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34kurt
3 years ago
13

If two circles are externally tangent, it is not possible to draw an external tangent line. True/False

Mathematics
2 answers:
charle [14.2K]3 years ago
7 0

Answer:

False

there is one tangent line that passes between the two circles, touching each at the point of tangency

kotykmax [81]3 years ago
6 0

Answer:

I think its true

Step-by-step explanation:

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Two friends leave there houses and go towards each others, one friend runs 8 miles a hour, the other runs 7 miles an hour, there
natka813 [3]

hi there the answer is 7:30 see the picture

3 0
3 years ago
Binomial Expansion/Pascal's triangle. Please help with all of number 5.
Mandarinka [93]
\begin{matrix}1\\1&1\\1&2&1\\1&3&3&1\\1&4&6&4&1\end{bmatrix}

The rows add up to 1,2,4,8,16, respectively. (Notice they're all powers of 2)

The sum of the numbers in row n is 2^{n-1}.

The last problem can be solved with the binomial theorem, but I'll assume you don't take that for granted. You can prove this claim by induction. When n=1,

(1+x)^1=1+x=\dbinom10+\dbinom11x

so the base case holds. Assume the claim holds for n=k, so that

(1+x)^k=\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k

Use this to show that it holds for n=k+1.

(1+x)^{k+1}=(1+x)(1+x)^k
(1+x)^{k+1}=(1+x)\left(\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k\right)
(1+x)^{k+1}=1+\left(\dbinom k0+\dbinom k1\right)x+\left(\dbinom k1+\dbinom k2\right)x^2+\cdots+\left(\dbinom k{k-2}+\dbinom k{k-1}\right)x^{k-1}+\left(\dbinom k{k-1}+\dbinom kk\right)x^k+x^{k+1}

Notice that

\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!}{\ell!(k-\ell)!}+\dfrac{k!}{(\ell+1)!(k-\ell-1)!}
\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)}{(\ell+1)!(k-\ell)!}+\dfrac{k!(k-\ell)}{(\ell+1)!(k-\ell)!}
\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)+k!(k-\ell)}{(\ell+1)!(k-\ell)!}
\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(k+1)}{(\ell+1)!(k-\ell)!}
\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{(k+1)!}{(\ell+1)!((k+1)-(\ell+1))!}
\dbinom k\ell+\dbinom k{\ell+1}=\dbinom{k+1}{\ell+1}

So you can write the expansion for n=k+1 as

(1+x)^{k+1}=1+\dbinom{k+1}1x+\dbinom{k+1}2x^2+\cdots+\dbinom{k+1}{k-1}x^{k-1}+\dbinom{k+1}kx^k+x^{k+1}

and since \dbinom{k+1}0=\dbinom{k+1}{k+1}=1, you have

(1+x)^{k+1}=\dbinom{k+1}0+\dbinom{k+1}1x+\cdots+\dbinom{k+1}kx^k+\dbinom{k+1}{k+1}x^{k+1}

and so the claim holds for n=k+1, thus proving the claim overall that

(1+x)^n=\dbinom n0+\dbinom n1x+\cdots+\dbinom n{n-1}x^{n-1}+\dbinom nnx^n

Setting x=1 gives

(1+1)^n=\dbinom n0+\dbinom n1+\cdots+\dbinom n{n-1}+\dbinom nn=2^n

which agrees with the result obtained for part (c).
4 0
3 years ago
Nora needs at least $115 for a new cell phone. She has already saved $45. She earns $7 an hour at her job. Write and solve an in
Volgvan

Answer:

she only needs to work ten hours

Explanation: if you subtract 45 from 115 you well end up with 70 a and if you work for ten hours 7 ×10=70 so you would have enough.

8 0
3 years ago
7. During a sale, the price of a pair of running shoes was marked down from $89.95 to $75.00. What was the
shutvik [7]

Answer:

16.6%

Step-by-step explanation:

(75/89.95)*100=83.37....

100-83.37...=16.62...

=16.6%

7 0
2 years ago
Andy wrote the following statement
Fantom [35]
Andy is incorrect because more then one triangle can be drawn with the given information
7 0
3 years ago
Read 2 more answers
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