PLZ HELP!!!!!!!!!!!!!!!!!!!!

Just question part d) pls

Vinil7 [7]

Answer:

fffffffffffffffffffffffffffffffffff

Step-by-step explanation:

fffffffffffffffff

5 0

2 years ago

Tasha wants to graph a line that is parallel to the line with equation y = 3x. Give two examples of linear equations that repres

Arturiano [62]

When two lines are <em><u>parallel, that means they have the same slope! </u></em>

The slope is the coefficient of x

In the equation y = 3x, <u>the slope is 3 </u>

An example of a line that is above y = 3x is ...

y = 3x + 2

The "+2" <em>shifts the line up </em>by 2 on the y axis

An examples of a line that is below y = 3x is ...

y = 3x - 3

The "-3" <em>shifts the line down</em> by 3 on the y axis

5 0

2 years ago

Solve for x. x-17=-57<br> i need help

seraphim [82]

X - 17 = -57
-17. -17
x = -74

hope this helps xo

7 0

2 years ago

The doubling period of a bacterial population is 20 20 minutes. At time t = 100 t=100 minutes, the bacterial population was 9000

Nuetrik [128]

Answer:

The initial population was 2810

The bacterial population after 5 hours will be 92335548

Step-by-step explanation:

The bacterial population growth formula is:

$P = P_0 \times e^{rt}$

where P is the population after time t, $P_0$ is the starting population, i.e. when t = 0, r is the rate of growth in % and t is time in hours

Data: The doubling period of a bacterial population is 20 minutes (1/3 hour). Replacing this information in the formula we get:

$2 P_0 = P_0 \times e^{r 1/3}$

$2 = e^{r \; 1/3}$

$ln 2 = r \; 1/3$

$ln 2 \times 3 = r$

$2.08 \% = r$

Data: At time t = 100 minutes (5/3 hours), the bacterial population was 90000. Replacing this information in the formula we get:

$90000 = P_0 \times e^{2.08 \; 5/3}$

$\frac{9000}{e^{2.08 \; 5/3}} = P_0$

$2810 = P_0$

Data: the initial population got above and t = 5 hours. Replacing this information in the formula we get:

$P = 2810 \times e^{2.08 \; 5}$

$P = 92335548$

3 0

2 years ago

Use the following information for problems 1 – 3. Suppose you sign a contract for an annual salary of $50,000 with a guaranteed

erik [133]

Initial salary = $50,000 .

Rate of raise = 5% each year.

Therefore, each next year salary would be 105% that is 1.05 times.

5% of 50,000 = 0.05 × 50000 = 2500.

Therefore raise is $2500 each year.

According to geometric sequence first term 50000 and common ratio 1.05.

Applying geometric sequence formula

$a_n = ar^{n-1}$

1) $a_n = 50000(1.05)^{n-1}$

2) In order to find salary in 5 years we need to plug n=5, we get

$a_5 = 50000(1.05)^{5-1}= 50000(1.05)^4$

= 50000(1.21550625)

3) In order to find the total salary in 10 years we need to apply sum of 10 terms formula of a geometric sequence.

$S_n = \frac{a(1-r^n}{1-r}$

Plugging n=10, a = 50000 and r= 1.05.

$S_10 = \frac{50000(1-(1.05)^{10}}{1-1.05}$

$S_10 = \frac{50000(0.050)^{10}}{0.05}$

= 628894.62678.

<h3>Therefore , you will have earned $ 628894.62678 in total salary by the end of your 10th year.</h3>

7 0

2 years ago