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3 years ago

Evaluate |6| - |-6|.

Mathematics

2 answers:

VashaNatasha [74]3 years ago

4 0

Answer:

Step-by-step explanation:

|6| - |-6|

|6| = 6

- |-6| is |-6| which is 6 and then add the - so its -6

so now we have

6 - 6 = 0

Elis [28]3 years ago

4 0

Answer:

Step-by-step explanation:

We are given subtraction of absolute numbers.

The absolute value of a number is always positive.

So,

l6l = 6

l -6 l = 6, The absolute value of a negative number is a positive number.

Therefore,

|6| - |-6|

We know that |6| = 6 and |-6| = 6.

So, 6 - 6 = 0

Hence, the answer is 0.

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Can anyone tell me what is 2/5 x 3/2 ?

Mars2501 [29]

Answer:

6/10

Step-by-step explanation:

2 × 3 = 6

5 × 2 = 10

6/10 = 3/5

4 0

2 years ago

Read 2 more answers

H(x)=1/8x^3-x^2<br><br> Over which interval does h have a positive average rate of change?

lara31 [8.8K]

Answer:

$(-\infty,0)\cup(\frac{16}{3},\infty)\\That \ is x\frac{16}{3}$

Step-by-step explanation:

$h(x)=\frac{1}{8}x^{3} -x^{2} \\\\Differentiate \ h(x) \ with \ respect \ to \ x\\h'(x)=\frac{3}{8}x^{2} -2x$

For positive rate of change $h'(x)>0$

$\frac{3}{8} x^{2} -2x>0\\x(\frac{3}{8}x-2)>0\\\\ When \ x0 \ (multiplication \ of \ two \ positives \ is \ positive)\\\\h'(x)>0 \ when \ x\frac{16}{3} \\\\$

6 0

3 years ago

Evaluate 2y²(x+y) when x = 1 and y = 5<br> Show working out

Harlamova29_29 [7]

Answer:

300

Step-by-step explanation:

2(5)²(1+5)

2(25)(6)

50(6)

300

4 0

3 years ago

Read 2 more answers

Neal is tiling his kitchen filler. Each square tile is 1 square foot. Neal uses 6 rows of tiles with 9 tiles in each row. What i

timofeeve [1]

Answer:

Step-by-step explanation:

9*6=54

6 rows of 9 titles

6 0

3 years ago

Solve using the quadratic formula 3x^2+11x+5=0

storchak [24]

Answer:

$3x^2 + 11x + 5 = 0 : x = \frac{-11 + \sqrt{61} }{6} , x = \frac{-11 - \sqrt{61} }{6}$

Decimal:

$(x = -0.53162..., x = -3.13504...)$

Step-by-step explanation:

$3x^2+11x+5=0$

Solve with the quadratic formula:

For a quadratic equation of the form $ax^2+bx+c=0$ the solutions are:

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

For $a=3,\:b=11,\:c=5:\quad x_{1,\:2}=\frac{-11\pm \sqrt{11^2-4\cdot \:3\cdot \:5}}{2\cdot \:3}$

$x=\frac{-11 + \sqrt{11^2-4\cdot \:3\cdot \:5}}{2\cdot \:3} : \frac{-11 + \sqrt{61} }{6}$

$x=\frac{-11 - \sqrt{11^2-4\cdot \:3\cdot \:5}}{2\cdot \:3} : \frac{-11 - \sqrt{61} }{6}$

The solutions to the quadratic equation are:

$x=\frac{-11+\sqrt{61}}{6},\:x=\frac{-11-\sqrt{61}}{6}$

Hope I helped. If so, may I get brainliest and a thanks?

Thank you, have a good day! =)

8 0

3 years ago