Solve for the missing variable. \( \left.\begin{array}{l}C=2 \pi r \\ \text { (Circumference of a circle) } \\ r=\square \\ \text { (Round to the nearest hundredth as needed.) }\end{array}\right] \) C=7.13, \( =3 \pi=3.14 \)
Solución de inteligencia artificial de Upstudy
Responder
Solución

¡Inicia sesión para desbloquear respuestas gratis!
Una plataforma de aprendizaje en la que confían millones de estudiantes y profesores reales.

Respondido por UpStudy AI y revisado por un tutor profesional
Mind Expander
To solve for the missing variable \( r \) in the equation \( C = 2 \pi r \), we start by substituting the circumference \( C = 7.13 \): \[ 7.13 = 2 \pi r \] Now, we will isolate \( r \): 1. Divide both sides by \( 2 \pi \): \[ r = \frac{7.13}{2 \pi} \] 2. Using \( \pi \approx 3.14 \): \[ r = \frac{7.13}{2 \times 3.14} \] 3. Calculate \( 2 \times 3.14 = 6.28 \): \[ r \approx \frac{7.13}{6.28} \approx 1.136 \] Rounded to the nearest hundredth, we have \( r \approx 1.14 \). So, the missing variable \( r \) is approximately \( 1.14 \). Finding the radius of a circle has real-world applications that are fun! For instance, knowing the radius helps you design a pizza or a garden bed, ensuring you have the right amount of space (and toppings) without wasting resources. So next time you're measuring for a round project, remember to use the formula for circumference — and enjoy your tasty results! Also, when dealing with formulas like this one, a common mistake is forgetting to use the correct value for \( \pi \). It's easy to get approximations mixed up! Always double-check your calculations, especially when dividing, as errors can lead to incorrect dimensions. Keeping a calculator handy can help you avoid these pitfalls. Happy calculating!
