a) six pens and five pencils cost $2.50 while three pens and two pencils cost $1.15 . what will be the cost of a pen and 4 pencils . please help me

Answers

Answer 1

Answer: x - pen
y - pencil

$6x+5y=2.5\n 3x+2y=1.15\n\n 6x+5y=2.5\n -6x-4y=-2.3\n --------\n y=\$0.2\n\n 6x+5\cdot0.2=2.5\n 6x+1=2.5\n 6x=1.5\n x=\$0.25\n \nx+4y=0.25+4\cdot0.2=\boxed{\$1.05 }$

Answer 2

Answer: Let pen's cost be x and pencil's cost be y.

6x + 5y = 2.50
3x + 2y = 1.15

Thus, 3x = 1.15 - 2y
Therefore, 6x = 2.30 - 4y

Substituting the value of 6x ;

(2.30 - 4y) + 5y = 2.50

or, 2.30 + y = 2.50

=> y = 0.20.

Substituting the value of y ;

6x = 2.30 - 4(0.20)
6x = 2.30 - 0.80 = 1.5
6x = 1.5
=> x = 0.25

Thus, 1 pen costs 0.25 $ and 1 pencil costs 0.20$.

Thus, cost of one pen is 0.25$.
Cost of 4 pencils is 0.80 $

0.25$ + 0.80$ = 1.05$

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Help please! I’m struggling in proving it

If there is 1000$ in the lockbox what year will it be worth 64,000$ at 24% interest?

Answers

Hey there!

First, set up an exponential equation that represents the rate at which your original amount, 1000, gains interest:

y = 1000(1.24)^x

Y represents the value after X years. 1.24 represents the rate at which the money gains interest, 1 + 0.24 (your 24% interest rate in decimal form). 1000 is your original amount.

Now, set this equation equal to 64000, graph y = 64000 and y = 1000(1.24)^x on a graphing calculator, and see where the two equations intersect in order to solve for x.

They intersect when x is about 19.334, as seen in the graph below (it is very zoomed in so that you can see where the two functions intersect). Therefore, it will be about 19 years after the year in which you deposited the 1000 dollars before the money is worth 64000 dollars.

What solid figure has 4 faces 6 vertices