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olga nikolaevna [1]
3 years ago
14

What should substitution should've ended used to rewrite x^8-3x^4+2=0 as a quadratic equation

Mathematics
1 answer:
iogann1982 [59]3 years ago
6 0

Step-by-step explanation:

is the answer something like x^12=-2

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Solve for x<br> questions 3 and 5!!
KIM [24]

Answer:

3. x = 1<x−2<71<x-2<7

         9<x+7<109<x+7<10

         7<x+4<12

5. x = 1<x−2<71<x-2<7

         9<x+7<109<x+7<10

         7<x+4<12

Step-by-step explanation:

Can I have brainliest?

4 0
3 years ago
Katy invests $18,000 at 14% simple interest for 1 year. How much is in the account at the end of the 1 year period? Answer: $___
Nitella [24]

Answer:

$20,520 at the end of the year

Step-by-step explanation:

A = P(1 + rt)

   A = Total Accrued Amount (principal + interest)

   P = Principal Amount

   I = Interest Amount

   r = Rate of Interest per year in decimal; r = R/100

   R = Rate of Interest per year as a percent; R = r * 100

   t = Time Period involved in months or years

P = $18,000

r = .14

t = 1

A = 18,000 (1+ ( .14 * 1 ) )

A = 18,000 (1 +.14)

A = 18,000 (1.14)

A = 20,520

$20,520 at the end of the year

8 0
3 years ago
PLEASE ANSWER ASAP LOOK AT THE PICTURE DUE IN 2 SECONDS !!!!!!!!!!!!!!
katrin2010 [14]
Answer: x> 30

Good luck !
3 0
3 years ago
Read 2 more answers
A movie theatre sells 20 orders of nachos, 25 candy bars, 40 boxes of popcorn and 45 sodas. Which ratio is the greatest?
Rus_ich [418]
<span>B. Number of boxes to popcorn to number of sodas  i hope  this helps you so much </span>
7 0
3 years ago
1 If a = p^1/3-p^-1/3<br>prove that: a^3 + 3a = p - 1/p​
alexandr402 [8]

Hello, please consider the following.

We know that

a = p^{\frac{1}{3}}-p^{-\frac{1}{3}}\\\\=p^{\frac{1}{3}}-\dfrac{1}{p^{\frac{1}{3}}}

And we can write that.

(p-\dfrac{1}{p})^3=(p-\dfrac{1}{p})(p^2-2+\dfrac{1}{p^2})\\\\=p^3-2p+\dfrac{1}{p}-p+\dfrac{2}{p}-\dfrac{1}{p^3}\\\\=p^3-\dfrac{1}{p^3}-3(p-\dfrac{1}{p})

It means that, by replacing p by p^{1/3}

(p^{1/3}-\dfrac{1}{p^{1/3}})^3=p-\dfrac{1}{p}-3(p^{1/3}-\dfrac{1}{p^{1/3}})\\\\\\\text{ So }\\\\a^3=p-\dfrac{1}{p}-3a\\\\\boxed{ a^3+3a=p-\dfrac{1}{p} }

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

4 0
3 years ago
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