All categories

3 years ago

How fast is a monkey running if it takes him 30 minutes to travel 240 meters

Mathematics

2 answers:

Zanzabum3 years ago

5 0

Answer:

the monkeys travels 8 meters per minute running .

speed=distance /time

Anna71 [15]3 years ago

3 0

Answer:

he's fast jk

Step-by-step explanation:

8 m/min because you have to find the unit rate at which he is running , and you would do that by taking 240 meters and then dividing it by 30 minutes, which would come out to 8

You might be interested in

The following triangle is equilaterial triangle. Solve for the unknown variable.

Iteru [2.4K]

Answer:

Step-by-step explanation:

Since the triangle is equilateral then the 3 sides are congruent, then

2x = 6 ( divide both sides by 2 )

x = 3

and

3y = 6 ( divide both sides by 3 )

y = 2

Thus x = 3 and y = 2 → d

8 0

3 years ago

Please help with all of them , thank you !

Lelu [443]

Answer:

A. 12x² + 15xy + 18x

B. - 12x² + 18xy + 24x

C. 10x² - 31x - 14

D. x² - 36

3a. x = 8

3b. x = 3

Step-by-step explanation:

A. 3x ( 4x + 5y +6 ) C. ( 2x - 7 ) ( 5x + 2 )

12x² + 15xy + 18x 10x² + 4x - 35x - 14

10x² - 31x - 14

B. 6x ( 3y - 2x + 4 )

18xy - 12x² + 24x D. ( x + 6 ) ( x - 6 )

-12x² + 18xy + 24x x² - 6x + 6x - 36

x² - 36

3a. 5/4x + 2 = 12 3b. 7/3x - 2 = 5/3x

5/4x = 12 - 2 -2 = 5/3x - 7/3x

5/4x = 10 -2 = - 2/3x

x = 10 · 4/5 -2/3x = -2

x = 8 x = 3

4 0

3 years ago

Find the value of x which makes the area of this triangle equal 72

Maslowich

The Answer To Your Question Would Be 12

8 0

3 years ago

The triangles are similar, find y

Luda [366]

Answer:

y = 3.6

x = 3.5

Step-by-step explanation:

The triangles are similar.

$\frac{3}{2.4} = \frac{x}{2.8} = \frac{4.5}{y}$

Find y:

$\frac{4.5}{y} = \frac{5}{4}$

$4.5 * 4 = 5y => 18 = 5y => y = 3.6$

Find x:

$\frac{x}{2.8} = \frac{5}{4}$

$2.8 * 5 = 4x => 14 = 4x => x = 3.5$

3 0

3 years ago

How do i solve this differential equation y'=2y-x over 2x-y

Tamiku [17]

$y'=\dfrac{2y-x}{2x-y}$

Let $y=xv$ , where $v=v(x)$ , so that $y'=xv'+v$ . Then the ODE is

$xv'+v=\dfrac{2xv-x}{2x-xv}$
$xv'=\dfrac{2x(v-1)}{x(2-v)}-v$
$xv'=\dfrac{2(v-1)}{2-v}-\dfrac{v(2-v)}{2-v}$
$xv'=\dfrac{v^2-2}{2-v}$

This is separable, so you have

$\dfrac{2-v}{v^2-2}\,\mathrm dv=\dfrac{\mathrm dx}x$

Integrate both sides, solving for $v$ if possible, then replace using $v=\dfrac yx$ and solve for $y$ if possible.

3 0

3 years ago