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77julia77 [94]
3 years ago
11

Classify each number below as a rational number or an irrational number.

Mathematics
1 answer:
ollegr [7]3 years ago
3 0

Answer: I think that the first one is irrational, the second one is rational, then the third one should be rational, the fourth one is irrational, and the fifth one is rational

Step-by-step explanation:

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Solve for x 4.72 = 3.25 + x
Anna [14]
Flip the situation:
X + 3.25 = 4.72

Subtract 3.25 from both sides
X + 3.25 - 3.25 = 4.72 - 3.25

X = 1.47
3 0
3 years ago
Ryker wants a new car. The dealership offers him a loan at 9% annual interest for 6 years. If Ryker wants to do some calculation
Contact [7]

The monthly interest rate is 0.75%

<u>Step-by-step explanation:</u>

The rate of interest= 9%

Total time = 6years

Interest rate per month = 9/12

= 0.75%

Monthly payment =cost of the car + 0.75% of cost of car

Interest per month is 0.75%

3 0
3 years ago
Help with this question ??
NNADVOKAT [17]

Answer: 12 tables minimum, 15 max.

Step-by-step explanation:

If you substitute 14 in an inequality

200c + 500t >= 8800

200(14) + 500t and solve for t, you get t must be at least 12.

C +T cannot exceed 29

So figure 14 +12 =26 so they could sell up to 15 tables and not go over 29.

You will have to enter the t values 12,13,14,15

5 0
3 years ago
Solve. 7/9 = b/b - 10
kifflom [539]

Answer:

No Solution

Step-by-step explanation:

There are no values of  b

that make the equation true.

No solution

8 0
3 years ago
What two rational expressions sum to 3x+4/x^2-6x+5?
Lubov Fominskaja [6]

Answer:

\frac{3x+4}{x^2-6x+5}=\frac{-7}{4(x-1)} +\frac{19}{4(x-5)}

Step-by-step explanation:

The given rational expression is:

\frac{3x+4}{x^{2}-6x+5} = \frac{3x+4}{(x-1)(x-5)}

We can use concept of Partial Fractions to solve this problem. Let,

\frac{3x+4}{(x-1)(x-5)}=\frac{A}{x-1} +\frac{B}{x-5}

Multiplying both sides by (x - 1)(x - 5), we get:

3x+4=A(x-5)+B(x-1)

Substituting x = 5, we get:

3(5)+4=A(5-5)+B(5-1)\\\\ 15+4=0+4B\\\\ 19=4B\\\\ B=\frac{19}{4}

Substituting x = 1, we get:

3(1)+4=A(1-5)+B(1-1)\\\\ 7=-4A\\\\ A=-\frac{7}{4}

Substituting the value of A and B, back in the original equation, we get:

\frac{3x+4}{x^2-6x+5}=\frac{-7}{4(x-1)} +\frac{19}{4(x-5)}

4 0
3 years ago
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