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igor_vitrenko [27]

3 years ago

11

Robert jogged 1 2/5 km.his brother jogged 2 1/2 km.Who jogged the longer distance ?how much longer?

1 answer:

jolli1 [7]3 years ago

5 0

His brother jogged more by 1 1/10 kilometers.

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Davita can jog 2000 feet in 4 minutes if she jogs at the same rate how many feet can she jog in 8 minutes

slavikrds [6]

The answer is 4,000. if she can run 2,000 feet in 4 minutes, multiply both by two and the answer is 4,000 feet in 8 minutes

7 0

3 years ago

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On the school football team, 2/12 of the students are first graders. In simplest form, what fraction of the students on the scho

almond37 [142]

I/6 you would just divide it

7 0

3 years ago

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Sergeeva-Olga [200]

Answer: A

solve the first equation for y
4x + y = -12
subtract both sides by 4x
y = -4x - 12

7 0

3 years ago

Find a number in the closed interval [1/2,3/2] such that the sum of the number and its reciprocal is

sattari [20]

Let's assume that number as x

so,

the sum of the number and its reciprocal is

$S(x)=x+\frac{1}{x}$

Firstly, we will find derivative

$S'(x)=1-\frac{1}{x^2}$

now, we can set it to 0

and then we can solve for x

$S'(x)=1-\frac{1}{x^2}=0$

$x=-1,x=1$

Since, only x=1 lies on [1/2,3/2]

so, we will consider only x=1

now, we can plug end values and x=1 into S

At x=1/2:

we can plug x=1/2

$S(\frac{1}{2})=\frac{1}{2} +\frac{1}{\frac{1}{2}}$

$S(\frac{1}{2})=\frac{1}{2} +2$

$S(\frac{1}{2})=\frac{5}{2}$

At x=1:

we can plug x=1

$S(1)=1 +\frac{1}{1}$

$S(1)=1 +1$

$S(1)=2$

At x=3/2:

we can plug x=3/2

$S(\frac{3}{2})=\frac{3}{2} +\frac{1}{\frac{3}{2}}$

$S(\frac{3}{2})=\frac{3}{2} +\frac{2}{3}$

$S(\frac{3}{2})=\frac{13}{6}$

(a)

Smallest value:

The smallest value among them

$S(1)=2$

So, a number is 1..........Answer

(b)

Largest value:

The largest value among them

$S(\frac{1}{2})=\frac{5}{2}$

So, a number is $\frac{1}{2}$ ........Answer

6 0

4 years ago

PLZ HELP! 20 POINTS!

agasfer [191]

Answer:

a) The error is that, the initial value is n=1 NOT n=3

b) The sum is $a_n=a_{n+1}+5=192$

c)The explicit formula is $a_n=5n+3$

The recursive formula is $a_n=a_{n+1}+5$ ,

Step-by-step explanation:

The given arithmetic series is 8 + 13 + ... + 43.

The first term is $a_1=8$ , the common difference is $d=13-8=5$

The nth term is given by:

$a_n=a_1+d(n-1)$

We substitute the values to get:

$a_n=8+5(n-1)\\a_n=8+5n-5\\\\a_n=3+5n$

To find how many terms are in the sequence we solve the equation:

$3+5n=43\\\implies 5n=43-3\\5n=40\\n=8$

The summation notation is $\sum_{n=1}^8(3+5n)$

The error the student made is in the initial value.

It should be n=1 NOT n=3

b) The sum of the arithmetic series is calculated using:

$S_n=\frac{n}{2}(a+l)$

We substitute o get:

$S_8=\frac{8}{2}(5+43)$

$S_8=4(48)$

$S_8=192$

c) The explicit formula we already calculated in a), which is $a_n=3+5n$

The recursive formula is given as:

$a_n=a_{n+1}+d$

We substitute d=5 to get:

$a_n=a_{n+1}+5$

6 0

3 years ago