Find the circumference of the given circle given

Ответ:

kimberlyvazquez1121

15.09.2022

Solving for x: three denominators: 4, 6, and 12.we can find the lcm of those denominatorswe can write each denominator as a product of prime factors $\begin{aligned} 4 & = 2 \cdot 2 = 2^2\\ 6 & = 2 \cdot 3 \\ 12 & = 2 \cdot 2 \cdot 3 =2^2 \cdot 3 \end{aligned}$ get the primes appearing the greatest number of times in any of the factors.we have unique factors of 2 and 3 highest power of 2 is 2² highest power of 3 is 3the lcd you get from multiplying those highest powers of prime lcd is 2² · 3 = 12the point here is that we want all the fractions to have the same denominator. then we can multiply both sides of the equation by that denominator to cancel it out. $\dfrac{x+7}{12} = \boxed{\dfrac{5}{6}} + \dfrac{x-4}{4}$ to make $\frac{5}{6}$ have denominator of 12, we need to multiply the numerator and denominator by 2 (because 6 · 2 = 12): $\begin{aligned} \dfrac{x+7}{12} & = \dfrac{5 \cdot 2}{6 \cdot 2} + \dfrac{x-4}{4} \\ \dfrac{x+7}{12} & = \dfrac{10}{12} + \dfrac{x-4}{4} \\ \dfrac{x+7}{12} & = \dfrac{10}{12} + \boxed{\dfrac{x-4}{4}} \end{aligned}$ to make $\frac{x-4}{4}$ have a denominator of 12, we need to multiply the numerator and denominator by 3 (because 4 · 3 = 12): $\begin{aligned} \dfrac{x+7}{12} & = \dfrac{10}{12} + \dfrac{3\cdot(x-4)}{3 \cdot 4} \\ \dfrac{x+7}{12} & = \dfrac{10}{12} + \dfrac{3x-12}{12} \end{aligned}$ multiply both sides of the entire equation by 12 so that it will cancel out all the denominators (distributes on the right-hand side) $\begin{aligned} 12\left(\dfrac{x+7}{12}\right) & = 12\left(\dfrac{10}{12} + \dfrac{3x-12}{12} \right) \\ \frac{x+7}{1} & = \dfrac{12 (10)}{12} + \dfrac{12(3x-12)}{12} \\ x + 7 & = 10 + 3x - 12 \end{aligned}$ 10 and -12 sum up to get -2 $x + 7 = 3x - 2$ get rid of x on left side by subtracting both sides by x. $\begin{aligned} 7 & = 3x - 2 - x\\ 7 & = 2x - 2 \end{aligned}$ add 2 to both sides $\begin{aligned} 7 + 2& = 2x \\ 9 & = 2x \end{aligned}$ divide both sides by 2 $x = \frac{9}{2}$ answer is x = 9/2======checking our answer: plug in x = 9/2 into the original equation and see if we get a true statement. $\begin{aligned} \frac{x+7}{12} & = \frac{5}{6} + \frac{x-4}{4} \\ \frac{\frac{9}{2}+7}{12} & = \frac{5}{6} + \frac{\frac{9}{2}-4}{4} \end{aligned}$ multiplying the entire numerator and denominator of the fraction $\frac{\frac{9}{2}+7}{12}$ to get rid of the little denominators of "2" $\begin{aligned} \frac{2\left(\frac{9}{2}+7\right)}{2(12)} & = \frac{5}{6} + \frac{2\left(\frac{9}{2}-4\right)}{2(4)} \\ \frac{9 + 14}{24} & = \frac{5}{6} + \frac{9 - 8}{8} \\ \frac{23}{24} & = \frac{5}{6} + \frac{1}{8} \end{aligned}$ the left-hand side has 24 in denominator; let us aim for thatget $\frac{5}{6}$ to have a denominator of 24 by multiplying num and denom by 4 (because 6 (denominator) · 4 = 24).get $\frac{1}{8}$ to have a denominator of 24 by multiplying num and denom by 3 (because 8 · 3 = 24). $\begin{aligned} \frac{23}{24} & = \frac{5 \cdot 4}{6 \cdot 4} + \frac{1 \cdot 3}{8 \cdot 3} \\ & = \frac{20}{24} + \frac{3}{24} \\ & = \frac{20 + 3}{24}\\ & = \frac{23}{24} \end{aligned}$ we get a true statement as 23/24 = 23/24.
Solve the linear equation step-by-step

0,0(0 оценок)

Find the circumference of the given circle given a diameter of 40 cm. Use 3.14 as an approximation of Pi.

Ответ:

Ответ:

Ask your question to the AI advisor

More tips

Answers on questions: Mathematics

Ask an AI advisor a question