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3 years ago

Translate the following into algebraic form: The difference of forty-seven and a number

Mathematics

1 answer:

JulsSmile [24]3 years ago

7 0

Answer:

Number - 47

Step-by-step explanation:

difference means subtract

Number - 47

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5. Find the 34th term in the arithmetic sequence 9, 6, 3, <br> A. 90<br> B. -90<br> C.-57<br> D. 108

mafiozo [28]

Answer:

B. -90

Step-by-step explanation:

a1=9; d=-3

a34=9-3(33-1)=-90

8 0

3 years ago

How many six digit even numbers can be formed that don't end in 0?

Fudgin [204]

1st digit: 9 possibilities (can’t have 0)
2nd digit: 10 possibilities
3rd digit:10
4th digit:10
5th digit:10
6th digit: 4 possibilities (it can only be even)
9•10•10•10•10•4 =360,000 possibilities

8 0

3 years ago

Can someone pls help me?

nignag [31]

Step-by-step explanation:

ofc , so that is that and the other is that

4 0

3 years ago

write the equation if the line in slope- intercept form that passes through (7,-4) with a slope of -1.

QveST [7]

9514 1404 393

Answer:

y = -x +3

Step-by-step explanation:

The point-slope form can be a useful place to start.

y -k = m(x -h) . . . . . line with slope m through point (h, k)

You require the line ...

y -(-4) = -1(x -7)

y = -x +7 -4 . . . . . . . . eliminate parentheses, add -4

y = -x +3 . . . . . . . . . slope-intercept form

7 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

2 years ago