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3 years ago

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Kirana buys boxes of crackers that each have the same cost, c. She represents the cost of 3 boxes of cheese crackers, 2 boxes of

poppy seed crackers, and 2 boxes of plain crackers using the expression 3c + 2c + 2c. What equivalent expression can represent the cost?

1 answer:

Vedmedyk [2.9K]3 years ago

4 0

The answer is 3c + 4c

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Alex17521 [72]

Answer:

A

Step-by-step explanation:

I think I am not sure I just started to do this in school...

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3 years ago

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DerKrebs [107]

-21 is your answer

it is reflected over the x axis, and the point is on 21 no?

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3 years ago

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Kurt spots a bird sitting at the top of a 40 foot tall telephone pole. If the angle of elevation from the ground where he is sta

Natalija [7]

Answer:

i guess its 24.03 ft

Step-by-step explanation:

let height of pole be perpendicular length and distance from pole be the base .so tan59=P/b..& b= 24.03ft

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3 years ago

I need help ASAP! It's urgent.. PLISSSSS

natali 33 [55]

Answer:

a) 6 mins

b) 70km/h

c) t= 45

Step-by-step explanation:

a) The bus stops from t=10 to t=16 minutes since the distance the busvtravelled remained constant at 15km

Duration

= 16 -10

= 6 minutes

b) Average speed

= total distance ÷ total time

Total time

= 24min

= (24÷60) hr

= 0.4 h

Average speed

= 28 ÷0.4

= 70 km/h

c) Average speed= total distance/ total time

Average speed

= 80km/h

= (80÷60) km/min

= 1⅓ km/min

1⅓= 28 ÷(t -24)

<em>since</em><em> </em><em>duration</em><em> </em><em>for</em><em> </em><em>return</em><em> </em><em>journey</em><em> </em><em>is</em><em> </em><em>from</em><em> </em><em>t</em><em>=</em><em>2</em><em>4</em><em> </em><em>mins</em><em> </em><em>to</em><em> </em><em>t</em><em> </em><em>mins</em><em>.</em>

$\frac{4}{3}$ (t -24)= 28

$\frac{4}{3}$ t - 32= 28

$\frac{4}{3}$ t= 32 +28

$\frac{4}{3}$ t= 60

t= $6 0\div \frac{4}{3}$

t= 45

*Here, I assume that this is a displacement- time graph, so the distance shown is the distance of the bus from the starting point because technically if it is a distance-time graph, the distance would still increase as the bus travels the 'return journey'.

Thus, distance is decreasing after t=24 and reaches zero at time= t mins so that is the return journey. (because when the bus returns back to starting point, displacement/ distance from starting point= 0km)

5 0

4 years ago

I need help with this question

Murrr4er [49]

The number of cows is given by

$14\cdot 2^{\frac{y}{5}}$

So, after $k$ years, the number of cows will be

$14\cdot 2^{\frac{y+k}{5}}$

We want this number to be twice as much as the original:

$14\cdot 2^{\frac{y+k}{5}} = 2(14\cdot 2^{\frac{y}{5}})$

First of all, we can cancel 14 from both sides:

$2^{\frac{y+k}{5}} = 2\cdot 2^{\frac{y}{5}}$

Finally, on the right hand side, we can use the exponent rule

$a^b\cdot a^c=a^{b+c}$

to get

$2^{\frac{y+k}{5}} = 2^{\frac{y}{5}+1}$

To solve this equation, we must impose that the two exponents are the same:

$\dfrac{y+k}{5} = \dfrac{y}{5}+1 \iff \dfrac{y+k}{5} = \dfrac{y+5}{5}$

And clearly this is true if and only if $k=5$ . So, it will take 5 years for the cow heard to double in number.

You can do the exact same steps to find the doubling time for the sheeps.

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3 years ago