1,3,5,7,...,(8th term)

saul85 [17]

$\frac{6}{15}$ = $\frac{2}{c}$ Multiply both sides by 15
6 = $\frac{30}{c}$ Multiply both sides by c
6c = 30 Divide both sides by 6
c = 5

7 0

3 years ago

When Jaylen started high school, 1,430 students attended his school. On the day of his graduation, only 1,001 students attended

e-lub [12.9K]

429 students decreased

8 0

3 years ago

Read 2 more answers

Identify the number of solutions. Y< -1/3x+5, y> -1/3x-1.

Elenna [48]

Answer: C. No solution. ( top shaded. Middle not. Bottom shaded.

Step-by-step explanation:

Given inequalities

y< -1/3x+5 and

y> -1/3x-1.

If we draw graph for y< -1/3x+5 equation, it would be a dotted line and we would shade down the dotted line because we have less than < symbol there.

Now, if we draw graph for second inequality y> -1/3x-1, it would also be a dotted line and we would shade up of dotted line because we have greater than > symbol there.

Now, we can see that slopes of both dotted lines are same that is -1/3.

So, there would not be any common shaded region.

Therefore, there would not by any solution of the system of inequalities.

And correct option is C option.

C. No solution. ( top shaded. Middle not. Bottom shaded.

8 0

3 years ago

the function f(x)= sqrt x is translated left 5 units and up 3 units to create the function g(x). what is the domain of g(x)?

Zinaida [17]

Answer:

x ≥ -5

Step-by-step explanation:

If we have a translation to left c units, we write " x + c " in the function, and

If we have a translation to right c units, we write " x - c" in the function

If we have vertical translation up b units, we "add b to the function", and

If we have vertical translation down b units, we "subtract b to the function"

The parent function is $f(x)=\sqrt{x}$

Since translation left 5 units and up 3 units, we can write:

$f(x)=\sqrt{x+5} + 3$

The domain is affected by the square root sign and we know the number under the square root CANNOT be negative, so we can say:

x + 5 ≥ 0

x ≥ -5

This is the domain.

3 0

3 years ago

Read 2 more answers

A ship leaves port at noon and has a bearing of S29oW. The ship sails at 20 knots. How many nautical miles south and how many na

ira [324]

Answer:

Approximately $58.2\; \text{nautical miles}$ (assuming that the bearing is ${\rm S$29^{\circ}$W}$ .)

Step-by-step explanation:

Let $v$ denote the speed of the ship, and let $t$ denote the duration of the trip. The magnitude of the displacement of this ship would be $v\, t$ .

Refer to the diagram attached. The direction ${\rm S$29^{\circ}$W}$ means $29^{\circ}$ west of south. Thus, start with the south direction and turn towards west (clockwise) by $29^{\circ}$ to find the direction of the displacement of the ship.

The hypothenuse of the right triangle in this diagram represents the displacement of the ship, with a length of $v\, t$ . The dashed horizontal line segment represents the distance that the ship has travelled to the west (which this question is asking for.) The angle opposite to that line segment is exactly $29^{\circ}$ .

Since the hypotenuse is of length $v\, t$ , the dashed line segment opposite to the $\theta = 29^{\circ}$ vertex would have a length of:

$\begin{aligned}& \text{opposite (to $\theta$)} \\ =\; & \text{hypotenuse} \times \frac{\text{opposite (to $\theta$)}}{\text{hypotenuse}} \\ =\; & \text{hypotenuse} \times \sin (\theta) \\ =\; & v\, t \, \sin(\theta) \\ =\; & v\, t\, \sin(29^{\circ})\end{aligned}$ .

Substitute in $\begin{aligned} v &= 20\; \frac{\text{nautical mile}}{\text{hour}}\end{aligned}$ and $t = 6\; \text{hour}$ :

$\begin{aligned} & v\, t\, \sin(29^{\circ}) \\ =\; & 20\; \frac{\text{nautical mile}}{\text{hour}} \times 6\; \text{hour} \times \sin(29^{\circ}) \\ \approx\; & 58.2\; \text{nautical mile}\end{aligned}$ .

7 0

2 years ago