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Ksivusya [100]
3 years ago
13

6/15 divided by 1/7

Mathematics
2 answers:
mylen [45]3 years ago
6 0
Question-
6/15 divided by 1/7

Answer-
Find the reciprocal of the divisor
Reciprocal of 1/7: 7/1Now, multiply it with the dividend
So, 6/15 ÷ 1/7 = 6/15 × 7/1= (6 × 7)/(15 × 1) = 42/15After reducing the fraction, the answer is 14/5In mixed form: 2 4/5
Anna71 [15]3 years ago
3 0
The answer is 2/35. 6x1=6 and 15x7=105. Then you have to divide both of them by 3. 
You final answer will be 2/35.
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One dollar and eighty cents for wayside hotel
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4 0
3 years ago
Will Mark Brainlest help please​
spayn [35]

Answer:

g(-1 )=-1 and g(2)+g(1)=7

Step-by-step explanation:

If g(x) = x^3+x^2-x-2 find g(-1)

if we find g(-1)

we substitute all the x's in the function with -1

-1^3+-1^2-(-1)-2

-1^3 = -1

-1^2 = 1

-1+1+1-2

(two minuses make a plus)

-1+1 = 0

0+1 = 1

1-2 = -1

if x=-1, g(-1) is -1

g(2)+g(1)

substitute the x's in the function with 2 and 1 and add your results

2^3+2^2-2-2

2^3 = 8, 2^2 = 4

8+4-2-2

8+4= 12, 12-2 = 10, 10-2 = 8

g(2)=8

g(1) now

1^3 + 1^2-1-2

1^3=1, 1^2 = 1

1+1-1-2

1+1 = 2, 2-1 = 1, 1-2 = -1

g(3) = -1

g(2) (which equals 8) + g(3) (which equals -1) =

8+(-1) = 7

g(2)+g(3)=7

5 0
3 years ago
Let a, b, c and x elements in the group G. In each of the following solve for x in terms of a, b, and c.
alina1380 [7]

Answer:

The answer is x=a^{-1}cb^{-1}.

Step-by-step explanation:

First, it is important to recall that the group law is not commutative in general, so we cannot assume it here. In order to solve the exercise we need to remember the axioms of group, specially the existence of the inverse element, i.e., for each element g\in G there exist another element, denoted by g^{-1} such that gg^{-1}=e, where e stands for the identity element of G.

So, given the equality axb=c we make a left multiplication by a^{-1} and we obtain:

a^{-1}axb =a^{-1}c.

But, a^{-1}axb = exb = xb. Hence, xb = a^{-1}c.

Now, in the equality xb = a^{-1}c we make a right multiplication by b^{-1}, and we obtain

xbb^{-1} = a^{-1}cb^{-1}.

Recall that bb^{-1}=e and xe=x. Therefore,

x=a^{-1}cb^{-1}.

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Answer: i think 23x

Step-by-step explanation: i might be incorrect

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