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guajiro [1.7K]
2 years ago
11

Simplify 14a over 15 Plus 2a + 36 over 5​

Mathematics
1 answer:
Yuliya22 [10]2 years ago
5 0

Answer:

(20a+108)/15

Step-by-step explanation:

First find LCM of the denominators. U will find it to be 15

Then calculate the number like this.

[(14a)×1/15×1]+[(2a+36)×3/5]

[(14a/15)]+[(6a+108)/15

(14a+6a+108)/15

(20a+108)/15

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jek_recluse [69]

18 km apart

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3 years ago
Pe lingNeed 2 yards of fabric for a craft project she has 3 feet already how many inches of fabric does she need still need
icang [17]

Answer:

36 inches

Step-by-step explanation:

12 inches equals 1 foot

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3 0
3 years ago
The solution to an inequality is graphed on the number line. What is another way to represent this solution set? O {x | X 4.5) O
lys-0071 [83]

Answer: Choice B  \{x | \ x \le 4.5\}

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

Explanation:

The endpoint is at 4.5 and this endpoint is a filled in circle. So we'll have "or equal to" as part of the inequality sign. This is because we are including the endpoint as part of the shaded solution region.

The other part of the inequality sign is "less than" because the shading is to the left of the endpoint. Any point in the shaded region is less than 4.5, or it could be equal to 4.5

Put another way: x is either 4.5 or smaller

We write that as x \le 4.5 which is read out as "x is less than or equal to 4.5"

Surrounding this in curly braces tells the reader we're dealing with a set of values. The first part "x |" means "x such that"

All together we end up with the answer \{x| \ x \le 4.5\} which translates to "x such that x is less than or equal to 4.5"

3 0
3 years ago
A rectangular box has a square base with an edge length of x cm and a height of h cm. The volume of the box is given by V = x^2h
tiny-mole [99]
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</span>Given : Volume of the rectangular box  = x²h 
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dV/dt = 192 
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3 0
2 years ago
The product of two is 450.the first number is half the second number
Bas_tet [7]

Answer:

I don't know you're options but the answer could be 30.

7 0
3 years ago
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