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dedylja [7]
3 years ago
15

PLEASE HELP!!!! Liner equations & graphs!!! Can someone please help me with this.

Mathematics
1 answer:
Alexxandr [17]3 years ago
4 0

Answer:

It would be A

Step-by-step explanation:

Since the starting amount was $11 you would look how much it was for 2 rides and subtract 17-11 than that is 6 so for two rides it is 6 dollars extra. So than for 1 ride you would divide 6 by 2 and get 3.

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12-5[2(8+5)-15]+2(7-10)³<br>show work!​
san4es73 [151]

Answer: -97

Step-by-step explanation:

3 0
3 years ago
The graph shows two polygons ABCD and A′B′C′D′.
stellarik [79]

Answer:

Translate ABCD down 1, then reflect it over the y-axis

and

Reflect ABCD over the y-axis, then translate down 1

Step-by-step explanation:

First of all, the two steps are the same, just in a different order

Second, if you look at the instructions you can see what would happen to the quadrilateral. Reflecting over the y-axis would make the shape flip horizontally (left to right), as shown in the image. This means the points on the left move to the right, and the points on the right move to the left. Top and bottom points stay the same. And the "translation" just means to slide. In both the answers, it says "translate 1 unit down", which just means "move 1 unit down", which is exactly what happens in the image

7 0
3 years ago
Read 2 more answers
$5000 grew to $9000 at the compound interest of 12% per annum what was the years taken for the investment​
snow_tiger [21]

Answer:

1year?hehehehehehehehehe

3 0
3 years ago
6. The angle of elevation from a small boat to the top of a lighthouse is 25°. If the top of the lighthouse is 150
Anna007 [38]

Answer:

321.68feet

Step-by-step explanation:

Given the following

Top of the house above the sea level  = 150feet = opposite

Angle of elevation = 25°

Required

Distance from the boat to the base of the house(adjacent)

Using the SOH CAH TOA identity

tan theta = opp/adj

tan 25 = 150/d

d = 150/tan25

d = 150/0.4663

d = 321.68

Hence the required distance is  321.68feet

3 0
3 years ago
What are expressions for MN and LN? Hint Construct the altitude from M to LN.
Nikitich [7]

The question is missing the figure. So, it is in the atachment.

Answer: MN = x\sqrt{2}  LN = \frac{x}{2}.(\sqrt{2} + \sqrt{6} )

Step-by-step explanation: The first figure in the attachment is the figure of the question. The second figure is a way to respond this question by tracing the altitude from M to LN as suggested. When an altitude is drawn, it forms a 90° angle with the base, as shown in the drawing. To determine the other angle, you have to remember that all internal angles of a triangle sums up to 180°.

For the triangle <u>on the left</u> of the altitude:

45+90+angle=180

angle = 45

For the triangle <u>on the right</u>:

30+90+angle=180

angle = 60

With the angles, use the Law of Sines, which is relates sides and angles, as follows:

\frac{a}{sinA} = \frac{b}{sinB} = \frac{c}{sinC}

For MN:

\frac{x}{sin(30)} = \frac{MN}{sin(45)}

MN = \frac{x.sen(45)}{sen(30)}

MN = x\sqrt{2}

For LN:

\frac{LN}{sen(105)} =\frac{x}{sin(30)}

LN = \frac{x.sin(105)}{sin(30)}

We can determine sin (105) as:

sin(105) = sin(45+60)

sin(105) = sin(45)cos(60) + cos(45)sin(60)

sin(105) = \frac{\sqrt{2} }{2}.\frac{1}{2} + \frac{\sqrt{2} }{2}.\frac{\sqrt{3} }{2}

sin(105) = \frac{\sqrt{2} }{4} + \frac{\sqrt{6} }{4}

LN = \frac{x.sin(105)}{sin(30)}

LN = x.(\frac{\sqrt{2} }{4} + \frac{\sqrt{6} }{4}  ) .2

LN = \frac{x}{2}.(\sqrt{2} + \sqrt{6} )

The expressions for:

MN = x\sqrt{2}

LN = \frac{x}{2}.(\sqrt{2} + \sqrt{6} )

6 0
3 years ago
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