a. Which pair of equations best models the relationship between c and a?
c = a − 5
c = a + 3
a = c + 5
a = 3c − 3
a = c − 5
a = 3c + 3
c = a + 5
c = a − 3
The pair of equations best models the relationship between c and a is option D : c = a + 5, c = a - 3
A system of equations is two or more equations that can be solved to get a unique solution. the power of the equation must be in one degree.
The given conditions are;
c is 5 more than variable a.
( c = a + 5)
c is also three less than variable a.
(c = a - 3)
Now, lets look at the answer choices,
c = a − 5
c = a + 3
Here, c is 5 less than "a". so it will be automatically disqualified.
a = c + 5
a = 3c − 3
So,
Simplified version :
c = a - 5
Here, c is 5 less than "a"..so it will be automatically disqualified.
a = c − 5
a = 3c + 3
Here also, we have to get "c" by itself in both top and bottom equation.
So,
Simplified version:
c = a + 5
Here, c is 5 more than "a"
c = (a - 3) / 3
thus, c is 3 less than "a" divided by 3 . So, this is not correct.
c = a + 5
c = a − 3
Here, c is 5 more than "A"
Also, c is 3 less than "a"
So, Which satisfies the given.
So, our answer is going to be the option D :
c = a + 5
c = a - 3
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x = 3
x = 8
x = 16
x = 24
Answer:
8.6666666
or
8.7
Step-by-step explanation:
To solve 4x5+6 divided by 3 using BIDMAS, multiply 4 by 5 to get 20, then add 6 to get 26. Finally, divide the sum by 3 to get 8.67.
To solve 4x5+6 divided by 3 using the order of operations known as BIDMAS, we start by performing the multiplication: 4x5=20. Then, we add the result to 6: 20+6=26. Finally, we divide the sum by 3: 26/3= 8.6667 (rounded to 4 decimal places) or 8.67 (rounded to 2 decimal places).
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Find h(height)
Between 2 and 3, but closer to 3
Between 1 and 2, but closer to 1
Between 1 and 2, but closer to 2
Answer:
Step-by-step explanation:
In words the place that the cube root of 63 would appear on the number line would be: between 3 and 4, but closer to 4
How to find the cube root
The cube root of 63 would be gotten from a calculator when we use this formula
The solution that we would have from the calculator would be 3.97. The value 3.97 is known to be greater than 3 by a lot but just a little less than 4 on the number line.
3.97 can be approximated to be 4. Hence we can say that In words the place that the cube root of 63 would appear on the number line would be: between 3 and 4, but closer to 4
Answer: A. between 2 and 3, but closer to 2.
Step-by-step explanation:
The cube root of 24 is a number that, when multiplied by itself three times, equals 24. To determine where this cube root would be plotted on a number line, we can compare it to other numbers.
Let's start by finding the perfect cubes of some numbers. The perfect cube of 2 is 2*2*2 = 8, and the perfect cube of 3 is 3*3*3 = 27. Since 24 is between 8 and 27, we know that the cube root of 24 would be between 2 and 3 on the number line.
Now, to determine if it's closer to 2 or 3, we can consider the difference between the cube of each of these numbers and 24. The cube of 2 is 2*2*2 = 8, and the difference between 8 and 24 is 24-8 = 16. The cube of 3 is 3*3*3 = 27, and the difference between 27 and 24 is 27-24 = 3.
Since 16 is greater than 3, we can conclude that the cube root of 24 is closer to 2 than to 3. Therefore, it would be plotted on the number line between 2 and 3, but closer to 2.
The expected value E[(1-x)(1-x)] is 1/4. This represents the average value of the function (1-x)(1-x) for the given probability distribution of x values.
We are given an indicator variable x with values 0 and 1. The probability of x = 0 is 1/4 and x = 1 is 3/4. We need to find the expected value E[(1-x)(1-x)].
Step 1: Determine the function we are working with.
We have the function (1-x)(1-x), which simplifies to (1-2x+x^2).
Step 2: Find the probabilities for each value of x.
For x = 0, the probability is 1/4.
For x = 1, the probability is 3/4.
Step 3: Compute the function values for each x value.
For x = 0, (1-2(0)+0^2) = 1.
For x = 1, (1-2(1)+1^2) = 0.
Step 4: Calculate the expected value.
E[(1-x)(1-x)] = (1)(1/4) + (0)(3/4) = 1/4.
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