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hram777 [196]
3 years ago
12

81^5=3^x What does “x” equal?

Mathematics
1 answer:
Lynna [10]3 years ago
7 0

Answer:

x =2 0

Step-by-step explanation:

{81}^{5}  =  {3}^{x}

We know that 3^4 = 81. So :

=  >  { ({3}^{4}) }^{5}  =  {3}^{x}

=  >  {3}^{4 \times 5}  =  {3}^{x}

=  >  {3}^{20}  =  {3}^{x}

Here both the Right hand side & Left hand side has same base i.e. have 3 as base in both the sides.

So , eliminating the base gives :-

x = 20

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Y=2x+6 and y=-3x+16 for solving the system
Butoxors [25]
Substitution

Step 1: Solve for x or y in either of the two equations. In the equations that you are given, y is already solved for you.

Step 2: Plug in 2x + 6 into the opposite equation.

   2x + 6 = -3x + 16
         - 6              - 6
------------------------------
   2x = -3x + 10
+ 3x  + 3x
------------------------------
   5x  =  10
 ------   ------
    5        5

x = 2

Step 3: Plug in the x value into either equation.

y = 2(2) + 6
y = 4 + 6
y = 10

The final answer gets expressed as a coordinate pair: (x,y) (2,10)

I hope this helps you :)




4 0
3 years ago
If f(x) = 1 + x2 and g(x) = 1-, which is the rule of function (f-)(x)?
STatiana [176]

Answer:

= -x^2 +x    

Step-by-step explanation:

(x) = 1 - x^2

g(x) = 1-x,

(f-g)(x) = 1-x^2 - ( 1-x)

Distribute the minus sign

          = 1-x^2 -1 +x

  Combine like terms

          = -x^2 +x    

8 0
3 years ago
Drag the expressions to the boxes to order them from least to greatest.
Lelechka [254]

Answer:

Step-by-step explanation:

more info sorry

4 0
3 years ago
The length of a rectangle is 4 more than its width. The area of the rectangle is 21m^2 . What is the measure of the width?
saw5 [17]
Make the length x+4 and the width x  and then make a formula x(x+4)=21
then u ll get x^2+4x-21=0  and then solve for x ..... the width is 3 

5 0
3 years ago
Solving Rational equations. LCD method. Show work. Image attached.
posledela

(k-2)(k-6)=k^2-2k-6k+12=k^2-8k+12

So in order to get all the fractions to have a common denominator, we need to multiply \dfrac k{k-2} by \dfrac{k-6}{k-6}, and \dfrac1{k-6} by \dfrac{k-2}{k-2}:

\dfrac k{k-2}\cdot\dfrac{k-6}{k-6}=\dfrac{k(k-6)}{(k-2)(k-6)}=\dfrac{k^2-6k}{k^2-8k+12}

\dfrac1{k-6}\cdot\dfrac{k-2}{k-2}=\dfrac{k-2}{(k-2)(k-6)}=\dfrac{k-2}{k^2-8k+12}

Now,

\dfrac4{k^2-8k+12}=\dfrac{(k^2-6k)+(k-2)}{k^2-8k+12}

As long as k\neq2 and k\neq6 (which we can't have because otherwise k^2-8k+12=0), we can cancel k^2-8k+12 in the denominators on both sides:

4=(k^2-6k)+(k-2)

4=k^2-5k-2

0=k^2-5k-6

We can factorize the right side:

0=(k-6)(k+1)

which tells us that k=6 and k=-1 are solutions.

4 0
4 years ago
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