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3 years ago

A car travels 45km in an hour. In each of the next two hours, it travels 78km. What is the average speed of the car.

Mathematics

1 answer:

romanna [79]3 years ago

7 0

Answer:

The average speed of car is 67 kilometers per hour.

Step-by-step explanation:

Given that:

Distance travelled in one hour = 45 km

Distance travelled in next 2 hours = 78*2 = 156 km

Total distance = 45+156 = 201

Total time = 1+2 = 3 hours

Average speed of the car = $\frac{Total\ distance}{Total\ time}$

Average speed of the car = $\frac{201}{3}$

Average speed of the car = 67 km/hr

Hence,

The average speed of car is 67 kilometers per hour.

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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).

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3 years ago

Can you guy help for this questing 1234

postnew [5]

Answer:

1. Median

2. Altitude

3. Altitude

4. Neither

Step-by-step explanation:

Median- splits the side into two congruent parts

Altitude- Makes a right angle

1. Splits the side. Median

2. Makes a right angle. Altitude

3. Makes a right angle. Altitude

4. Doesn't make a right angle or split the sides. Neither

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Show that the maximum rate of change, with respect to radius, of the volume of a deflating balloon is four times the sphere's in

dmitriy555 [2]

Step-by-step explanation:

Let's say R is the initial radius of the sphere, and r is the radius at time t.

The volume of the sphere at time t is:

V = 4/3 π r³

Taking derivative with respect to radius:

dV/dr = 4π r²

This is a maximum when r is a maximum, which is when r = R.

(dV/dr)max = 4π R²

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schepotkina [342]

Answer:

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Step-by-step explanation:

There is no data in the image provided. To better help you, I'm assuming we have an arbitrary value of

$BC=6\sqrt{2}$

and the triangle B0C is right

Relations in the Circle

The diameter (D) is twice the radius (r) and the radius is the distance measured from the center of the circle to any point of the circumference.

Since the triangle B0C has a right angle, BD is the hypotenuse and 0B=0C=r

Applying Pythagoras's theorem:

$BC^2=r^2+r^2=2r^2$

Thus we have

$2r^2=(6\sqrt{2})^2$

$2r^2=72$

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Ronch [10]

Answer:

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Step-by-step explanation:

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