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Fiesta28 [93]
1 year ago
15

5/8=n/n-9Solve each equation

Mathematics
1 answer:
QveST [7]1 year ago
8 0
\begin{gathered} \frac{5}{8}=\frac{n}{n-9}(\text{cross multiplying it)} \\ 5(n-9)=8n \\ 5n-45=8n \\ 5n-8n=45 \\ -3n=45 \\ n=-15 \end{gathered}

So, the answer to the equation is n=-15

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Answer:

it is 7x-4

Step-by-step explanation:

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3 years ago
draw a graph representing a real-world linear function and write an equation for the graph. describe what the graph represents.​
nikklg [1K]

Answer:

slope intersept form is y=mx+b

Step-by-step explanation:

We have 4 ways of solving one-step equations: Adding, Substracting, multiplication and division. If we add the same number to both sides of an equation, both sides will remain equal. If we subtract the same number from both sides of an equation, both sides will remain equal.

 If you pay 30 dollars a month for your cell phone and 10 cents per minute of usage the monthly cost of using your cell phone would be a linear equation of a function, C, the monthly cost based on the number of minutes you use monthly.

7 0
3 years ago
Here’s a graph of a linear function. write the equation that describes that function
sergeinik [125]

Answer:

y=1/2x+4

Step-by-step explanation:

Slope is 1/2. (m)

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Slope intercept form is Y=mx+b.

8 0
4 years ago
For each part, give a relation that satisfies the condition. a. Reflexive and symmetric but not transitive b. Reflexive and tran
Vesnalui [34]

Answer:

For the set X = {a, b, c}, the following three relations satisfy the required conditions in (a), (b) and (c) respectively.

(a) R = {(a,a), (b,b), (c, c), (a, b), (b, a), (b, c), (c, b)} is reflexive and symmetric but not transitive .

(b) R = {(a, a), (b, b), (c, c), (a, b)} is reflexive and transitive but not symmetric .

(c) R = {(a,a), (a, b), (b, a)} is symmetric and transitive but not reflexive .

Step-by-step explanation:

Before, we go on to check these relations for the desired properties, let us define what it means for a relation to be reflexive, symmetric or transitive.

Given a relation R on a set X,

R is said to be reflexive if for every a \in X, (a,a) \in R.

R is said to be symmetric if for every (a, b) \in R, (b, a) \in R.

R is said to be transitive if (a, b) \in R and (b, c) \in R, then (a, c) \in R.

(a) Let R = {(a,a), (b,b), (c, c), (a, b), (b, a), (b, c), (c, b)}.

Reflexive: (a, a), (b, b), (c, c) \in R

Therefore, R is reflexive.

Symmetric: (a, b) \in R \implies (b, a) \in R

Therefore R is symmetric.

Transitive: (a, b) \in R \ and \ (b, c) \in R but but (a,c) is not in  R.

Therefore, R is not transitive.

Therefore, R is reflexive and symmetric but not transitive .

(b) R = {(a, a), (b, b), (c, c), (a, b)}

Reflexive: (a, a), (b, b) \ and \ (c, c) \in R

Therefore, R is reflexive.

Symmetric: (a, b) \in R \ but \ (b, a) \not \in R

Therefore R is not symmetric.

Transitive: (a, a), (a, b) \in R and (a, b) \in R.

Therefore, R is transitive.

Therefore, R is reflexive and transitive but not symmetric .

(c) R = {(a,a), (a, b), (b, a)}

Reflexive: (a, a) \in R but (b, b) and (c, c) are not in R

R must contain all ordered pairs of the form (x, x) for all x in R to be considered reflexive.

Therefore, R is not reflexive.

Symmetric: (a, b) \in R and (b, a) \in R

Therefore R is symmetric.

Transitive: (a, a), (a, b) \in R and (a, b) \in R.

Therefore, R is transitive.

Therefore, R is symmetric and transitive but not reflexive .

4 0
3 years ago
What will be the lowest common denominator for the fractions 2/3 and 4/5?
dusya [7]
To find the lowest common denominator between the fractions, what you need to do is simply write out the multiples of both 3 and 5 and find what multiple is common between both:

3,6,9,12,15
5,10,15

The multiple that is common and is the lowest in both would be 15.

The solution thus is B.15, it is the LCD for the fractions of 2/3 and 4/5.
4 0
3 years ago
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