All categories

3 years ago

A number that is relatively prime to the given number 32

Mathematics

1 answer:

makvit [3.9K]3 years ago

8 0

Answer:

32 and 49 are relatively prime because

hey have no common prime factors, that is, if their greatest common factor is 1.

Step-by-step explanation:

You might be interested in

Five hundred plus five hundred fifty

JulsSmile [24]

Answer:

1,050

Step-by-step explanation:

500

550

1,050 You add the numbers getting your total

4 0

3 years ago

A local University accepted 2,300 students out of 4,500 applicants for admission. What was the acceptance rate, expressed as a p

Mademuasel [1]

2300 out of 4500 were chosen

2300 / 4500 = 0.5111

0.51 = 51%

51% is the acceptance rate.

4 0

2 years ago

Read 2 more answers

Please answer! Will mark brainliest. <br><br> -3|5x-2|>4

Mrac [35]

Answer:

Step-by-step explanation:

eact from : x= 2, 6/5

decial from : x= 2, -1.2

mixed number from :x= 2, - 1 1/5

6 0

3 years ago

Read 2 more answers

Solve. Good luck! Please do not try to google this.

Elena L [17]

$\frac{x^{2} + x - 2}{6x^{2} - 3x} = \sqrt{2x} + \frac{3x^{2}}{2}$
$\frac{x^{2} + 2x - x - 2}{3x(x) - 3x(1)} = \frac{2\sqrt{2x}}{2} + \frac{3x^{2}}{2}$
$\frac{x(x) + x(2) - 1(x) - 1(2)}{3x(x - 1)} = \frac{2\sqrt{2x} + 3x^{2}}{2}$
$\frac{x(x + 2) - 1(x + 2)}{3x(2x - 1)} = \frac{2\sqrt{2x} + 3x^{2}}{2}$
$\frac{(x - 1)(x + 2)}{3x(2x - 1)} = \frac{2\sqrt{2x} + 3x^{2}}{2}$
$\frac{(x - 1)(x + 2)}{3x(2x - 1)} = \frac{2\sqrt{2x} + 3x^{2}}{2}$
$3x(2x - 1)(2\sqrt{2x} + 3x^{2}) = 2(x + 2)(x - 1)$
$3x(2x(2\sqrt{2x} + 3x^{2}) - 1(2\sqrt{2x} + 3x^{2}) = 2(x(x - 1) + 2(x - 1))$
$3x(2x(2\sqrt{2x}) + 2x(3x^{2}) - 1(2\sqrt{2x}) - 1(3x^{2})) = 2(x(x) - x(1) + 2(x) - 2(1)$
$3x(4x\sqrt{2x} + 6x^{3} - 2\sqrt{2x} - 3x^{2}) = 2(x^{2} - x + 2x - 2)$
$3x(4x\sqrt{2x} - 2\sqrt{2x} + 6x^{3} - 3x^{2}) = 2(x^{2} + x - 2)$
$3x(4x\sqrt{2x}) - 3x(2\sqrt{2x}) + 3x(6x^{3}) - 3x(3x^{2}) = 2(x^{2}) + 2(x) - 2(2)$
$12x^{2}\sqrt{2x} - 6x\sqrt{2x} + 18x^{4} - 9x^{3} = 2x^{2} + 2x - 4$
$12x^{2}\sqrt{2x} - 6x\sqrt{2x} = -18x^{4} + 9x^{3} + 2x^{2} + 2x - 4$
$6x\sqrt{2x}(2x) - 6x\sqrt{2x}(1) = -9x^{3}(2x) - 9x^{3}(-1) + 2(x^{2}) + 2(x) - 2(2)$
$6x\sqrt{2x}(2x - 1) = -9x^{3}(2x - 1) + 2(x^{2} + x - 2)$

5 0

3 years ago

The lateral area of a right cylinder which has a base diameter of 12 cm and a height of 8 cm is?

garik1379 [7]

Answer:

$A = 301.59\ cm^2$ or $A=96\pi\ cm^2$