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Dvinal [7]
2 years ago
7

Evaluate the function

Mathematics
1 answer:
Sedaia [141]2 years ago
4 0

f(0) =  {4}^{3 \times 0  + 3}  =  {4}^{0 + 3}  =  {4}^{3}  = 64
The answer is above.
Additional note 1:
{4}^{3}  = 4 \times 4 \times 4 = 16 \times 4 = 64
Additional note 2:
In the answer I filled in "0" at the place where "n" was. This is because the question tells us n=0
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What are the equations that have the same solutions as 3x-12=24 between
Effectus [21]

Answer:

ddqsc

Step-by-step explanation:

rdfgrrffferrrrrddr

5 0
2 years ago
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HELLPPPPPPPPPPPPPPPPPPPPPPPPP
navik [9.2K]

Answer:

more the letter x is often used in algebra to mean a value that is not yet known is is calld a variable or sometimes an unknown. in x+2=7,x is a variable but we can work out its value if we try

5 0
2 years ago
Find the least common multiple for 6(x+1)^3(x-4)^2 and 10(x+1)^8(x-4)^5
dolphi86 [110]

Answer:

30(x+1)^8 (x-4)^5

Step-by-step explanation

<h3>6(x+1)^{3}(x-4))^{2}=2X3 X (x+1)^3(x-4)^2\\10(x+1)^8(x-4)^5)= 2X5X(x+1)^8 (x-4)^5\\</h3>

In order to find the Least common multiple we write each of the factor only once from each of the expression and for the common expression we take their LCM as maximum of the exponent in  all expressions

as here in the question exponent of (x+1) are 3 and 8 so we take exponent 8

likewise for (x-4) we shall take maximum of 2 and 5 which is 5

so our expression for Least common multiple will be

2X3X5 X (x+1)^8 (x-4)^5

30(x+1)^8 (x-4)^5

7 0
3 years ago
Help please use picture
Katarina [22]
I think it is 22, but I did it in my head so i might be way off

What you do is you “take apart” the net so it has individual rectangles, then you multiply the 2 sides... you do that for each rectangle then add it all up
4 0
3 years ago
if a right triangle has one side measuring 3√2 and another side measuring 2√3, what is the length of the hypotenuse?
lina2011 [118]

Answer:

<h2>\sqrt{30}</h2>

Step-by-step explanation:

Given,

Perpendicular ( p ) = 3√2

Base ( b ) = 2√3

Hypotenuse ( h ) = ?

Now, let's find the length of the hypotenuse:

Using Pythagoras theorem:

{h}^{2}  =  {p}^{2}  +  {b}^{2}

plug the values

{h}^{2}  =  {(3 \sqrt{2} )}^{2}  +  {(2 \sqrt{3} )}^{2}

To raise a product to a power, raise each factor to that power

{h}^{2}  = 9 \times 2 + 4 \times 3

Multiply the numbers

{h}^{2}  = 18 + 12

Add the numbers

{h }^{2}  = 30

Take the square root of both sides of the equation

h =  \sqrt{30}

Hope this helps...

Best regards!!

6 0
3 years ago
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