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s2008m [1.1K]
2 years ago
15

c) The cost of 1 pen is Rs 10 and the cost of 1 pencil is Rs 8. Find the total cost of 5 pens and 4 pencils.​

Mathematics
2 answers:
otez555 [7]2 years ago
8 0

Answer:

82Rs

Step-by-step explanation:

let p be pen

let n be pencil

1p=10Rs

1n=8Rs

10×5=50

8×4=32

50+32=82Rs

Setler [38]2 years ago
3 0

Answer:

cost of 1 pen = ₹10

cost of 5 pens = 5×10 = ₹ 50

cost of 1 pencil = ₹8

cost of 4 pencil = 8×4 = ₹ 32

Total cost = 50 + 32 = ₹82

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What is the value of the expression below when z = • 3?<br> 622 + 2z + 10
inn [45]

Answer:

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Step-by-step explanation:

Z = .3 right? It looks weird on my screen

Anyway, if z is .3 than this is how to solve

First you plug in 0.3 where the z would go, making the expression 622 + 2(0.3) + 10. Then you solve! (below)

622 + 2(0.3) + 10.

622 + 0.6 + 10.

622 + 10 + 0.6

632 + 0.6

632.6

I hope this helped! If you have any questions, feel free to message me :)

3 0
2 years ago
Plz help i can’t figure it out
Ivan

Answer:

-(x-10)=7

-x+10=7

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5 0
3 years ago
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I'm having trouble with this<br><br> -9x + 4y = 55<br> -11x + 7y = 82
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Answer:

Step-by-step explanation:

99x - 44y = -605

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(-3, 7)

7 0
3 years ago
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The nth term of a sequence is represented by 3n4−10/6n4+7 . What is the limit of the the nth term as n becomes increasingly larg
Cerrena [4.2K]

Answer:

<h3>1/2</h3>

Step-by-step explanation:

Given the limit of the function expressed as;

\lim_{n \to \infty} \frac{3n^4-10}{6n^4+7} \\

To take the limit, we nee to divide through by the highest power of n first

\lim_{n \to \infty} \frac{3n^4/n^4-10/n^2}{6n^4/n^4+7/n^4}\\= \lim_{n \to \infty} \frac{3-10/n^2}{6+7/n^4}\\= \frac{3-10/(\infty)^2}{6+7/(\infty)^4}\\= \frac{3-0}{6-0}\\= \frac{3}{6}\\ = \frac{1}{2}\\

Hence the limit of the the nth term as n becomes increasingly large is 1/2

3 0
3 years ago
Help! open the attachment below!
gregori [183]

Answer:

yes it is

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