Solve the logarithmic equation. Round to the nearest ten-thousandth if necessary. log(x+9)-log x=3

Doomtown is 300 miles due west of Sagebrush and Joshua is due west of Doomtown. At 9 a.m. Mr. Archer leaves Sagebrush for Joshua

e-lub [12.9K]

Refer to the diagram shown below.

Let the distance that Sassoon travels is d miles.
Let his average speed be x mph.
The time of travel is from 1 p.m. to 4 p.m., which is 3 hours.
Therefore
3x = d (1)

The distance that Archer travels is (d + 300) miles.
The time of travel is from 9 a.m. to 4 p.m., which is 7 hours.
The average traveling speed is (x + 20) mph, therefore
7(x + 20) = d + 300
That is,
7x + 140 = d + 300
7x = d + 160 (2)

Subtract (1) from (2).
7x - 3x = d + 160 - d
4x = 160
x = 40 mph (Sassoon's average speed)
x+20 = 60 mph (Archer's average speed)
From (1), obtain d = 120 mi

Answer:
Archer's speed is 60 mph.
Sassoon's speed is 40 mph.

8 0

3 years ago

Please help I’ll give you guys brainiest!<br><br><br> Thank youuuuuuu :))

mariarad [96]

Answer:

look at graph

3 0

3 years ago

Since at t=0, n(t)=n0, and at t=[infinity], n(t)=0, there must be some time between zero and infinity at which exactly half of t

Ostrovityanka [42]

An expression for this timet-half will be given as ln(2) / λ ≈ 0.693 / λ

<h3>What is an expression?</h3>

Expression in maths is defined as the collection of the numbers variables and functions by using signs like addition, subtraction, multiplication and division.

The complete question is:

Suppose a radioactive sample initially contains N0unstable nuclei. These nuclei will decay into stable nuclei, and as they do, the number of unstable nuclei that remain, N(t), will decrease with time. Although there is no way for us to predict exactly when any one nucleus will decay, we can write down an expression for the total number of unstable nuclei that remain after a time t:

N(t)=No e−λt,

where λ is known as the decay constant. Note that at t=0, N(t)=No, the original number of unstable nuclei. N(t) decreases exponentially with time, and as t approaches infinity, the number of unstable nuclei that remain approaches zero.

Part (A) Since at t=0, N(t)=No, and at t=∞, N(t)=0, there must be some time between zero and infinity at which exactly half of the original number of nuclei remain. Find an expression for this time, t half.

Express your answer in terms of N0 and/or λ.

1) Equation given:

← I used α instead of λ just for editing facility..

Where No is the initial number of nuclei.

2) Half of the initial number of nuclei: N (t-half) = No / 2

So, replace in the given equation:

$N9t) = Ne^{-\lambda t}$

3) Solving for α (remember α is λ)

$N_{t-half}=\dfrac{N_o}{2}=N_oe^{-\alpha t}$

$\dfrac{1}{2} = e^{-\alpha t}2=e^{-alpha t}\alpha t=In(2)$

αt ≈ 0.693

⇒ t = ln (2) / α ≈ 0.693 / α ← final answer when you change α for λ

Therefore expression for this timet-half will be given as ln(2) / λ ≈ 0.693 / λ

To know more about an expression follow

brainly.com/question/723406

#SPJ4

3 0

2 years ago

What is the factorization of the trinomial below -x^2+x+42

aksik [14]

For the given trinomial

$a=-1, b=1 ,c=42$

We need to multiply a by c
$-1 \times 42 = - 42$

We split the middle term with these two factors
$7,-6$
To obtain,

$- {x}^{2} + 7x -6x + 42$
We now factor

$- {x}^{2} + x + 42 \: = x( - x + 7) + 6( - x + 7)$
This implies

$- {x}^{2} + x + 42: = (x + 6)(7 - x)$

4 0

4 years ago

Read 2 more answers

What is the probability of getting a number less than four on a fair die and a tail on a coin flip?

Elanso [62]

Probability of getting a number less than 4 is 3/6 or 1/2.
Probability of getting a tail is 1/2

1/2*1/2= 1/4 or 25% or 0.25

3 0

3 years ago