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3 years ago

How do you reduce improper fractions?

Mathematics

1 answer:

sasho [114]3 years ago

6 0

To reduce it even more, u can divide 32 by 3 and get 10 2/3;10.6

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1.) (-6) (-7) =

Jobisdone [24]

$\left(-6\right)\left(-7\right)=42$

$46=2\quad :\quad \mathrm{False}$

$\left(-9\right)+25=16$

$\left(13\right)\left(-3\right)=-39$

$14+3=17$

5 0

3 years ago

Read 2 more answers

How do you divide fractions?

Zanzabum

2/8 divided by 1/4. Steps: 1. Division in math actually means multiplication. You just have to multiply by the reciprocal of the second fraction (reciprocal= denominator and numerator are switched). 2. Do 2/8 times 4/1. 3. 2 times 4= 8 and 8 times 1= 8 4. Your answer would be 8/8 which equals 1. This is how you divide fractions

4 0

3 years ago

Which of the following is a correct solutions to the following quadratic?<br> 2x2 - 6x + 5 = 0

astraxan [27]

Answer:

-1/6

Step-by-step explanation:

4-6x+5=0

-6x+5=-4

-6x=-4+5

-6x=1

6x=-1

x=-1/6

6 0

4 years ago

FIND THE VALUE OF Y.

Alex17521 [72]

Answer:

y = 3 $\sqrt{5}$

Step-by-step explanation:

According to Euclidian theorem:

y^2 = 3*(12+3)

y^2 = 45 find the root of both sides

y = 3 $\sqrt{5}$

6 0

2 years ago

Let C(n, k) = the number of k-membered subsets of an n-membered set. Find (a) C(6, k) for k = 0,1,2,...,6 (b) C(7, k) for k = 0,

vladimir1956 [14]

Answer:

(a) $C(6,0) = 1$ , $C(6,1) = 6$ , $C(6,2) = 15$ , $C(6,3) = 20$ , $C(6,4) = 15$ , $C(6,5) = 6$ , $C(6,6) = 1$ .

(b) $C(7,0) = 1$ , $C(7,1) = 7$ , $C(7,2) = 21$ , $C(7,3) = 35$ , $C(7,4) = 35$ , $C(7,5) = 21$ , $C(7,6) = 7$ , $C(7,7)=1$ .

Step-by-step explanation:

In this exercise we only need to recall the formula for C(n,k):

$C(n,k) = \frac{n!}{k!(n-k)!}$

where the symbol $n!$ is the factorial and means

$n! = 1\cdot 2\cdot 3\cdot 4\cdtos (n-1)\cdot n$ .

By convention 0!=1. The most important property of the factorial is $n!=(n-1)!\cdot n$ , for example 3!=1*2*3=6.

(a) The explanations to the solutions is just the calculations.

$C(6,0) = \frac{6!}{0!(6-0)!} = \frac{6!}{6!} = 1$

$C(6,1) = \frac{6!}{1!(6-1)!} = \frac{6!}{5!} = \frac{5!\cdot 6}{5!} = 6$

$C(6,2) = \frac{6!}{2!(6-2)!} = \frac{6!}{2\cdot 4!} = \frac{5!\cdot 6}{2\cdot 4!} = \frac{4!\cdot 5\cdot 6}{2\cdot 4!} = \frac{5\cdot 6}{2} = 15$

$C(6,3) = \frac{6!}{3!(6-3)!} = \frac{6!}{3!\cdot 3!} = \frac{5!\cdot 6}{6\cdot 6} = \frac{5!}{6} = \frac{120}{6} = 20$

$C(6,4) = \frac{6!}{4!(6-4)!} = \frac{6!}{4!\cdot 2!} = frac{5!\cdot 6}{2\cdot 4!} = \frac{4!\cdot 5\cdot 6}{2\cdot 4!} = \frac{5\cdot 6}{2} = 15$

$C(6,5) = \frac{6!}{5!(6-5)!} = \frac{6!}{5!} = \frac{5!\cdot 6}{5!} = 6$

$C(6,6) = \frac{6!}{6!(6-6)!} = \frac{6!}{6!} = 1$ .

(b) The explanations to the solutions is just the calculations.

$C(7,0) = \frac{7!}{0!(7-0)!} = \frac{7!}{7!} = 1$

$C(7,1) = \frac{7!}{1!(7-1)!} = \frac{7!}{6!} = \frac{6!\cdot 7}{6!} = 7$

$C(7,2) = \frac{7!}{2!(7-2)!} = \frac{7!}{2\cdot 5!} = \frac{6!\cdot 7}{2\cdot 5!} = \frac{5!\cdot 6\cdot 7}{2\cdot 5!} = \frac{6\cdot 7}{2} = 21$

$C(7,3) = \frac{7!}{3!(7-3)!} = \frac{7!}{3!\cdot 4!} = \frac{6!\cdot 7}{6\cdot 4!} = \frac{5!\cdot 6\cdot 7}{6\cdot 4!} = \frac{120\cdot 7}{24} = 35$

$C(7,4) = \frac{7!}{4!(7-4)!} = \frac{6!\cdot 7}{4!\cdot 3!} = frac{5!\cdot 6\cdot 7}{4!\cdot 6} = \frac{120\cdot 7}{24} = 35$

$C(7,5) = \frac{7!}{5!(7-2)!} = \frac{7!}{5!\cdot 2!} = 21$

$C(7,6) = \frac{7!}{6!(7-6)!} = \frac{7!}{6!} = \frac{6!\cdot 7}{6!} = 7$

$C(7,7) = \frac{7!}{7!(7-7)!} = \frac{7!}{7!} = 1$

For all the calculations just recall that 4! =24 and 5!=120.

6 0

3 years ago

How do you reduce improper fractions?​

How do you reduce improper fractions?