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3 years ago

A + 1/a= p find a^3 + 1/a^3(1) p^3 + 3p(2) 3p(3) p^3 - 3p

Mathematics

1 answer:

REY [17]3 years ago

4 0

9514 1404 393

Answer:

(3) p^3 -3p

Step-by-step explanation:

(a +1/a) = p . . . . . . . given

(a +1/a)^3 = p^3 . . . . . . . . cube both sides

a^3 +3a^2(1/a) +3a(1/a)^2 +(1/a)^3 = p^3 . . . . . . expand

(a^3 +1/a^3) +3(a +1/a) = p^3 . . . . . . . . . . simplify, group

(a^3 +1/a^3) +3p = p^3 . . . . . . . . . . substitute p for a+1/a

(a^3 +1/a^3) = p^3 -3p . . . . . . subtract 3p from both sides

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Three times my age in 5 years is 54. How old am i now.

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Answer:

D. 13

Step-by-step explanation:

54 ÷ 3 = 18 (your age in 5 years)

18 - 5 = 13 (your age now)

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2 years ago

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the Answer is

4.3213647e+12

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3 years ago

Complete the table of values

Agata [3.3K]

Both problems give you a function in the second column and the x-values. To find out the values of a through f, you need to plug in those x-values into the function and simplify!

You need to know three exponent rules to simplify these expressions:
1) The negative exponent rule says that when a base has a negative exponent, flip the base onto the other side of the fraction to make it into a positive exponent. For example, $3^{-2} =
\frac{1}{3^{2} }$ .
2) Raising a fraction to a power is the same as separately raising the numerator and denominator to that power. For example, $(\frac{3}{4}) ^{3} = \frac{ 3^{3} }{4^{3} }$ .
3) The zero exponent rule says that any number raised to zero is 1. For example, $3^{0} = 1$ .


Back to the Problem:
Problem 1
The x-values are in the left column. The title of the right column tells you that the function is $y = 4^{-x}$ . The x-values are:
1) x = 0
Plug this into $y = 4^{-x}$ to find letter a:
$y = 4^{-x}\\
y = 4^{-0}\\
y = 4^{0}\\
y = 1$

2) x = 2
Plug this into $y = 4^{-x}$ to find letter b:
$y = 4^{-x}\\ 
y = 4^{-2}\\ 
y = \frac{1}{4^{2}} \\ 
y= \frac{1}{16}$

3) x = 4
Plug this into $y = 4^{-x}$ to find letter c:
$y = 4^{-x}\\ 
y = 4^{-4}\\ 
y = \frac{1}{4^{4}} \\ 
y= \frac{1}{256}$


Problem 2
The x-values are in the left column. The title of the right column tells you that the function is $y = (\frac{2}{3})^x$ . The x-values are:
1) x = 0
Plug this into $y = (\frac{2}{3})^x$ to find letter d:
$y = (\frac{2}{3})^x\\
y = (\frac{2}{3})^0\\
y = 1$

2) x = 2
Plug this into $y = (\frac{2}{3})^x$ to find letter e:
$y = (\frac{2}{3})^x\\ y = (\frac{2}{3})^2\\ y = \frac{2^2}{3^2}\\
y = \frac{4}{9}$

3) x = 4
Plug this into $y = (\frac{2}{3})^x$ to find letter f:
$y = (\frac{2}{3})^x\\ y = (\frac{2}{3})^4\\ y = \frac{2^4}{3^4}\\ y = \frac{16}{81}$

-------

Answers:
a = 1
b = $\frac{1}{16}$ 
c = $\frac{1}{256}$
d = 1
e = $\frac{4}{9}$
f = $\frac{16}{81}$

5 0

4 years ago

‼️WILL LIST BRAINLIEST ‼️Carlos went to an Astros game. He spent $25.50 each on two tickets, $10.00 for parking, and $27.75 on f

algol13

Answer:

$88.75

Step-by-step explanation:

since he bought two tickets for $25.5 each you have to do 2(25.50) and since he also spent $10 and $27.75 you have to add that as well.

y=2(25.50)+10+27.75

y=51+10+27.75

y=61+27.75

y=88.75

3 0

3 years ago

The median for the given set of six ordered values is 28.5

Alika [10]

X=the missing value

the median is the middle number, but in this case there are two which happens to be 25 and x
so the formula is

25+x/2=31.5
use algebra to solve
25+x=63
x=63-25
x= 38

Therefore, the missing value is 38.

Hope this helped. Let me know if you have any further questions:)

6 0

3 years ago

A + 1/a= p find a^3 + 1/a^3(1) p^3 + 3p(2) 3p(3) p^3 - 3p​

A + 1/a= p find a^3 + 1/a^3(1) p^3 + 3p(2) 3p(3) p^3 - 3p