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Advocard [28]
3 years ago
5

Write (3y)^2 without exponents

Mathematics
2 answers:
krek1111 [17]3 years ago
7 0
There are two ways of doing this

3y x 3y
Or
9y
dlinn [17]3 years ago
7 0
" ^ " indicates exponentiation.  " ^2 " indicates squaring, or multiplying the base by itself.  Thus, w^2 is the same as w times w.

What is (3y) times (3y)?
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Find the value of x from the following trigonometrical equation :
Yuliya22 [10]

Answer:

x = 2/3

Step-by-step explanation:

sin²30° + cos²30° + x.tan²60° = cot²30°​

=> (1/2)² + (√3/2)² + x(√3)² = (√3)²

=> 1/4 + 3/4 + 3x = 3

=> 3x = 2

=> x = 2/3

6 0
2 years ago
Scenario: A new Swimming pool is opening up this summer in the Bronx. Students that are 13 - 18 years old can use the pool, gym
MA_775_DIABLO [31]
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3 0
3 years ago
1. |4n-15|=|n|<br> 2. |2c+8|=|10c| <br> can you please show the work for both
Alja [10]

Problem 1

<h3>Answers: n = 3 or n = 5</h3>

---------------

Work Shown:

|4n - 15| = |n|\\\\4n - 15 = n \ \text{ or } \ 4n-15 = -n\\\\-15 = n-4n \ \text{ or } \ -15 = -n-4n\\\\-15 = -3n \ \text{ or } \ -15 = -5n\\\\-3n = -15 \ \text{ or } \ -5n = -15\\\\n = \frac{-15}{-3} \ \text{ or } \ n = \frac{-15}{-5}\\\\n = 5 \ \text{ or } \ n = 3\\\\

==============================================

Problem 2

<h3>Answers: c = 1 or c = -2/3</h3>

-----------------------

Work Shown:

|2c+8| = |10c|\\\\2c+8 = 10c \ \text{ or } \ 2c+8 = -10c\\\\8 = 10c-2c \ \text{ or } \ 8 = -10c-2c\\\\8 = 8c \ \text{ or } \ 8 = -12c\\\\8c = 8 \ \text{ or } \ -12c = 8\\\\c = \frac{8}{8} \ \text{ or } \ c = \frac{8}{-12}\\\\c = 1 \ \text{ or } \ c = -\frac{2}{3}

8 0
3 years ago
A rectangle has a side length of (x+2) meters and a width of (2x+3) meters.
Kaylis [27]

Step-by-step explanation:

area=length x width

91=(x+2)(2x+3)

91=x^2+5x+6

x^2+5x-85=0

x=14.2 orx=-24.1

7 0
3 years ago
Find the limit of the function by using direct substitution.
Pavel [41]

ANSWER

lim_{x \to \: 0}( {x}^{2} - 1)  =  - 1

EXPLANATION

The given limit is

lim_{x \to \: 0}( {x}^{2} - 1)

To evaluate this limit by direct substitution,

We put x=0 in the function.

This implies that that ,

lim_{x \to \: 0}( {x}^{2} - 1)  =  {0}^{2}  - 1

This simplifies to,

lim_{x \to \: 0}( {x}^{2} - 1)  = 0 - 1

lim_{x \to \: 0}( {x}^{2} - 1)  =  - 1

This means that as x-values approach zero, the function approaches -1.

7 0
3 years ago
Read 2 more answers
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