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saul85 [17]
3 years ago
5

The graph that best represents y = e x + 1 is what?

Mathematics
1 answer:
velikii [3]3 years ago
4 0
I'ts 5x, 5x and then x = 1 is 2 and then y = e equals 3 and then 2x + 3 = 5x
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Basher collected 4 1/3 basket of peaches and Orchard if each basket holds 21
miv72 [106K]
To do this you times 3 by 4 and add it on to the numerator of the fraction, to turn it into a top heavy fraction:
3*4 = 12
1+12/3 = 13/3

Then you multiply the numerator by 21 to work out what 21 times the fraction is:
13*21/3 = 273/3

Then you can divide 273 by 3 to get the final answer:
273/3 = 91

He will have 91 peaches overall.
Hope this helps! :)
6 0
3 years ago
Please help me ASAP. For homework due right now
sveta [45]

Answer:

at least 20 rows.

Step-by-step explanation:

when you add the students, teachers, and chaperones, it comes to 159. then you divide 159 by the number of seats each row has, so 159/8 = 19.875, you would have to round to the nearest whole number so 20 should be correct.

5 0
3 years ago
Annette plans to visit an amusement park where she must pay for admission and
natulia [17]

Answer:

...

Step-by-step explanation:

7 0
3 years ago
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I need help finding out what AC is!!
iragen [17]
I think it’s b.5. Hope this helps. :)
5 0
3 years ago
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A person on a runway sees a plane approaching. The angle of elevation from the runway to the plane is 11.1° . The altitude of th
Gnoma [55]

Answer:

The horizontal distance from the plane to the person on the runway is 20408.16 ft.

Step-by-step explanation:

Consider the figure below,

Where AB represent altitude of the plane is 4000 ft above the ground , C represents the runner.  The angle of elevation from the runway to the plane is 11.1°

BC is the horizontal distance from the plane to the person on the runway.

We have to find distance BC,

Using trigonometric ratio,

\tan\theta=\frac{Perpendicular}{base}

Here, \theta=11.1^{\circ} ,Perpendicular AB = 4000

\tan\theta=\frac{AB}{BC}

\tan 11.1^{\circ} =\frac{4000}{BC}

Solving for BC, we get,

BC=\frac{4000}{\tan 11.1^{\circ} }

BC=\frac{4000}{0.196} (approx)

BC=20408.16(approx)  

Thus, the horizontal distance from the plane to the person on the runway is 20408.16 ft

8 0
3 years ago
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