Answer:
3
Step-by-step explanation:
Rewriting this sentence as an algebraic equation, we get:
3^10 = 3^7 * x^3.
Dividing both sides by 3^7 simplifies the problem:
3^3 = x^3
Here this x corresponds to the 3 in 3^3. Thus, the solution is x = 3.
Using exponent rules and logarithms, the given equation can be simplified and solved to find that x equals 3.
The subject of the question falls into the field of Mathematics, specifically exponent rules, which are used in High School level algebra. In the mathematics problem provided ('3 to the 10th power = 3 to the 7th power times x to the 3rd power'), we need to solve for x using the knowledge of exponents.
First, it's helpful to know that when two expressions with the same base (in this case, 3) are set equal to each other, their exponents must also be equal. Since we know that 3 to the 7th power times x to the 3rd power equals 3 to the 10th power, we can break this down further. This leads to the equation 7 + 3log(x) = 10, where 'log' represents the logarithm.
By simplifying the equation, we get 3log(x) = 10-7, so 3log(x) = 3. Therefore, log(x)=3/3, hence log(x)=1. If we apply anti logarithm on both sides, we'll find that x = 3 to the power of 1, so x = 3. In other words, 3 to the 3rd power will equal3 to the 10th power divided by 3 to the 7th power. Hence, the value of x is 3.
#SPJ3
B. 3/16
C. 5/16
D. 7/16
(A) First, write an equation you can use to answer this question. Use x as your variable.
The equation is
(B) Solve your equation in part (A) to find the number of labor hours needed to do the job.
Answer: The number of labor hours was __________________.
Equation: 53x + 460 = 1096
Solution 1096 - 460 = 53x
636/53=x
12=x
12 hours
AND 5x -8 >=40
Solve for x please
Answer:
First Inequality:
Second Inequality:
Step-by-step explanation:
Step 1: Solve for x in the first inequality
Answer:
Step 2: Solve for x in the second inequality
Answer:
Answer:
That's just -6
Step-by-step explanation:
The 19 is smaller then the 25 so you gotta go back to the negatives, so do 25 - 19 but add a negative.