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2 years ago

Expression that is equivalent to 5^6

Mathematics

2 answers:

Marta_Voda [28]2 years ago

8 0

Answer:

5^6 = 5*5*5*5*5*5 = 15625

Step-by-step explanation:

definition of exponents

cluponka [151]2 years ago

6 0

Answer:

5 to the 6th power is 5*5*5*5*5*5

Step-by-step explanation:

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Darina [25.2K]

Hey there! I'm happy to help!

Step 2 is incorrect. She added 4 and 2 first, but you are supposed to multiply and divide first. Then she multiplied 6 and 6, which is 36. Then she divided 36 and 2, which is 18, and then she subtracted 7, giving her 11. Step 2 is what threw the whole thing off.

Here is what Sandy should have done.

STEP 1: 6·4+2÷2-7

STEP 2: 24+1-7

STEP 3: 25-7

STEP 4: 18

I hope that this helps! Have a wonderful day! :D

3 0

2 years ago

Read 2 more answers

Given a function I and a subset A of its domain, let I(A) represent the range of lover the set A; that is, I(A) = {I(x) : x E A}

Ede4ka [16]

Step-by-step explanation:

(a)

Using the definition given from the problem

$f(A) = \{x^2 \, : \, x \in [0,2]\} = [0,4]\\f(B) = \{x^2 \, : \, x \in [1,4]\} = [1,16]\\f(A) \cap f(B) = [1,4] = f(A \cap B)\\$

Therefore it is true for intersection. Now for union, we have that

$A \cup B = [0,4]\\f(A\cup B ) = [0,16]\\f(A) = [0,4]\\f(B)= [1,16]\\f(A) \cup f(B) = [0,16]$

Therefore, for this case, it would be true that $f(A\cup B) = f(A)\cup f(B)$ .

(b)

1 is not a set.

(c)

To begin with

$A\cap B \subset A,B$

Therefore

$g(A\cap B) \subset g(A) \cap g(B)$

Now, given an element of $g(A) \cap g(B)$ it will belong to both sets, therefore it also belongs to $g(A\cap B)$ , and you would have that

$g(A)\cap g(B) \subset g(A \cap B)$ , therefore $g(A)\cap g(B) = g(A \cap B)$ .

(d)

To begin with $A,B \subset A \cup B$ , therefore

$g(A) \cup g(b) \subset g(A\cup B)$

7 0

3 years ago

Please help! Thank you!

KIM [24]

The correct answer is: [B]: " (2, 5) ".
__________________________________________
Given:
__________________________________________
-5x + y = -5 ;
  -4x + 2y = 2 .
___________________________________________
Consider the first equation:
___________________________
-5x + y = -5 ; ↔ y + (-5x) = -5 ;

↔ y - 5x = -5 ; Add "5x" to each side of the equation; to isolate "y" on one side of the equation; and to solve in terms of "y".
_____________________________________________
   y - 5x + 5x = -5 + 5x

   y = -5 + 5x ; ↔ y = 5x - 5 ;
____________________________________________
Now, take our second equation:
______________________________
     -4x + 2y = 2 ; and plug in "(5x - 5)" for "y" ; and solve for "x" :
_____________________________________________________
         -4x + 2(5x - 5) = 2 ;
______________________________________________________
Note, 2(5x - 5) = 2(5x) - 2(5) = 10x - 10 ;
__________________________________________
So:   -4x + 10x - 10 = 2 ;

On the left-hand side of the equation, combine the "like terms" ;

-4x +10x = 6x ; and rewrite:

6x - 10 = 2 ;

Now, add "10" to each side of the equation:

6x - 10 + 10 = 2 + 10 ;

to get:

           6x = 12 ; Now, divide EACH side of the equation by "6" ; to isolate "x" on one side of the equation; and to solve for "x" ;

           6x/6 = 12 / 6 ;

                 x = 2 ;
_________________________________
Now, take our first given equation; and plug our solved value for "x" ; which is "2" ; and solve for "y" ;
_____________________________________
-5x + y = -5 ;

-5(2) + y = -5 ;

-10 + y = -5 ; ↔
                              y - 10 = -5 ;

Add "10" to each side of the equation; to isolate "y" on one side of the equation; and to solve for "y" ;

         y - 10 + 10 = -5 + 10 ;

                   y = 5 .
_____________________________
So, we have, x = 2 ; and y = 5 .
____________________________
Now, let us check our work by plugging in "2" for "x" and "5" for "y" in BOTH the original first and second equations:
______________________________
first equation:

-5x + y = -5 ;

-5(2) + 5 =? -5?

-10 + 5 =? -5 ? YES!
______________________
second equation:

-4x + 2y = 2 ;

-4(2) + 2(5) =? 2 ?

-8 + 10 =? 2 ? Yes!
_______________________________________________________
So, the answer is:
___________________________________________________________
          x = 2 , y = 5 ; or, "(2, 5)" ; which is: "Answer choice: [B] " .
___________________________________________________________

3 0

3 years ago

What is the area of a parallelogram if is length is x+1 and height is x+8?

Irina-Kira [14]

Answer:

8 square units

Step-by-step explanation:

x + 1 = 0

x = -1

x + 8 = 0

x = -8

Area = (-1) * (-8) = 8 square units

3 0

2 years ago

Suppose the null hypothesis, H0, is: Darrell has enough money in his bank account to purchase a new television. What is the Type

seropon [69]

<h2>Answer with explanation:</h2>

In statistics, The Type II error occurs when the null hypothesis is false, but fails to be rejected.

Given : Suppose the null hypothesis, $H_0$ , is: Darrell has enough money in his bank account to purchase a new television.

Then , Type II error in this scenario will be when the null hypothesis is false, but fails to be rejected.

i.e. Darrell has not enough money in his bank account to purchase a new television but fails to be rejected.

3 0

2 years ago