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2 years ago

A^2+B^2=C^2 B=2. 1 C=2.9

1 answer:

3 0

Are you being asked to solve for A? B?
$A^{2} + B ^{2} = C^{2}$
$A ^{2} +B ^{2} =(2.9) ^{2}$
$A ^{2} + B^{2} =8.41$

Solve for A:
$A^2=8.41-B^2$
Take a plus of minus square root
A= $\sqrt{-b^2+8.41}$
A= $- \sqrt{-b^2+8.41}$

Solve B:
$A^2+B^2=8.41$
$B^2=-A^2+8.41$
Take plus or minus square roots
$B= \sqrt{A^2+8.41}$
$B= -\sqrt{A^2+8.41}$

You might be interested in

What is each power of i with its multiplicative inverse.

pentagon [3]

Answer:

1) multiplicative inverse of i = -i

2) Multiplicative inverse of i^2 = -1

3) Multiplicative inverse of i^3 = i

4) Multiplicative inverse of i^4 = 1

Step-by-step explanation:

We have to find multiplicative inverse of each of the following.

1) i

The multiplicative inverse is 1/i

if i is in the denominator we find their conjugate

$=1/i * i/i\\=i/i^2\\=We\,\, know\,\, that\,\, i^2 = -1\\=i/(-1)\\= -i$

So, multiplicative inverse of i = -i

2) i^2

The multiplicative inverse is 1/i^2

We know that i^2 = -1

1/-1 = -1

so, Multiplicative inverse of i^2 = -1

3) i^3

The multiplicative inverse is 1/i^3

We know that i^2 = -1

and i^3 = i.i^2

$1/i^3\\=1/i.i^2 \\=1/i(-1)\\=-1/i * i/i\\=-i/i^2\\= -i/-1\\= i$

so, Multiplicative inverse of i^3 = i

4) i^4

The multiplicative inverse is 1/i^4

We know that i^2 = -1

and i^4 = i^2.i^2

$=1/i^2.i^2\\=1/(-1)(-1)\\=1/1\\=1$

so, Multiplicative inverse of i^4 = 1

3 0

3 years ago

Solve (1/81)^x*1/243=(1/9)^−3−1 by rewriting each side with a common base.

elena55 [62]

Answer:

$x=(243)log_{\frac{1}{81}}[(\frac{1}{81})-1]$

Step-by-step explanation:

you have the following formula:

$(\frac{1}{81})^{\frac{x}{243}}=(\frac{1}{9})^{-3}-1$

To solve this equation you use the following properties:

$log_aa^x=x$

Thne, by using this propwerty in the equation (1) you obtain for x

$log_{(\frac{1}{81})}(\frac{1}{81})^{\frac{x}{243}}=log_{\frac{1}{81}}[(\frac{1}{81})-1]\\\\\frac{x}{243}=log_{\frac{1}{81}}[(\frac{1}{81})-1]\\\\x=(243)log_{\frac{1}{81}}[(\frac{1}{81})-1]$