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4 years ago

Item 1 Evaluate 2xy when x=−14 and y=3 . explain your answer

Mathematics

2 answers:

meriva4 years ago

8 0

2xy

2(-14)(3)

-28*3

-84

juin [17]4 years ago

6 0

Answer:

$-84$

Step-by-step explanation:

We have been given an expression $2xy$ . We are asked to evaluate our given expression, when $x=-14$ and $y=3$ .

Upon substituting $x=-14$ and $y=3$ in our given expression, we will get:

$2(-14)(3)$

$6(-14)$

$-84$

Therefore, the value of our expression at given values is $-84$ .

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Answer:a

Step-by-step explanation:

A is the correct answer

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3 years ago

Margaret is making strawberry milkshakes for the kids party. The recipe calls for of a cup of strawberry syrup to make 10 milksh

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3 years ago

Read 2 more answers

Let f(x)= 2x-3/9 and g(x)=9x+3/2 Find (f o g)(x) Find (f o g)(x)

svlad2 [7]

These seem tough, but they are not that bad.
(f o g)(x)

That means to take function g when x = x (It just means that you don't need to do anything with the function) and substitute it into function f wherever there is a x value. (So basically g becomes what x equals)
Here is a visual representation of it.

(f o g)(x) = 2(9x + (3/2)) - (3/9)

We need to distribute the 2 to everything inside the parenthesis.

(f o g)(x) = 18x + (6/2) - (3/9)

Let's simplify the fraction on the right and left, then find a common denominator so we can add / subtract the two fractions.

(f o g)(x) = 18x + (3) - (1/3)
(f o g)(x) = 18x + (9/3) - (1/3)

Combine like terms.

(f o g)(x) = 18x + (8/3)

Our final answer is:
(f o g)(x) = 18x + 8/3

8 0

3 years ago

the function f(x)= x1/3 is transformed to get function h. h(x)= (2x)1/3+5 which statements are true about function h

Bess [88]

The transformation of a function may involve any change. The function f(x) is vertically stretched and shifted 5 units upwards to form h(x).

<h3>How does the transformation of a function happen?</h3>

The transformation of a function may involve any change.

Usually, these can be shifted horizontally (by transforming inputs) or vertically (by transforming output), stretched (multiplying outputs or inputs) etc.

If the original function is y = f(x), assuming the horizontal axis is the input axis and the vertical is for outputs, then:

Horizontal shift (also called phase shift):

Left shift by c units, y=f(x+c) (same output, but c units earlier)
Right shift by c units, y=f(x-c)(same output, but c units late)

Vertical shift

Up by d units: y = f(x) + d
Down by d units: y = f(x) - d

Stretching:

Vertical stretch by a factor k: y = k \times f(x)
Horizontal stretch by a factor k: y = f(\dfrac{x}{k})

The function f(x)=x^(1/3) is transformed to form the function of h(x)=(2x)^(1/3)+5. Therefore, the transformation made to the function is,

Vertically stretched by a factor of 2^(1/3) ⇒ 2^(1/3) × x^(1/3) = (2x)^(1/3)

Up by 5 units ⇒ (2x)^(1/3) + 5

Hence, the function f(x) is vertically stretched and shifted 5 units upwards to form h(x).

Learn more about Transforming functions:

brainly.com/question/17006186

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2 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:

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

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

$\mathbf{Total = 302400}$