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3 years ago

2,16,68,208,508,1062,1986 what’s the next number

Mathematics

1 answer:

lilavasa [31]3 years ago

4 0

2,16,68,208,508,1062,1987

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Multiply and simplify. 2x4yx⋅6xy2z3

OLEGan [10]

Answer:

The given expression $2x^4yx \times 6xy^2z^3$ on multiplying is $12x^6y^3z^3$

Step-by-step explanation:

Consider the given two expressions $2x^4yx$ and $6xy^2z^3$

We have multiply both expressions,

$2x^4yx \times 6xy^2z^3$

To multiply two terms first multiply constant numbers that is 6 × 2 = 12

For x , y and z apply property of exponent,

$x^a\cdot x^b=x^{a+b}$

Then power of x together will be,

$x^4\cdot x\cdot x=x^{4+1+1}=x^6$

Similarly for y powers,

$y \cdot y^2=y^{1+2}=y^3$

Since first term do not have any expression for z so it will remain same.

Thus, the given expression on multiplying become,

$2x^4yx \times 6xy^2z^3$ is $12x^6y^3z^3$

7 0

3 years ago

Read 2 more answers

At a local pizza place, the cost for 3 small pizzas and 3 medium pizzas is $51 The cost for 5 small and 2 medium is $55 How much

faltersainse [42]

Answer:

Step-by-step explanation:

3 0

2 years ago

What is the slope of 9 and 180

Kipish [7]

Answer:

a slope of 9 and 180 would just be 9 and 180 over 1

Step-by-step explanation:

8 0

2 years ago

PLZ ANSWER NUMBERS 8 AND 9. SHOWS YOUR WORK TOO!!!!

ASHA 777 [7]

8. 3x-4>38
3x>38+4
3x>42
x>42/3
x>14
So the answer is A. x>14

9. It's A. Two less than four times a number is 12. Because 4x (four times) and also it says 2 less than 4 times a number...

3 0

3 years ago

In a cohort of 35 graduating students, there are three different prizes to be awarded. If no student can receive more than one p

sweet [91]

We have been given in a cohort of 35 graduating students, there are three different prizes to be awarded. We are asked that in how many different ways could the prizes be awarded, if no student can receive more than one prize.

To solve this problem we will use permutations.

$_{r}^{n}\textrm{P}={_{3}^{35}\textrm{P}}$

We know that formula for permutations is given as

$_{r}^{n}\textrm{P}=\frac{n!}{(n-r)!}$

On substituting the given values in the formula we get,

${_{3}^{35}\textrm{P}}=\frac{35!}{(35-3)!}=\frac{35!}{32!}$

$=\frac{35\cdot 34\cdot 33\cdot 32!}{32!}\\
\\
=35\cdot 34\cdot 33=39270$

Therefore, there are 39270 ways in which prizes can be awarded.

6 0

3 years ago