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Akimi4 [234]
3 years ago
10

Which expression is equivalent to the given expression? (ab²2)³/b⁵ A.a⁴/b B.a³ C.a³/b D.a³b

Mathematics
1 answer:
sveta [45]3 years ago
4 0

Answer:

gh/b6

Step-by-step explanation:

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How much heat in kJ is released when nitrogen and hydrogen react as below to form 100.g ammonia heat equals-92.2kj
tresset_1 [31]

100g of ammonia releases 271.18 kJ of heat.

<u>Explanation:</u>

Given:

N₂ reacts with H₂

The reaction is:

2N₂ + 3H₂ → 2NH₃ + 92.2 kJ

According to the balanced reaction:

2 X 17g of ammonia releases 92.2 kJ of heat

34g of ammonia releases 92.2 kJ of heat

100g of ammonia will release = \frac{92.2}{34} X 100 kJ of heat

                                                 = 271.18 kJ

Therefore, 100g of ammonia releases 271.18 kJ of heat.

8 0
3 years ago
A mini basketball court has an area of 500
aivan3 [116]

Answer:

Step-by-step explanation:

let x,y be the sides of court.

2(x+y)=90

x+y=90/2=45

y=45-x

xy=500

x(45-x)=500

45x-x²=500

x²-45x+500=0

x²-25x-20x+500=0

x(x-25)-20(x-25)=0

(x-25)(x-20)=0

x=25,20

when x=25

y=45-x=45-25=20

when x=20

y=45-20=25

sides are 25 ft,20 ft.

8 0
3 years ago
Read 2 more answers
What multiplies to give you 15 and adds to give you -8?
Nitella [24]

Answer:

The system of equations is:

        x + y = 15

           xy = 15

Step-by-step explanation:

solve the first for y and substitute in the second:

            y = 15-x

      x(15-x) = 15

     15x - x� = 15

15x - x� - 15 = 0

-x� + 15x - 15 = 0

x� - 15x + 15 = 0

7 0
2 years ago
Read 2 more answers
2 less than the product of 5 and n
Deffense [45]

The product of 5 and n is written as 5n.

Two less than this product is 5n-2

7 0
3 years ago
Solve the following equation. X cubed minus 6X squared plus 6X equals zero
anyanavicka [17]

We have to solve this equation:

x^3-6x^2+6x=0

Third degree polynomials like this one are not easily solved, but this one has a root at x = 0. The let us factorize this polynomial as x times a second degree polynomial:

\begin{gathered} x^3-6x^2+6x=0 \\ x(x^2-6x+6)=0 \end{gathered}

Now we can find the roots of the quadratic polynomial as:

\begin{gathered} x=\frac{-(-6)\pm\sqrt[]{(-6)^2-4\cdot1\cdot6}}{2\cdot1} \\ x=\frac{6\pm\sqrt[]{36-24}}{2} \\ x=\frac{6\pm\sqrt[]{12}}{2} \\ x=\frac{6\pm\sqrt[]{4\cdot3}}{2} \\ x=\frac{6\pm2\sqrt[]{3}}{2} \\ x=3\pm\sqrt[]{3} \\ x_1=3-\sqrt[]{3} \\ x_2=3+\sqrt[]{3} \end{gathered}

Then, the solutions to the equation are:

x = 0

x = 3 - √3

x = 3 + √3

4 0
1 year ago
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