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Levart [38]
3 years ago
14

-2x - 5 = 9 Please answer quick!

Mathematics
1 answer:
Mice21 [21]3 years ago
5 0

Answer:

X = -7

Step-by-step explanation:

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The value of an investment A (in dollars) after t years is given by the function A(t) = A0ekt. If it takes 10 years for an inves
Ludmilka [50]

Answer:

20 years.

Step-by-step explanation:

We have been given a formula A(t)=A_0\cdot e^{kt}, which represents the  value of an investment A (in dollars) after t years.

Substitute the given values:

\$3,000=\$1,000\cdot e^{k*10}

Let us solve for k.

\frac{\$3,000}{\$1,000}=\frac{\$1,000\cdot e^{k*10}}{\$1,000}

3=e^{k*10}

Take natural log of both sides:

\text{ln}(3)=\text{ln}(e^{k*10})

Using property \text{ln}(a^b)=b\cdot \text{ln}(a), we will get:

\text{ln}(3)=10k\cdot\text{ln}(e)

We know that \text{ln}(e)=1, so

\text{ln}(3)=10k\cdot 1

\text{ln}(3)=10k

\frac{\text{ln}(3)}{10}=\frac{10k}{10}

\frac{\text{ln}(3)}{10}=k

\$9,000=\$1,000\cdot e^{\frac{\text{ln}(3)}{10}*t}

Dividing both sides by 1000, we will get:

9=e^{\frac{\text{ln}(3)}{10}*t}

Take natural log of both sides:

\text{ln}(9)=\text{ln}(e^{\frac{\text{ln}(3)}{10}*t)

\text{ln}(9)=\frac{\text{ln}(3)}{10}*t\cdot\text{ln}(e)

\text{ln}(9)=\frac{\text{ln}(3)}{10}*t\cdot1

10*\text{ln}(9)=10*\frac{\text{ln}(3)}{10}*t

10\text{ln}(9)=\text{ln}(3)*t

10\text{ln}(3^2)=\text{ln}(3)*t

2\cdot 10\text{ln}(3)=\text{ln}(3)*t

20\text{ln}(3)=\text{ln}(3)*t

Divide both sides by \text{ln}(3):

\frac{20\text{ln}(3)}{\text{ln}(3)}=\frac{\text{ln}(3)*t}{\text{ln}(3)}

20=t

Therefore, it will take 20 years for the investment to be $9,000.

5 0
3 years ago
When you send out a resume, the probability of being called for an interview is 0. 40. what is the probability that the first in
Kipish [7]

The probability that the first interview occurs on the fifth resume that you sent is 0.2592.

According to the given question.

The probability of being called for an interview, p = 0.40.

So, the probability of not being called for an interview, q = 1 - 0.40 = 0.60

As we know that binomial distribution summarizes the number of trials, or observations when each trial has the same probability of attaining one particular value. The binomial distribution determines the probability of observing a specified number of successful outcomes in a specified number of trials.

Therefore, the probability that the first interview occurs on the fifth resume that you send out

= ^{5} C_{1} (0.40)^{1} (0.60)^{4}

= 5(0.40)(0.1296)

= 0.2592

Hence, the probability that the first interview occurs on the fifth resume that you sent is 0.2592.

Find out more information about probability here:

brainly.com/question/14210034

#SPJ4

3 0
2 years ago
When does a compound inequality with OR have a solution that is the entire number line? Can the solution be bounded to one side?
dalvyx [7]

Answer:

Concept: Algebraic Analysis

  1. \geqslant   \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \leqslant
  2. Those two symbols indicate one solution
  3. <  \:  \:  \:  \:  \:  \:  \:  >
  4. Those symbols indicate a range of values
  5. x < 15 < x
  6. Indicates two bounded answers
  7. In an inequality there is always a solution
6 0
3 years ago
Given the formula for the perimeter of a rectangle where I represents the length and w represents the width 2(l+w)
s344n2d4d5 [400]
2 represents area Becuase of the area formula
8 0
3 years ago
Find how many kilometers they traveled by
Murrr4er [49]

Answer:

Step-by-step explanation:

Biking is x and bussing is y. We know that the total distance is 325. Thus, the first equation in this system is

x + y = 325 (which says that the km traveled by bike plus the km traveled by bus totaled 325 km).

We also know that the pair biked 75 km more than they rode, giving us the second equation in our system:

x = y + 75. Now we use simple substitution and plug y + 75 in for x in the first equation:

(y + 75) + y = 325 and

2y + 75 = 325 and

2y = 250 so

y = 125 km. They rode the bus for 125 km and biked for 325 - 125 which is 200. And the difference is 75 km, as it should be!

7 0
3 years ago
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