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9966 [12]
3 years ago
13

1/2(4x-2)-2/3(6x+9<4

Mathematics
1 answer:
babymother [125]3 years ago
6 0
\frac{1}{2}(4x - 2) - \frac{2}{3}(6x + 9) < 4   Use the Distributive Property
2x - 1 - 4x + 6 < 4   Combine like terms (-1 and 6)
    2x + 5 - 4x < 4   Combine (2x and -4x)
          -2x + 5 < 4   Subtract 5 from both sides
                -2x < -1   Divide both sides by -2, don't forget to flip the inequality                                    sign after dividing by a negative
                    x > \frac{1}{2}
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If the sum of the even integers between 1 and k, inclusive, is equal to 2k, what is the value of k?
Alisiya [41]
If k is odd, then

\displaystyle\sum_{n=1}^{\lfloor k/2\rfloor}2n=2\dfrac{\left\lfloor\frac k2\right\rfloor\left(\left\lfloor\frac k2\right\rfloor+1\right)}2=\left\lfloor\dfrac k2\right\rfloor^2+\left\lfloor\dfrac k2\right\rfloor

while if k is even, then the sum would be

\displaystyle\sum_{n=1}^{k/2}2n=2\dfrac{\frac k2\left(\frac k2+1\right)}2=\dfrac{k^2+2k}4

The latter case is easier to solve:

\dfrac{k^2+2k}4=2k\implies k^2-6k=k(k-6)=0

which means k=6.

In the odd case, instead of considering the above equation we can consider the partial sums. If k is odd, then the sum of the even integers between 1 and k would be

S=2+4+6+\cdots+(k-5)+(k-3)+(k-1)

Now consider the partial sum up to the second-to-last term,

S^*=2+4+6+\cdots+(k-5)+(k-3)

Subtracting this from the previous partial sum, we have

S-S^*=k-1

We're given that the sums must add to 2k, which means

S=2k
S^*=2(k-2)

But taking the differences now yields

S-S^*=2k-2(k-2)=4

and there is only one k for which k-1=4; namely, k=5. However, the sum of the even integers between 1 and 5 is 2+4=6, whereas 2k=10\neq6. So there are no solutions to this over the odd integers.
5 0
3 years ago
What is the value of the expression below?<br><br> 1 3/4 divided by 1/2 minus (1 1/2)^3
kogti [31]

Answer:

1/8

Step-by-step explanation:

Simplify the following:

(1 + 3/4)/(1/2) - (1/2 + 1)^3

Hint: | Write (1 + 3/4)/(1/2) as a single fraction.

Multiply the numerator of (1 + 3/4)/(1/2) by the reciprocal of the denominator. (1 + 3/4)/(1/2) = ((1 + 3/4)×2)/1:

(3/4 + 1) 2 - (1/2 + 1)^3

Hint: | Put the fractions in 1 + 1/2 over a common denominator.

Put 1 + 1/2 over the common denominator 2. 1 + 1/2 = 2/2 + 1/2:

(1 + 3/4) 2 - (2/2 + 1/2)^3

Hint: | Add the fractions over a common denominator to a single fraction.

2/2 + 1/2 = (2 + 1)/2:

(1 + 3/4) 2 - ((2 + 1)/2)^3

Hint: | Evaluate 2 + 1.

2 + 1 = 3:

(1 + 3/4) 2 - (3/2)^3

Hint: | Put the fractions in 1 + 3/4 over a common denominator.

Put 1 + 3/4 over the common denominator 4. 1 + 3/4 = 4/4 + 3/4:

4/4 + 3/4 2 - (3/2)^3

Hint: | Add the fractions over a common denominator to a single fraction.

4/4 + 3/4 = (4 + 3)/4:

(4 + 3)/4×2 - (3/2)^3

Hint: | Evaluate 4 + 3.

4 + 3 = 7:

7/4×2 - (3/2)^3

Hint: | Express 7/4×2 as a single fraction.

7/4×2 = (7×2)/4:

(7×2)/4 - (3/2)^3

Hint: | In (7×2)/4, divide 4 in the denominator by 2 in the numerator.

2/4 = 2/(2×2) = 1/2:

7/2 - (3/2)^3

Hint: | Simplify (3/2)^3 using the rule (p/q)^n = p^n/q^n.

(3/2)^3 = 3^3/2^3:

7/2 - 3^3/2^3

Hint: | In order to evaluate 3^3 express 3^3 as 3×3^2.

3^3 = 3×3^2:

7/2 - (3×3^2)/2^3

Hint: | In order to evaluate 2^3 express 2^3 as 2×2^2.

2^3 = 2×2^2:

7/2 - (3×3^2)/(2×2^2)

Hint: | Evaluate 2^2.

2^2 = 4:

7/2 - (3×3^2)/(2×4)

Hint: | Evaluate 3^2.

3^2 = 9:

7/2 - (3×9)/(2×4)

Hint: | Multiply 2 and 4 together.

2×4 = 8:

7/2 - (3×9)/8

Hint: | Multiply 3 and 9 together.

3×9 = 27:

7/2 - 27/8

Hint: | Put the fractions in 7/2 - 27/8 over a common denominator.

Put 7/2 - 27/8 over the common denominator 8. 7/2 - 27/8 = (4×7)/8 - 27/8:

(4×7)/8 - 27/8

Hint: | Multiply 4 and 7 together.

4×7 = 28:

28/8 - 27/8

Hint: | Subtract the fractions over a common denominator to a single fraction.

28/8 - 27/8 = (28 - 27)/8:

(28 - 27)/8

Hint: | Subtract 27 from 28.

| 2 | 8

- | 2 | 7

| 0 | 1:

Answer: 1/8

4 0
3 years ago
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