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3 years ago

the graph of y= 1/x is translated 3 units up and 2 units to the left. what is an equation of the translated graph

Mathematics

1 answer:

wel3 years ago

8 0

-Move 2 units left:y=(1/x +2)
-Move 3 units up:y= (1/x +2)+3
So the equation is y= (1/x +2)+3

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nika2105 [10]

You are doing the distributive property.
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3 years ago

1. What is the volume, in cubic centimeters, of the sphere shown below? Use 3.14<br> for Pi

prohojiy [21]

Answer:

113.04 cm³

Step-by-step explanation:

diameter = 6cm

so radius = 3cm

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= 4 * 28.26

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3 years ago

H(x) = x2<br> j(x) = 2x2-5<br> find h(x) - j(x)

yKpoI14uk [10]

Answer:

Step-by-step explanation:

h(x) - j(x) => $x^2-(2x^2-5)$

The negative sign out in front of the set of parenthesis changes the signs of everything inside the parenthesis, so now we have, without parenthesis:

$x^2-2x^2+5$ which simplifies down to

$-x^2+5$

4 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 $t = 6\; \text{hour}$ :

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

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2 years ago

What do these circles have in common?

Mnenie [13.5K]

They are all circles

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3 years ago

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