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3 years ago

Compute 35/6 Enter your answer using remainder notation, if necessary

2 answers:

7 0

ANSWER - 5 5/6

WORKING OUT
35 ÷ 6 = 5 remainder 5
Which means there are 5 wholes and 5 parts of a whole in 35/6.
So you’d write down the the whole number (5) down and with the remainder (5) you’d put into a fraction. With the same initial denominator (denominator = lower number)
So...
5 and 5/6

Aleksandr [31]3 years ago

4 0

Answer:

Im pretty sure it's 5.83

Step-by-step explanation:

Hope this helped! Pls mark brainliest thx.

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For which of the following graphs would NOT make sense to connect the points?

jeka94

Answer:

i believe d. I could be wrong

Step-by-step explanation:

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3 years ago

Find the zero of f(x) = x^5-12x^2+32x

Alexeev081 [22]

The zeroes of f(x) = x^5-12x^2+32x can be found by factoring,

f(x) = x^5-12x^2+32x=(x-8)(x-4)

By the zero product theorem, (x-8)=0 or (x-4)=0 which means
x=8 or x=4.

So the zeroes of f(x) are S={4,8}

8 0

3 years ago

2. In how many ways can 3 different novels, 2 different mathematics books and 5 different chemistry books be arranged on a books

insens350 [35]

The number of ways of the books can be arranged are illustrations of permutations.

When the books are arranged in any order, the number of arrangements is 3628800
When the mathematics book must not be together, the number of arrangements is 2903040
When the novels must be together, and the chemistry books must be together, the number of arrangements is 17280
When the mathematics books must be together, and the novels must not be together, the number of arrangements is 302400

The given parameters are:

$\mathbf{Novels = 3}$

$\mathbf{Mathematics = 2}$

$\mathbf{Chemistry = 5}$

(a) The books in any order

First, we calculate the total number of books

$\mathbf{n = Novels + Mathematics + Chemistry}$

$\mathbf{n = 3 + 2 + 5}$

$\mathbf{n = 10}$

The number of arrangement is n!:

So, we have:

$\mathbf{n! = 10!}$

$\mathbf{n! = 3628800}$

(b) The mathematics book, not together

There are 2 mathematics books.

If the mathematics books, must be together

The number of arrangements is:

$\mathbf{Maths\ together = 2 \times 9!}$

Using the complement rule, we have:

$\mathbf{Maths\ not\ together = Total - Maths\ together}$

This gives

$\mathbf{Maths\ not\ together = 3628800 - 2 \times 9!}$

$\mathbf{Maths\ not\ together = 2903040}$

(c) The novels must be together and the chemistry books, together

We have:

$\mathbf{Novels = 3}$

$\mathbf{Chemistry = 5}$

First, arrange the novels in:

$\mathbf{Novels = 3!\ ways}$

Next, arrange the chemistry books in:

$\mathbf{Chemistry = 5!\ ways}$

Now, the 5 chemistry books will be taken as 1; the novels will also be taken as 1.

Literally, the number of books now is:

$\mathbf{n =Mathematics + 1 + 1}$

$\mathbf{n =2 + 1 + 1}$

$\mathbf{n =4}$

So, the number of arrangements is:

$\mathbf{Arrangements = n! \times 3! \times 5!}$

$\mathbf{Arrangements = 4! \times 3! \times 5!}$

$\mathbf{Arrangements = 17280}$

(d) The mathematics must be together and the chemistry books, not together

We have:

$\mathbf{Mathematics = 2}$

$\mathbf{Novels = 3}$

$\mathbf{Chemistry = 5}$

First, arrange the mathematics in:

$\mathbf{Mathematics = 2!}$

Literally, the number of chemistry and mathematics now is:

$\mathbf{n =Chemistry + 1}$

$\mathbf{n =5 + 1}$

$\mathbf{n =6}$

So, the number of arrangements of these books is:

$\mathbf{Arrangements = n! \times 2!}$

$\mathbf{Arrangements = 6! \times 2!}$

Now, there are 7 spaces between the chemistry and mathematics books.

For the 3 novels not to be together, the number of arrangement is:

$\mathbf{Arrangements = ^7P_3}$

So, the total arrangement is:

$\mathbf{Total = 6! \times 2!\times ^7P_3}$

$\mathbf{Total = 6! \times 2!\times 210}$

$\mathbf{Total = 302400}$

Read more about permutations at:

brainly.com/question/1216161

8 0

2 years ago

Mark Brainliest what is 2/3 + 7/8 Please help

Pachacha [2.7K]

1 13/24 and that's it

3 0

3 years ago

Read 2 more answers

Find the equation of the line that is parallel to the line x + 5y = 10 and passes through the point (1,3).

Nataliya [291]

Answer:

Step-by-step explanation:

for two lines to be parallel they MUST have the same slope but have different Y intercepts ( Y axis crossing values)

parallel lines will never intersect each other

first I use a graphing calcuator to solve the problem

I like to use y intercept form better, but you don't have use it

I will try to solve it directly without going through the y intercept equation

x + 5y = 10 in y intercept form y = mx + b is

5y = -x + 10 divide both sides by 5

y = -x/5 + 10/5

y = -x/5 + 2

Now I need to find a parallel line that passes through the x =1 and y = 3 point recall this line will have the same slope = m = -1/5

and a different Y axis crossing point 'b' that we don't know

y = -x/5 + b y = 3 when x = 1 so slove for 'b'

3 = -1/5 + b

3 = -0.2 + b add -0.2 to both sides

3.2 = b

y = -x/5 + 3.2 this is answer in y intercept form

if you multiply both sides by 5

5y = -x + 16 add x to both sides

x + 5y = 16 answer in standard form

THE EASIER WAY TO SOLVE THE PROBLEM IS ......

x + 5y = 10 find a line parallel that passes through x = 1, y = 3

x + 5y = Z Z is an unknown value plug in x and y and solve for Z

1 + 5(3) = Z

Z = 16 so the parallel line in standard form is

x + 5y = 16

same as before, my y intercept method was correct, but not worth the effort

6 0

3 years ago