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3 years ago

How to divide 43.75 and 3.5

Mathematics

1 answer:

Alona [7]3 years ago

4 0

$\dfrac{43.75}{3.5} = 12.5.$ <span>
</span>

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Step-by-step explanation:

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2 years ago

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Jack's last six scores are 75,78,83,80,79,84. If he scores a 90 on the next test, by how much will his mean test score increase?

kondaur [170]

Answer:

1.46

Step-by-step explanation:

First find the the mean score before he made a 90.

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His mean test score increased by 1.46.

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3 years ago

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3 years ago

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Trigonometry, please help thank you

alina1380 [7]

Step-by-step explanation:

<h2>Length of x is 98.2 m</h2><h2 /><h2>Step-by-step explanation:</h2><h2 /><h2>Step 1:</h2><h2 /><h2>Use the trigonometric ratio tan 27° to find the common side of both the right angled triangles.</h2><h2 /><h2>tan 27° = opposite side/adjacent side =</h2><h2 /><h2>opposite side/9</h2><h2 /><h2>Opposite side = 9 tan 27° 9 x 3.27 =</h2><h2 /><h2>-29.46 m</h2><h2 /><h2>Step 2:</h2><h2 /><h2>Use this side and trigonometric ratio cosine to find the value of x.</h2><h2 /><h2>cos 49° = adjacent side/x = -29.46/x</h2><h2 /><h2>x = -29.46/cos 49° -29.46/0.30</h2><h2 /><h2>= 98.2 m (negative value neglected)</h2><h2 /><h2><em><u>Please</u></em><em><u> </u></em><em><u>mark</u></em><em><u> </u></em><em><u> </u></em><em><u>me </u></em><em><u>Brainlist.</u></em></h2>

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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