago

Read 2 more answers

If ABCD is a parallelogram then what is the measure of < A - <C

iragen [17]

Answer:

The answer is 0.

4 0

3 years ago

Does there exist a di↵erentiable function g : [0, 1] R such that g'(x) = f(x) for all x 2 [0, 1]? Justify your answer

agasfer [191]

Answer:

No; Because g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Step-by-step explanation:

Assuming: the function is $f(x)=x^{2}$ in [0,1]

And rewriting it for the sake of clarity:

Does there exist a differentiable function g : [0, 1] →R such that g'(x) = f(x) for all g(x)=x² ∈ [0, 1]? Justify your answer

1) A function is considered to be differentiable if, and only if both derivatives (right and left ones) do exist and have the same value. In this case, for the Domain [0,1]:

$g'(0)=g'(1)$

2) Examining it, the Domain for this set is smaller than the Real Set, since it is [0,1]

The limit to the left

$g(x)=x^{2}\\g'(x)=2x\\ g'(0)=2(0) \Rightarrow g'(0)=0$

$g(x)=x^{2}\\g'(x)=2x\\ g'(1)=2(1) \Rightarrow g'(1)=2$

g'(x)=f(x) then g'(0)=f(0) and g'(1)=f(1)

3) Since g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Because this is the same as to calculate the limit from the left and right side, of g(x).

$f'(c)=\lim_{x\rightarrow c}\left [\frac{f(b)-f(a)}{b-a} \right ]\\\\g'(0)=\lim_{x\rightarrow 0}\left [\frac{g(b)-g(a)}{b-a} \right ]\\\\g'(1)=\lim_{x\rightarrow 1}\left [\frac{g(b)-g(a)}{b-a} \right ]$

This is what the Bilateral Theorem says:

$\lim_{x\rightarrow c^{-}}f(x)=L\Leftrightarrow \lim_{x\rightarrow c^{+}}f(x)=L\:and\:\lim_{x\rightarrow c^{-}}f(x)=L$

4 0

3 years ago