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Nimfa-mama [501]
3 years ago
7

-

Mathematics
1 answer:
tankabanditka [31]3 years ago
7 0

Answer:

Step-by-step explanation:

3C + 8T = 58 ----- (eq1)

5C + 2T = 23 ------(eq2)

Multiply equation (1) by 5 and equation (2) by 3

15C + 40T = 290 -----(eq3)

15C + 6T = 69 ----------(eq4)

Subtract equation (4) from (3)

34T = 221

T = 221/34

T = $6.5

Therefore 1 Table will cost $6 and 5 cents

Solve for (C) in equation (1)

3C + 8T = 58

3C + 8(6.5) = 58

3C + 52 = 58

3C = 58-52

3C = 6

C = 6/3

C = $2

1 Chair will cost $2

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Decompose 2/6 into 4 equal lengths
anyanavicka [17]
I'm not 100% sure, but I'm pretty sure it's 1/12, 1/12, 1/12, 1/12
4 0
3 years ago
Represent this statement as an equation: The product of a number and 4 plus 2 is 14.
DiKsa [7]

Answer:

4x+2=14

Step-by-step explanation:

8 0
3 years ago
C=3<br> c2 + 5 <br><br><br><br><br><br><br><br><br> ..............
vaieri [72.5K]

Answer:

3x2+5=11

Step-by-step explanation:

8 0
3 years ago
How many different linear arrangements are there of the letters a, b,c, d, e for which: (a a is last in line? (b a is before d?
inna [77]
A) Since a is last in line, we can disregard a, and concentrate on the remaining letters.
Let's start by drawing out a representation:

_ _ _ _ a

Since the other letters don't matter, then the number of ways simply becomes 4! = 24 ways

b) Since a is before d, we need to account for all of the possible cases.

Case 1: a d _ _ _ 
Case 2: a _ d _ _
Case 3: a _ _ d _
Case 4: a _ _ _ d

Let's start with case 1.
Since there are four different arrangements they can make, we also need to account for the remaining 4 letters.
\text{Case 1: } 4 \cdot 4!

Now, for case 2:
Let's group the three terms together. They can appear in: 3 spaces.
\text{Case 2: } 3 \cdot 4!

Case 3:
Exactly, the same process. Account for how many times this can happen, and multiply by 4!, since there are no specifics for the remaining letters.
\text{Case 3: } 2 \cdot 4!

\text{Case 4: } 1 \cdot 4!

\text{Total arrangements}: 4 \cdot 4! + 3 \cdot 4! + 2 \cdot 4! + 1 \cdot 4! = 240

c) Let's start by dealing with the restrictions.
By visually representing it, then we can see some obvious patterns.

a b c _ _

We know that this isn't the only arrangement that they can make.
From the previous question, we know that they can also sit in these positions:

_ a b c _
_ _ a b c

So, we have three possible arrangements. Now, we can say:
a c b _ _ or c a b _ _
and they are together.

In fact, they can swap in 3! ways. Thus, we need to account for these extra 3! and 2! (since the d and e can swap as well).

\text{Total arrangements: } 3 \cdot 3! \cdot 2! = 36
7 0
3 years ago
What is the least common denominator (LCD) of 12 and 23?
leva [86]

Answer: none of those, 276...?

Step-by-step explanation:

You need to know the least common denominator (LCD) of 12 and 23 if you want to add or subtract two fractions with 12 and 23 as denominators.

The least common denominator, also called lowest common denominator (LCD), of 12 and 23 is 276.

Here is a math problem example where you need to know the LCD of 12 and 23 to solve:

3/12 + 2/23 = ?

Step 1) Take the LCD and divide each denominator by it as follows:

276/12 = 23

276/23 = 12

Step 2) Multiply each nominator with the respective answers from Step 1:

3 x 23 = 69

2 x 12 = 24

Step 3) Put it all together to solve the problem:

69/276 + 24/276 = 93/276

= 3/12 + 2/23 = 93/276

It's that easy! Once again, the lowest common denominator (LCD) of 12 and 23 is as follows:

276

7 0
3 years ago
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