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DENIUS [597]
3 years ago
12

after sandy picked 3 lemons she wanted to share them with her fellow classmates if sandy wants to give 3tenth of a lemon to each

of her classmates then how how many classmates will get some lemon
Mathematics
1 answer:
Amanda [17]3 years ago
8 0
3 lemons can support up to 30 classmates assuming each only gets 3 tenths of a lemon
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Sally has 3 nickels, 9 dimes, and 18 quarters
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3 years ago
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Y=2/5x+2 how do you graph it on a graph
ANTONII [103]
That is how you graph y=2/5x+2 since 2/5 is the slope and 2 is the y intercept.

8 0
3 years ago
Next
AveGali [126]

Answer:

C. Robbie's glider

Step-by-step explanation:

P.S -The exact question is -

Given - Melissa and Robbie are flying remote control gliders.

The altitude of Melissa's glider, h(s), in feet, is modeled by this function, where sis time, in seconds, after launch.

m(s) = 0.4(s³ - 11s² + 31s – 1)

The altitude of Robbie's glider is modeled by function r, where sis time, in seconds, after launch.

To find - Which glider reaches the greater maximum altitude in the first 6 seconds after launch?

A. Neither glider reaches a maximum altitude on the given interval.

В. Melissa's glider

C. Robbie's glider

D. Both gliders reach the same altitude on the given interval.

Proof -

At t = 6 sec,

Melisa glider is

h(6) = 0.4((0.6)³ - 11(0.6)² + 31(0.6) – 1)

      = 0.4( 0.216 - 3.96 + 18.6 - 1 )

      = 0.4(13.856) = 5.5424

⇒h(6) = 5.5424

∴ we get

At t = 6 seconds, Melissa glider is at the altitude of 5.54 feet

Now,

From the figure, we can see that,

At t = 6 seconds, Robbie's glider is approximately at an altitude of 13 feet

Now,

As 13 > 5.5

So,

Robbie's glider is at maximum height in 6 seconds after the launch.

So,

The correct option is - C. Robbie's glider

8 0
2 years ago
I'm not sure how to solve this. please help​
zepelin [54]

Answer:

The awnser is 2 and 3!!!!!!

8 0
2 years ago
If f(x) = x + 4 and g(x)=x^2-1, what is m(g o f)(x)?
Andreas93 [3]

Until now, given a function  f(x), you would plug a number or another variable in for x. You could even get fancy and plug in an entire expression for x. For example, given  f(x) = 2x + 3, you could find f(y2 – 1) by plugging y2 – 1 in for x to get f(y2 – 1) = 2(y2 – 1) + 3 = 2y2 – 2 + 3 = 2y2 + 1.

In function composition, you're plugging entire functions in for the x. In other words, you're always getting "fancy". But let's start simple. Instead of dealing with functions as formulas, let's deal with functions as sets of (x, y) points:

Let f = {(–2, 3), (–1, 1), (0, 0), (1, –1), (2, –3)} and  

let g = {(–3, 1), (–1, –2), (0, 2), (2, 2), (3, 1)}.  

 

Find (i) f (1), (ii) g(–1), and (iii) (g o f )(1).

(i) This type of  exercise is meant to emphasize that the (x, y) points are really (x, f (x)) points. To find  f (1), I need to find the (x, y) point in the set of (x, f (x)) points that has a first coordinate of x = 1. Then f (1) is the y-value of that point. In this case, the point with x = 1 is (1, –1), so:

8 0
3 years ago
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