For this case we have the following situation:
The monthly cost of a gym in the city is $ 3. The initial registration is $ 25. Write a mathematical expression that models the problem.
So we have to define variables:
x: number of months
y: total cost
The mathematical expression that models the problem is:
Answer:
1/2 foot
Step-by-step explanation:
1 x 1 = 1
1/2 = 1/2
We will solve this using a system of equations. The first part tells us that building a is 190 feet shorter than building b. Our first equation, then, is b=190+a. The second part tells us that the addition of the two buildings' heights is 1480. So our second equation is a + b = 1480. The first equation is already solved for b, so let's sub that value into the second equation for b: a+(190+a)=1480. 2a + 190 = 1480 and 2a = 1290. That means that building a is 645 feet tall. Building b is 190 feet taller, so b = 190 + 645, which is 835.
We have that the probability of Janice rolling a 2 or 5 on her 7th toss of the dice is

From the question we are told that
Janice rolled either a 2 or a 5 on the last 6 rolls of the die
Janice rolling a 2 or 5 on her 7th toss of the dice

Where

Janice roll the standard die 7th time the probability of getting 2 or 5 is

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Answer:
C) Both
Step-by-step explanation:
The given equation is:

To solve the given equation, we can use the Zero Product Property according to which if the product <em>A.B = 0</em>, then either A = 0 OR B = 0.
Using this property:

So, Erik's solution strategy would work.
Now, let us discuss about Caleb's solution strategy:
Multiply
i.e.
= 
So, the equation becomes:

Comparing this equation to standard quadratic equation:

a = 3, b = -10, c = -8
So, this can be solved using the quadratic formula.


The answer is same from both the approaches.
So, the correct answer is:
C) Both