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3 years ago

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6 people sharing 40 lbs of gold evenly. How many do they get? Then subtract 1 2/3 from the result of the problem above & sim

plify completely. Please help cause it's not accepting what I have

2 answers:

r-ruslan [8.4K]3 years ago

7 0

Answer: 6

Step-by-step explanation:

Given Data:

Number of people = 6

Amount of Gold to be shared = 40lbs

Subtract 2/3 from your answer.

Therefore

If six people are to share 40lbs of gold evenly

= 40lbs / 6

Divide through by 2

= 20lbs / 3

= 20lbs / 3 - 2/3

= 18/3

= 6

Each person would get approximately 6lbs of gold

umka21 [38]3 years ago

6 0

Answer:

1. Each individual gets $6\frac{2}{3}lb$ of gold

2. After subtracting by $1\frac{2}{3}$ , the result is 5

Step-by-step explanation:

Given

Size of Gold = 40 lbs

Number of people = 6

Required

1. Proportion of each individual

2. Subtract $1\frac{2}{3}$ from each proportion

1. Proportion of each individual

To calculate how many each individuals get (in other words, the proportion of each individual), we simply divide the total size of gold by the number of people

Mathematically,

$Proportion = \frac{Total}{People}$

$Proportion = \frac{40}{6}$

Reduce fraction to lowest term by dividing numerator and denominator by 2

$Proportion = \frac{20}{3}$

Convert to mixed fraction

$Proportion = 6\frac{2}{3}$

Hence each individual gets $6\frac{2}{3}lb$ of gold

2. Subtract $1\frac{2}{3}$ from each proportion

This is represented by

$6\frac{2}{3} - 1\frac{2}{3}$

Convert both fractions to improper fractions

= $\frac{20}{3} - \frac{5}{3}$

Find LCM of both fractions

= $\frac{20 - 5}{3}$

Subtract

= $\frac{15}{3}$

= 5

Hence, after subtracting by $1\frac{2}{3}$ , the result is 5

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Daniel [21]

Answer:

Step-by-step explanation:

As the statement is ‘‘if and only if’’ we need to prove two implications

$f : X \rightarrow Y$ is surjective implies there exists a function $h : Y \rightarrow X$ such that $f\circ h = 1_Y$ .
If there exists a function $h : Y \rightarrow X$ such that $f\circ h = 1_Y$ , then $f : X \rightarrow Y$ is surjective

Let us start by the first implication.

Our hypothesis is that the function $f : X \rightarrow Y$ is surjective. From this we know that for every $y\in Y$ there exist, at least, one $x\in X$ such that $y=f(x)$ .

Now, define the sets $X_y = \{x\in X: y=f(x)\}$ . Notice that the set $X_y$ is the pre-image of the element $y$ . Also, from the fact that $f$ is a function we deduce that $X_{y_1}\cap X_{y_2}=\emptyset$ , and because $f$ the sets $X_y$ are no empty.

From each set $X_y$ choose only one element $x_y$ , and notice that $f(x_y)=y$ .

So, we can define the function $h:Y\rightarrow X$ as $h(y)=x_y$ . It is no difficult to conclude that $f\circ h(y) = f(x_y)=y$ . With this we have that $f\circ h=1_Y$ , and the prove is complete.

Now, let us prove the second implication.

We have that there exists a function $h:Y\rightarrow X$ such that $f\circ h=1_Y$ .

Take an element $y\in Y$ , then $f\circ h(y)=y$ . Now, write $x=h(y)$ and notice that $x\in X$ . Also, with this we have that $f(x)=y$ .

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3 years ago

Algebra 2 question need help.

MrMuchimi

Answer: Choice C

h(x) = -x^4 + 2x^3 + 3x^2 + 4x + 5

===========================================

Explanation:

When reflecting the function f(x) over the y axis, we replace every x with -x and simplify like so

f(x) = -x^4 - 2x^3 + 3x^2 - 4x + 5

f(-x) = -(-x)^4 - 2(-x)^3 + 3(-x)^2 - 4(-x) + 5

f(-x) = -x^4 + 2x^3 + 3x^2 + 4x + 5

h(x) = -x^4 + 2x^3 + 3x^2 + 4x + 5

Note the sign changes that occur for the terms that have odd exponents (the terms -2x^3 and -4x become +2x^3 and +4x); while the even exponent terms keep the same sign.

The reason why we replace every x with -x is because of the examples mentioned below

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Examples:

The point (1,2) moves to (-1,2) after a y axis reflection

Similarly, (-5,7) moves to (5,7) after a y axis reflection.

As you can see, the y coordinate stays the same but the x coordinate flips in sign from negative to positive or vice versa. This is the direct reason for the replacement of every x with -x.

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Read 2 more answers

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Shtirlitz [24]

Answer:

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