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3 years ago

1. 6/5/2/5 A)2 4/5 B) 3 C)1 2/5 D) 12/25

Mathematics

2 answers:

snow_lady [41]3 years ago

8 0

Answer: 3

Do Keep Change Flip (KCF)

Keep: 6/5

Change: ÷ into ×

Flip: 2/5 into 5/2

Your new problem: 6/5×5/2

Multiply

$6/5*5/2=30/10\\30/10=3$

Nady [450]3 years ago

6 0

So the right answer is 3.

Look at the attached picture ⤴

Hope it will help u...

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Ayudaaaaa porfavor Convertir las unidades de medida 8 000 m = _______ km 4 000 g = _______ kg 3 000 m = ________ km 7 m = ______

jek_recluse [69]

Answer:

a) $x = 8\,km$ , b) $x = 4\,kg$ , c) $x = 3\,km$ , d) $x = 700\,cm$ , e) $x = 3\,cm$ , f) $x = 100\,mm$ , g) $x = 7\,mm$ , h) $x = 10000\,mL$ , i) $x = 8\,m$ , j) $x = 6000\,g$

Step-by-step explanation:

a) 1 kilómetro equivale a 1000 metros, entonces obtenemos el valor requerido mediante la siguiente regla de tres simple:

$x = 8000\,m \times \frac{1\,km}{1000\,m}$

$x = 8\,km$

b) 1 kilogramo equivale a 1000 gramos, entonces obtenemos el valor requerido mediante la siguiente regla de tres simple:

$x = 4000\,g \times \frac{1\,kg}{1000\,g}$

$x = 4\,kg$

c) 1 kilómetro equivale a 1000 metros, entonces obtenemos el valor requerido mediante la siguiente regla de tres simple:

$x = 3000\,m \times \frac{1\,km}{1000\,m}$

$x = 3\,km$

d) 1 metro equivale a 100 centímetros, entonces obtenemos el valor requerido mediante la siguiente regla de tres simple:

$x = 7\,m \times \frac{100\,cm}{1\,m}$

$x = 700\,cm$

e) 1 centímetro equivale a 10 milímetros, entonces obtenemos el valor requerido mediante la siguiente regla de tres simple:

$x = 30\,mm \times \frac{1\,cm}{10\,mm}$

$x = 3\,cm$

f) 1 centímetro equivale a 10 milímetros, entonces obtenemos el valor requerido mediante la siguiente regla de tres simple:

$x = 10\,cm \times \frac{10\,mm}{1\,cm}$

$x = 100\,mm$

g) 1 centímetro equivale a 10 milímetros, entonces obtenemos el valor requerido mediante la siguiente regla de tres simple:

$x = 70\,mm\times \frac{1\,cm}{10\,mm}$

$x = 7\,mm$

h) 1 litro equivale a 1000 mililitros, entonces obtenemos el valor requerido mediante la siguiente regla de tres simple:

$x = 10\,L\times \frac{1000\,mL}{1\,L}$

$x = 10000\,mL$

i) 1 metro equivale a 100 centímetros, entonces obtenemos el valor requerido mediante la siguiente regla de tres simple:

$x = 800\,cm \times \frac{1\,m}{100\,cm}$

$x = 8\,m$

j) 1 kilogramo equivale a 1000 gramos, entonces obtenemos el valor requerido mediante la siguiente regla de tres simple:

$x = 6\,kg\times \frac{1000\,g}{1\,kg}$

$x = 6000\,g$

8 0

3 years ago

-5(5c-3b-1) distribute property

andrew11 [14]

Answer:-25c+15b+5 ..add it up then boom you'll get your answer

Step-by-step explanation:

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3 years ago

The high school tennis team is selling tennis balls to raise $500 for new equipment. They sell the balls for $4 each. A case h

Citrus2011 [14]

Answer:

1.75 cases need to be sold

Step-by-step explanation:

The complete question is

The high school tennis team is selling tennis balls to raise $500 for new equipment.They sell the balls for $2 each.

4).How many cases of tennis balls does the tennis team have to sell in order to raise enough money for new equipment.

Solution

Price of each ball = $4

No. of balls in each package = 3

Total number of balls that need to sold in order to raise $500 is

$\frac{500}{4} = 125$

one package = 3balls

Hence no of packages to be sold

$\frac{125}{3} = 41$ packages and 2 balls or total 42 packages

1 case has 24 packages , thus total number of cases to be sold = $\frac{42}{24} = \frac{14}{8} = \frac{7}{4} = 1.75$

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3 years ago

The quotient of 20 and a number will be 10

Dimas [21]

Answer:

Step-by-step explanation:

lets think of the missing number as x so any number times 10 should equal 20 so x=2

and your equation should look like this--------> 20÷2=10

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3 years ago

For an integer n ≥ 0, show that (n/2) - (-n/2) = n. Use greatest integer function

mars1129 [50]

The greatest integer function returns the largest integer smaller than the number you provide it. That is, if n ≤ x < n + 1, where n is an integer, then the "greatest integer of x" is [x] = n.

• Let n be even. Then we can write n = 2k for some integer k ≥ 0. Now,

[n/2] = [k] = k

while

[-n/2] = [-k] = -k

so that

[n/2] - [-n/2] = 2k = n

• Let n be odd. Then n = 2k + 1 for some integer k ≥ 0. Every odd integer occurs between two even integers, so that

n - 1 < n < n + 1

or equivalently,

2k < n < 2k + 2

so that

k < n/2 < k + 1

It follows that [n/2] = k.

Similarly, if we negative the previous inequality, we have

-k > -n/2 > -(k + 1), or -k - 1 < -n/2 < -k

which means [-n/2] = -k - 1.

So we make the same conclusion,

[n/2] - [-n/2] = k - (-k - 1) = 2k + 1 = n

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2 years ago