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3 years ago

A first-order reaction will yield a straight line when time is plotted on the x-axis and _______ is plotted on the y-axis.

1 answer:

6 0

In reaction kinetics, the equation that would best describe the relationship between time and concentration is,
dCA/dt = k or
dCa = k(dt)
Then, we integrate both sides of the equation to yield.
Ca - Cao = kt
The first-order reaction will yield a straight line when time is plotted against the concentration.

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At a movie theatre adult tickets cost $10 and child ticjets cost $6. Which equation can be used to find s, the number of dollars

timurjin [86]

Answer:

s = 10k + 30

Step-by-step explanation:

Cost of Adults ticket = $10 per adult

Cost of Ticket = $6 per child

Number of adults = k

Number of Children = 5

s = Total cost

Total cost = (Cost per adult * number of adults) + (Cost per child * Number of children)

s = ($10 * k) + ($6 * 5)

s = $10k + $30

s = 10k + 30

4 0

2 years ago

Simplify (6^-4)^6 <br> Please help me

shutvik [7]

Answer:

6^ -24

Step-by-step explanation:

We know that a^b^c = a^ (b*c)

(6^-4)^6 = 6^ (-4*6) = 6^ -24

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3 years ago

Find the value of: (ii) [(1/4) -1 + (1/3) -2 - (1/2) -3 ] ÷ 5

Anastaziya [24]

Answer:

-71/6

Step-by-step explanation:

[(1/4) -1 + (1/3) -2 - (1/2) -3 ] ÷ 5

[(3/12)-(12/12)+(4/12)-(24/12)-(6/12)-(36/12)] ÷ 5

[-71/12]/5

-71/60

6 0

3 years ago

What is the form of the two squares identity?

zvonat [6]

Answer:

Step-by-step explanation:

Consider all options:

A. False, because

$(a^2+b^2)(c^2+d^2)=a^2c^2+a^2d^2+b^2c^2+b^2d^2\\ \\(ab-cd)^2+(ac+bd)^2=a^2b^2-2abcd+c^2d^2+a^2c^2+2abcd+b^2d^2=\\ =a^2b^2+c^2d^2+a^2c^2+b^2d^2\\ \\a^2c^2+a^2d^2+b^2c^2+b^2d^2\neq a^2b^2+c^2d^2+a^2c^2+b^2d^2$

B. False, because

$(a^2-b^2)(c^2+d^2)=a^2c^2+a^2d^2-b^2c^2-b^2d^2\\ \\(ac-bd)^2-(ad+bc)^2=a^2c^2-2abcd+b^2d^2+a^2d^2+2abcd+b^2c^2=\\=a^2c^2+b^2d^2+a^2d^2+b^2c^2\\ \\a^2c^2+a^2d^2-b^2c^2-b^2d^2\neq a^2c^2+b^2d^2+a^2d^2+b^2c^2$

C. False, because

$(a^2+b^2)(c^2-d^2)=a^2c^2-a^2d^2+b^2c^2-b^2d^2\\ \\(ac+bd)^2-(ad+bc)^2=a^2c^2+2abcd+b^2d^2-a^2d^2-2abcd-b^2c^2=\\=a^2c^2+b^2d^2-a^2d^2-b^2c^2\\ \\a^2c^2-a^2d^2+b^2c^2-b^2d^2\neq a^2c^2+b^2d^2-a^2d^2-b^2c^2$

D. True, because

$(a^2+b^2)(c^2+d^2)=a^2c^2+a^2d^2+b^2c^2+b^2d^2\\ \\(ac-bd)^2-(ad+bc)^2=a^2c^2-2abcd+b^2d^2+a^2d^2+2abcd+b^2c^2=\\=a^2c^2+b^2d^2+a^2d^2+b^2c^2\\ \\a^2c^2+a^2d^2+b^2c^2+b^2d^2= a^2c^2+b^2d^2+a^2d^2+b^2c^2$

6 0

3 years ago

What is the answer to 18x-5 < 3(6x-2) ?

Marina86 [1]

Answer:

No solution

Step-by-step explanation:

18x - 5 < 3(6x - 2) Distribute the 3 to (6x - 2)

18x - 5 < 18x - 6

-18x -18x Subtract 18x from both sides

-5 < -6 There is no solution because -5 is not less than -6.

If this answer is correct, please make me Brainliest!

5 0

3 years ago