Evaluate using the order of operations. 54 + (9.2

ExtremeBDS [4]

More information please

4 0

3 years ago

Algebra 2, I need help!!! Solve x^2 + 6x + 7 = 0. If you are going to comment in here please know the answer, this is so serious

docker41 [41]

Answer:

Third option

Step-by-step explanation:

We can't factor this so we need to use the quadratic formula which states that when ax² + bx + c = 0, x = (-b ± √(b² - 4ac)) / 2a. However, we notice that b (which is 6) is even, so we can use the special quadratic formula which states that when ax² + bx + c = 0 and b is even, x = (-b' ± √(b'² - ac)) / a where b' = b / 2. In this case, a = 1, b' = 3 and c = 7 so:

x = (-3 ± √(3² - 1 * 7)) / 1 = -3 ± √2

6 0

3 years ago

2x + 4y = 14 and 3x – 4y = -39 solve by elimination

gizmo_the_mogwai [7]

Answer: x=-5 and y=6

Step-by-step explanation: i hope i did it right tell if i didnt

5 0

3 years ago

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

3 years ago

Pls help with mathh!!!

sergeinik [125]

Answer: x<5

Step-by-step explanation:

If the circle was dark then it would be x is less than or equal to 5. But here, since the circle is open, that means that 5 is not a possible solution for x. The answer is x<5. Hope this helps!

6 0

3 years ago

Evaluate using the order of operations. 54 + (9.2 - 1.413)