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3 years ago

What is the equivalent logarithmic for of ∛8 =2

1 answer:

8 0

If a^b = c, then log(base a)(c) = b. It becomes a question - to what power bust we raise a to get c?

In this case a = 8, b = (1/3), c = 2. Therefore:

log(base 8)(2) = (1/3)

I hope this helps :)

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Each of three coins has two sides, heads and tails. Represent the heads or tails status of each coin by a logical variable (A fo

jeka94

Answer:

(a) F(A, B, C) = AB'C' + A'BC' + A'B'C

(b) F(A, B, C) = (A' + B' + C')(A' + B + C)(A + B' + C)(A + B + C')(A + B + C)

Step-by-step explanation:

(a) If F(A, B, C) = 1 iff exactly one of the coins is heads, then either

A is heads and the others are tails (AB'C')

B is heads and the others are tails (A'BC')

C is heads and the others are tails (A'B'C)

Hence, as a minterm expansion,

F(A, B, C) = AB'C' + A'BC' + A'B'C

(b) To get the corresponding maxterm expansion, we convert to binary.

$F(A, B, C) = \sum (100, 010, 001) = \sum(4,2,1)$

The maxterm is the product of the complements.

$F(A, B, C) = \prod (0, 3, 5, 6, 7) = \prod(000, 011, 101, 110, 111)$

Expanding,

F(A, B, C) = (A' + B' + C')(A' + B + C)(A + B' + C)(A + B + C')(A + B + C)

3 0

3 years ago

Subtract- a(b-5) from b(5-a)

TiliK225 [7]

Answer:

5b - 5a

Step-by-step explanation:

b(5 - a) = 5b - ab

- a(b - 5) = - ab + 5a

Subtracting gives

5b - ab - (- ab + 5a) ← distribute parenthesis by - 1

= 5b - ab + ab - 5a ← collect like terms

= 5b - 5a

3 0

3 years ago

Evaluate the following. 4x4+6divided by2-5

LenKa [72]

Factor the numerator and denominator and cancel the common factors.
4x^4 - 2

8 0

3 years ago

Read 2 more answers

Frederick works in a grocery store and also mows lawns . He earns $ 10 per hour at the grocery store and $ 8.50 per hour mowing

Mariana [72]

Answer:

just multiply 35 by x and then 10 by however many lawns he mowed (if that makes sense)

Step-by-step explanation:

8 0

3 years ago

Divide.<br> a.1 − 2i/2i<br> b. 5 − 2i/5 + 2i<br> c.√3 − 2i/−2 − √3i

Umnica [9.8K]

Answer:

(a) \frac{-1i}{2}-1[/tex]

(b) $\frac{20i+21}{29}$

Step-by-step explanation:

We have to perform division

(a) $\frac{1-2i}{2i}$

So after division

$\frac{1-2i}{2i}=\frac{1}{2i}-\frac{2i}{2i}=\frac{-1i}{2}-1$

(b) We have given expression $\frac{5-2i}{5+2i}$

After rationalizing $\frac{5-2i}{5+2i}\times \frac{5-2i}{5-2i}=\frac{(5-2i)^2}{25+4}=\frac{25+4i^2+20i}{29}=\frac{20i+21}{29}$

After rationalizing

$\frac{\sqrt{3}-2i}{-2-\sqrt{3i}}\times \frac{-2+\sqrt{3i}}{-2+\sqrt{3i}}=\frac{-2\sqrt{3}+7i+2\sqrt{3}}{7}=\frac{7i}{7}=i$

5 0

4 years ago