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Wittaler [7]
3 years ago
9

WILL GIVE BRAINLIEST TO FIRST CORRECT ANSWER!!!!!!!!!!!!!!!!!!!!

Mathematics
1 answer:
Gekata [30.6K]3 years ago
8 0
The equilibrium constant for the reaction 2NO(g)+Br2⇌2NOBr(g) is Kc=2.0 ×10⁻² at certain temperature.
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7W + 3 = -4W - 85 <br><br>a. -11<br>b. 11<br>c. 8<br>d. -8<br>​
mylen [45]
-8
Step 1) Subtract 3 from both sides of the equation
Step 2) Add 4W to both sides of the equation
Step 3) Divide both sides of the equation by the same factor
6 0
2 years ago
Help i will give brainliest
ycow [4]
0.329, 127.5, and -89 is rational. Square root of 101 is irrational.
4 0
3 years ago
Must a function this is decreasing over a given interval always be negative over the same interval? Explain
aleksklad [387]

It is certainly possible for a function decreasing over a certain interval to be negative, but no rule that says it must be. On the other hand, where the function is decreasing, the rate of change of the function must be negative.


4 0
3 years ago
Read 2 more answers
What is the true solution to the logarithmic equation below?
Oliga [24]

Answer:

  • x = 2

Step-by-step explanation:

<u>Given:</u>

  • log ₄ (2x) = 1

<u>Solving for x:</u>

  • 2x = 4¹
  • 2x = 4
  • x = 2
3 0
2 years ago
What is the area of a rectangle that has side lengths of 3/4 yard and 5/6 yard
Ahat [919]

We are asked to find the area of the rectangle that has side lengths of 3/4 yard and 5/6 yard.

We know that area of rectangle is width times length.

To find the area of the given rectangle we will multiply both side lengths as:

\text{Area of rectangle}=\text{Width}\times \text{Length}

\text{Area of rectangle}=\frac{3}{4}\text{ yard}\times\frac{5}{6}\text{ yard}

\text{Area of rectangle}=\frac{3}{4}\times\frac{5}{6}\text{ yard}^2

\text{Area of rectangle}=\frac{1}{4}\times\frac{5}{2}\text{ yard}^2

\text{Area of rectangle}=\frac{1\times5}{4\times2}\text{ yard}^2

\text{Area of rectangle}=\frac{5}{8}\text{ yard}^2

Therefore, the area of the given rectangle would be \frac{5}{8} square yards.

5 0
3 years ago
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