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3 years ago

12

In preparation for wedding egg and bought 60 candles. He paid $0.37 for each candle his sister bought 170 candles at the same pr

ice how much more money in dollars did Aidan sister spend?

1 answer:

pshichka [43]3 years ago

5 0

She spent:

170 * 0.37 = $62.90

He spent:

60 * 0.37 = $22.20

So the difference is:

62.90 - 22.20 = <u>$40.70</u>

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3 years ago

What is the radius of a circle with circumference 56 inches

pochemuha

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The answer is 8.91 or 8.9

Step-by-step explanation:

SInce the circumference is 56, you'll divide it by pi or 3.14. It's as if you're doing it backwards, but not always. After you divide it by 3.14, you'll get your diameter, or in this case, 17.8. Divide the diameter by 2 because the diameter is twice the radius. Eventually your final answer will be 8.91 or 8.9

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2 years ago

I barely ask questions but I really need help with #3 and a little with 4. On number four I got 1 1/20 which doesn’t seem right.

uranmaximum [27]

Let r, g and b represent red, green and blue.

r+g+b = 74

r=g-1

b=r+g

Again, r+g+b = 74. Let's substitutte r+g for b: r+g+(r+g) = 74.

Next, let's eliminate r. Use r=g-1. Then g-1 + g + g-1 + g = 74

Combining the g terms, 4g - 2 = 74 => 4g = 76 => g = 19

Recall that r=g-1
and
b=r+g

Find r. If r=g-1, and g=19, then r = 19-1=18

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3 years ago

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3 years ago

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 $span(S_1)=span(S_2)$ .

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3 years ago