3 years ago

A box of highlighters originally cost

Mathematics

1 answer:

Vanyuwa [196]3 years ago

3 0

Answer:

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Solve the following equation:<br><br> 3 x one half

Kamila [148]

<h3>hello!</h3>

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Let's simplify the expression:-

$\bigstar{\boxed{\frac{3}{1} *\frac{1}{2}}$

Multiply 3 times 1 and 1 times 2:-

$\bigstar{\boxed{\pmb{\frac{3}{2} }}$

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Hope everything is clear; if you need any clarification/explanation, kindly let me know, and I'll comment and/or edit my answer :)

6 0

2 years ago

4 b a 1. Joseph plans to give a bag of candy to each of the 22 students in his class. Each bag of candy will cost $2.87 after ta

zavuch27 [327]

Answer:

Step-by-step explanation:

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2 years ago

A package of tickets for 4 home games cost $180 . What proportion can you write to find what a 12 -game package cost if all indi

Amiraneli [1.4K]

You would write it like 4/180.
Then the other one would be
12/x.
To find the individual price you need to divide 180 by 4. You then get 45.
So now to find how many 12 tickets would be, multiply 12 by 45.
The answer would then be $540. Pricey!

Hope this helps!

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3 years ago

40k+36=? <br> ivbuivbr vpui34vb[ervb[uir3bv34

kobusy [5.1K]

Answer:

Step-by-step explanation:

$40k = - 36$

$k = - \frac{9}{10}$

$40( - \frac{9}{10} ) + 36 = 0$

$- 36 + 36 = 0$

4 0

2 years ago

A tank contains 200 liters of fluid in which 50 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pu

mel-nik [20]

At the start, the tank contains A(0) = 50 g of salt.

Salt flows in at a rate of

(1 g/L) * (5 L/min) = 5 g/min

and flows out at a rate of

(A(t)/200 g/L) * (5 L/min) = A(t)/40 g/min

so that the amount of salt in the tank at time t changes according to

A'(t) = 5 - A(t)/40

Solve the ODE for A(t):

A'(t) + A(t)/40 = 5

e^(t/40) A'(t) + e^(t/40)/40 A(t) = 5e^(t/40)

(e^(t/40) A(t))' = 5e^(t/40)

e^(t/40) A(t) = 200e^(t/40) + C

A(t) = 200 + Ce^(-t/40)

Given that A(0) = 50, we find

50 = 200 + C ==> C = -150

so that the amount of salt in the tank at time t is

A(t) = 200 - 150 e^(-t/40)

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3 years ago