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3 years ago

Please help me with this problemln -3x + ln 4 = ln 58

Mathematics

2 answers:

allsm [11]3 years ago

7 0

9514 1404 393

Answer:

x = -29/6 = -4 5/6

Step-by-step explanation:

The equation simplifies to ...

ln(-3x·4) = ln(58)

Taking antilogs, the equation becomes ...

(-3x)(4) = 58

Dividing by -12 shows us the value of x.

x = -58/12

x = -29/6 = -4 5/6

The applicable rule of logarithms is ...

ln(ab) = ln(a) +ln(b)

Artist 52 [7]3 years ago

5 0

Answer:

7x - 4x = 21

3x = 21

x = 21÷3

x = 7

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Which statement about convergent infinite

balandron [24]

Answer:

B) Convergent infinite geometric series sum to a single value

Step-by-step explanation:

An infinite geometric series of the form $\[ \sum_{r=1}^{\infty} ar^{n-1}$ converges if $|r|$ , so the summation of the series will result in a single value if that is true.

8 0

3 years ago

Try the following example. Remember to use the hints if you need them. Solve for x. 2.9×10−7=(x)(x)2(4.6)(4.6)2(3)2 Choose the c

galben [10]

Answer:

$2.3\times 10^{-2}$

Step-by-step explanation:

Given equation,

$2.9\times 10^{-7}=\frac{(x)(x)^2(4.6)}{(4.6)^2(3)^2}$

$2.9\times 10^{-7}=\frac{x^3}{9\times 4.6}$

$(\because a^m = \frac{1}{a^{-m}})$

By cross multiplication,

$2.9\times 9\times 4.6\times 10^{-7}=x^3$

$120.06\times 10^{-7}=x^3$

$\implies x = (120.06\times 10^{-7})^\frac{1}{3}$

$=0.0228980999294$

$\approx 0.023$

$=2.3\times 10^{-2}$

7 0

3 years ago

In a certain city, the daily consumption of water (in millions of liters) follows approximately a gamma distribution with α = 2

Ne4ueva [31]

Answer:

The probability that on any given day the water supply is inadequate $0.1991$

Step-by-step explanation:

Given

α = 2 and β = 3

As per Gamma distribution Function

$P(X>9)= 1-P(X\leq 9)\\$

Expanding the function and putting the given values, we get -

$[tex]1 - \int\limits^9_0 {f(x;2,3)} \, dx \\1- \int\limits^0_0 {\frac{1}{9}xe^{\frac{-x}{3} } \, dx\\\\= 1- \frac{1}{9} [x(-3e^{\frac{-x}{3}}) -\int\limits {(-3e^{\frac{-x}{3}})} \, dx]^9_0= 1- 1/9 [x(-3e^{\frac{-x}{3}}) -9e^{\frac{-x}{3}})} \, dx]^9_0\\1-((\frac{-1}{3} *9*e^({\frac{-9}{3}})- e^(\frac{-9}{3}))- ((\frac{-1}{3} *0*e^(\frac{-9}{3})-e^{\frac{-0}{3}})\\1-((-3e^{\frac{-9}{3} }-e^{\frac{-9}{3}}-(0-1))\\1-(1-4e^{-3})\\1-(1-0.1991)\\1-0.8009\\0.1991$