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Elis [28]
3 years ago
5

Please help me .. I really need help with this ASAP ​

Mathematics
1 answer:
morpeh [17]3 years ago
4 0

Given:

Number of flower pots = 6

To find:

The number of ways of the gardener to arrange the flower pots.

Solution:

Number of ways to arrange n items is n!.

So, the number of ways to arrange 6 pots is:

6!=6\times 5\times 4\times 3\times 2\times 1

6!=720

Therefore, there are total 720 ways of the gardener to arrange the flowerpots.

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4
zloy xaker [14]

Answer: A.  3x^{12} - 2x + x^5

Step-by-step explanation:

  • The degree in a polynomial is the greatest exponent of any term in that expression.

A. 3x^{12} - 2x + x^5

Here, highest exponent = 12

Degree of polynomial =12

B. 8x-200

Here, highest exponent = 1

Degree of polynomial =1

C. 15x^3 + 3x^2 - 10

Here, highest exponent = 3

Degree of polynomial =3

D. 150x^3

Here, highest exponent = 3

Degree of polynomial =3

Option A. has the polynomial with the highest degree as 12.

4 0
3 years ago
Read 2 more answers
I will put a answer up in 3 hours i have 30,000 points so be ready
algol [13]

Answer:

we need more people like you

Step-by-step explanation:

thank you!

5 0
3 years ago
Read 2 more answers
-2-3(5x+7)<br> May someone be able to help me with this equation
Xelga [282]

Answer: -15x - 23

Step-by-step explanation:

There u go- Lily ^_^

6 0
3 years ago
For what values of the variable, do the following fractions exist: y^2-1/y+y/y-3
ale4655 [162]

Answer:

Remember that the division by zero is not defined, this is the criteria that we will use in this case.

1) \frac{y^2 - 1}{y}  + \frac{y}{y - 3}

So the fractions are defined such that the denominator is never zero.

For the first fraction, the denominator is zero when y = 0

and for the second fraction, the denominator is zero when y = 3

Then the fractions exist for all real values except for y = 0 or y = 3

we can write this as:

R / {0} U { 3}

(the set of all real numbers except the elements 0 and 3)

2) \frac{b + 4}{b^2 + 7}

Let's see the values of b such that the denominator is zero:

b^2 + 7 = 0

b^2 = -7

b = √-7

This is a complex value, assuming that b can only be a real number, there is no value of b such that the denominator is zero, then the fraction is defined for every real number.

The allowed values are R, the set of all real numbers.

3) \frac{a}{a*(a - 1) - 1}

Again, we need to find the value of a such that the denominator is zero.

a*(a - 1) - 1 = a^2 - a - 1

So we need to solve:

a^2 - a - 1 = 0

We can use the Bhaskara's formula, the two values of a are given by:

a = \frac{-(-1) \pm \sqrt{(-1)^2 + 4*1*(-1)}  }{2*1}  = \frac{1 \pm \sqrt{5} }{2}

Then the two values of a that are not allowed are:

a = (1 + √5)/2

a = (1 - √5)/2

Then the allowed values of a are:

R / {(1 + √5)/2} U {(1 - √5)/2}

7 0
3 years ago
I will mark brainiest whoever answers this correctly
Paladinen [302]

Answer:

k = 3.9

Step-by-step explanation:

Since the three possible outcomes are given. Therefore, the sum of all the probabilities of the three possible outcomes will be equal to one:

\frac{7}{13} +\frac{1}{k}+ \frac{1}{2k} = 1 \\\\\frac{1}{k} \frac{1}{2k} =1-\frac{7}{13}\\\\\frac{3}{2k} \frac{5}{13}\\\\2k = \frac{(13)(3)}{5}\\\\k = \frac{39}{(5)(2)}\\

<u>k = 3.9</u>

6 0
3 years ago
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