4^3 · 4^4 ------------ 4^9 A) 416 B) 42 C) 1/4^2 D) 1/4

A baseball player throws a ball from second base to home plate. How far does the player throw the ball? Include a diagram showin

12345 [234]

84.85 feet
Explanation:

7 0

3 years ago

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Which of the following gives all of the sets that contain √9?

romanna [79]

Soldiers for the nation of Gilead. “Blessed Be the Fruit”: The standard greeting amongst Gilead residents. The traditional reply is “May the Lord open.” “Under His Eye”: The Gileadean equivalent of “Aloha” — it works as both a hello and a goodbye.

7 0

3 years ago

Which formula does NOT describe the sequence 8, 6, 3, –1, –6, … ?

Lapatulllka [165]

To get an explicit formula, we need to find an expression which gives the n-th term without having to compute earlier terms in the sequence. Looking at the numbers, and from the recursive formula, we see that the sequence is built by subtracting n from the previous term. This is similar to the triangular number sequence 1,3,6,10,15,... which has the explicit formula a_n = n(n+1)/2. In our case we are subtracting n from the previous term, so we multiply by -1/2 instead of 1/2. However, we also need to add a constant term to reproduce the numbers of the sequence. We can write a_1 = -1(2)/2 + c = 8. Therefore, c = 9.
So the explict formula is:
a_n = -n(n+1)/2 + 9

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3 years ago

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Find the smallest 4 digit number such that when divided by 35, 42 or 63 remainder is always 5

alex41 [277]

The smallest such number is 1055.

We want to find $x$ such that

$\begin{cases}x\equiv5\pmod{35}\\x\equiv5\pmod{42}\\x\equiv5\pmod{63}\end{cases}$

The moduli are not coprime, so we expand the system as follows in preparation for using the Chinese remainder theorem.

$x\equiv5\pmod{35}\implies\begin{cases}x\equiv5\equiv0\pmod5\\x\equiv5\pmod7\end{cases}$

$x\equiv5\pmod{42}\implies\begin{cases}x\equiv5\equiv1\pmod2\\x\equiv5\equiv2\pmod3\\x\equiv5\pmod7\end{cases}$

$x\equiv5\pmod{63}\implies\begin{cases}x\equiv5\equiv2\pmod 3\\x\equiv5\pmod7\end{cases}$

Taking everything together, we end up with the system

$\begin{cases}x\equiv1\pmod2\\x\equiv2\pmod3\\x\equiv0\pmod5\\x\equiv5\pmod7\end{cases}$

Now the moduli are coprime and we can apply the CRT.

We start with

$x=3\cdot5\cdot7+2\cdot5\cdot7+2\cdot3\cdot7+2\cdot3\cdot5$

Then taken modulo 2, 3, 5, and 7, all but the first, second, third, or last (respectively) terms will vanish.

Taken modulo 2, we end up with

$x\equiv3\cdot5\cdot7\equiv105\equiv1\pmod2$

which means the first term is fine and doesn't require adjustment.

Taken modulo 3, we have

$x\equiv2\cdot5\cdot7\equiv70\equiv1\pmod3$

We want a remainder of 2, so we just need to multiply the second term by 2.

Taken modulo 5, we have

$x\equiv2\cdot3\cdot7\equiv42\equiv2\pmod5$

We want a remainder of 0, so we can just multiply this term by 0.

Taken modulo 7, we have

$x\equiv2\cdot3\cdot5\equiv30\equiv2\pmod7$

We want a remainder of 5, so we multiply by the inverse of 2 modulo 7, then by 5. Since $2\cdot4\equiv8\equiv1\pmod7$ , the inverse of 2 is 4.

So, we have to adjust $x$ to

$x=3\cdot5\cdot7+2^2\cdot5\cdot7+0+2^3\cdot3\cdot5^2=845$

and from the CRT we find

$x\equiv845\pmod2\cdot3\cdot5\cdot7\implies x\equiv5\pmod{210}$

so that the general solution $x=210n+5$ for all integers $n$ .

We want a 4 digit solution, so we want

$210n+5\ge1000\implies210n\ge995\implies n\ge\dfrac{995}{210}\approx4.7\implies n=5$

which gives $x=210\cdot5+5=1055$ .

5 0

3 years ago

Ravi buys pounds of flour. How many ounces of flour does he buy?

Umnica [9.8K]

There is 16 ounces in each pound, how ever many pounds it is multiply 16 by the number of pounds,

Step-by-step explanation:

how ever many pounds it is multiply 16 by the number of pounds,

5 0

3 years ago