Find the area of a circle with a circumference of 16

1 answer:

3 0

The formula is
$\frac{ c^{2} }{4 \pi }$

replace c with 16 and solve
$\frac{ 16^{2} }{4 \pi }$

your answer will be
A≈20.37

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A biology class examined some flowers in a local meadow. They saw 50 flowers, of which 40% were perennials. How many perennial f

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Answer:20

Step-by-step explanation:

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3 years ago

Someone please help me answer this!!

dybincka [34]

Answer:

17.9 units is the answer

Step-by-step explanation:

points are (-7,-2) and (9,6)

Pythagorean theorem can be written as,

$d=\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \\\\d=\sqrt{9-(-7))^2 + (6-(-2))^2}\\\\d=\sqrt{(16)^2+(8)^2} \\\\d=\sqrt{256+64} \\\\d=\sqrt{320} \\\\d=17.88$

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3 years ago

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Answer:

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2 years ago

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Show that if S1 and S2 are subsets of a vector space V such that S1 c S2 then span(S1) c span(S2). In particular, if S1 c S2 the

klemol [59]

Answer:

See proof below

Step-by-step explanation:

Assume that V is a vector space over the field F (take F=R,C if you prefer).

Let $x\in span(S_1)$ . Then, we can write x as a linear combination of elements of s1, that is, there exist $v_1,v_2,\cdots,v_k \in S_1$ and $a_1,a_2,\cdots,a_k\in F$ such that $x=a_1v_1+a_2v_2+\cdots+a_kv_k$ . Now, $S_1\subseteq S_2$ then for all $y\in S_1$ we have that $y\in S_2$ . In particular, taking $y=v_j$ with $j=1,2,\cdots,k$ we have that $v_j\in S_2$ . Then, x is a linear combination of vectors in S2, therefore $x\in span(S_2)$ . We conclude that $span(S_1)\subseteq span(S_2)$ .

If, additionally $S_2\subseteq S_1$ then reversing the roles of S1 and S2 in the previous proof, $span(S_2)\subseteq span(S_1)$ . Then $span(S_1)\subseteq span(S_2)\subseteq span(S_1)$ , therefore $span(S_1)=span(S_2)$ .