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Scorpion4ik [409]

2 years ago

9

isabella and jorge are placing markers along a cross-country running route. They need to place the 1st marker 50m from the start

and place the other 50m to the end of the 2km route.

1 answer:

DochEvi [55]2 years ago

4 0

Answer:

<h3>102Km route I think</h3>

Step-by-step explanation:

i hope it helps :)

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during a sale , a shop allows discount of the marked price of clothing . What will a costumer pay for a dress with a marked pric

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On a certain hot summer's day, 477 people used the public swimming pool. The daily prices are 1.50 for children and 2.25 for ad

Akimi4 [234]

Answer:

212 children, and 265 adults

Step-by-step explanation:

To find the number of children and adults, we can set up a systems of equations.

x= number of children

y= number of adults

Equation 1: Price

1.50x+2.25y=914.25

Equation 2: Total number of people

x+y=477

Now, let's solve the equation using substitution.

Rearrange the second equation to solve for one variable.

x+y=477

x=477-y

Now plug x equals into the first equation, and solve for y.

1.50x+2.25y=914.25

1.50(477-y)+2.25y=914.25

715.5-1.50y+2.25y=914.25

715.5+0.75y=914.25

0.75y=198.75

y=265

We just solved for the number of adults. Now let's plug y equals into the second equation to find the number of children.

x+y=477

x+265=477

x=212

7 0

2 years ago

Jacky read at the rate of 15 pages per hour. Johnny read at the rate of 1/3 of a pages per minute. Who was reading faster? By ho

o-na [289]

We can compare by converting Johnny's reading time into the same unit of Jacky's, hours.

There are 60 minutes in an hour, so use this:

$\frac{1}{3} * \frac{60}{1} = 20$

15 < 20, so Johnny read faster. This is 5 pages faster per hour than Jacky.

3 0

3 years ago

Find a compact form for generating functions of the sequence 1, 8,27,... , k^3

pantera1 [17]

This sequence has generating function

$F(x)=\displaystyle\sum_{k\ge0}k^3x^k$

(if we include $k=0$ for a moment)

Recall that for $|x|$ , we have

$\displaystyle\frac1{1-x}=\sum_{k\ge0}x^k$

Take the derivative to get

$\displaystyle\frac1{(1-x)^2}=\sum_{k\ge0}kx^{k-1}=\frac1x\sum_{k\ge0}kx^k$

$\implies\dfrac x{(1-x)^2}=\displaystyle\sum_{k\ge0}kx^k$

Take the derivative again:

$\displaystyle\frac{(1-x)^2+2x(1-x)}{(1-x)^4}=\sum_{k\ge0}k^2x^{k-1}=\frac1x\sum_{k\ge0}k^2x^k$

$\implies\displaystyle\frac{x+x^2}{(1-x)^3}=\sum_{k\ge0}k^2x^k$

Take the derivative one more time:

$\displaystyle\frac{(1+2x)(1-x)^3+3(x+x^2)(1-x)^2}{(1-x)^6}=\sum_{k\ge0}k^3x^{k-1}=\frac1x\sum_{k\ge0}k^3x^k$

$\implies\displaystyle\frac{x+4x^3+x^3}{(1-x)^4}=\sum_{k\ge0}k^3x^k$

so we have

$\boxed{F(x)=\dfrac{x+4x^3+x^3}{(1-x)^4}}$

5 0

3 years ago