V= pi r^2=3.14*r^2
our V=14,542 m^3
14,524=3.14*r^2 divide both sides by 3.14
4625.47=r^2
r =
r=68
Answer:
its B
Step-by-step explanation:
you just have to add (89+82+41+86+40)÷5 = 72
try them all and the highest would be team 2
<span>Minimal residual cancer cells may not be detected in surgical margins of oral squamous cell carcinoma (OSCC) with routine histological examination. Using molecular markers at surgical margins can be helpful. We attempted to evaluate the MMP-9 and E-cadherin expression in OSCC samples and tumor-free surgical margins and association with clinicopathological factors. We examined E-cadherin and MMP-9 expression in 58 OSCCs including 19 grade I, 21 grade II and 18 grade III with histological tumor-free surgical margins by immunohistochemistry. Specimens were also divided in two groups: 19 samples as an early and 39 as an advanced stage. For E-cadherin in OSCCs and surgical margins, significant difference was observed between poor and moderate tumor differentiation. Different stages of OSCC demonstrated significant differences with higher expression in early stage tumors. For surgical margins, 82.1% of advanced and 84.2% of early stage samples demonstrated immunoreactivity. Both OSCC samples and surgical margins demonstrated significant differences for MMP-9 between stages with higher immunoreactivity in advanced stage, whereas there were not differences between different grades in surgical margins. E-cadherin and MMP-9 expression at histologically negative surgical margins shows the significance of these markers for prognostic values in OSCC patients with E-cadherin being the preferred predictor.</span>
<h3>
Answer:</h3>
3 ft
<h3>
Step-by-step explanation:</h3>
The statue's height is 1.5 times the length of its shadow, so we expect the same relationship for the globe.
... 1.5 × 2 ft = 3 ft
_____
<em>Comment on the problem</em>
As a practical matter, with the sun high enough in the sky to cast a shadow shorter than the object's height, it will be quite difficult to measure the length of the shadow of the point at the top of the globe. The shadow of other parts of the globe will interfere.
