All categories

3 years ago

All points of the step function f(x) are graphed.

Mathematics

2 answers:

Bond [772]3 years ago

4 0

Number 4 or {x| 2 < x ≤ 4}

Step-by-step explanation:

{x| 2 < x ≤ 4} is the only one that fits the peramiters on the graph no other one starts at any point stated

tino4ka555 [31]3 years ago

4 0

Answer:

A or {x| –4 < x ≤ 4}

Step-by-step explanation:

it goes from -4 to 4 on the x axis

You might be interested in

An encyclopedia has eight volumes. In how many ways can the eight volumes be replaced on the shelf? Explain.

aev [14]

Imagine there are 8 spots on the shelf. Replace the volumes one by one. The first volume to be replaced could go in any one of the eight spots. The second volume to be replaced could then go in any one of the seven remaining spots. The third volume to be replaced could then go in any one of the six remaining spots. etc So the total number of ways the eight volumes could be replaced = 8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 40,320

7 0

3 years ago

The table shows the syrup production of four states:

Olin [163]

Answer:

Step-by-step explanation:

The table shows the syrup production of four states :

State Syrup Production (liters) Total

Maine 1.10 × 106 116.60

New Hampshire 3.14 × 105 329.70

New York 9.65 × 105 1,013.25

Vermont 1.89 × 106 200.34

Now order the states from least to greatest.

1. Maine

2. Vermont

3. New Hampshire

4. New York

4 0

3 years ago

Read 2 more answers

(a) the number 561 factors as 3 · 11 · 17. first use fermat's little theorem to prove that a561 ≡ a (mod 3), a561 ≡ a (mod 11),

Vitek1552 [10]

LFT says that for any prime modulus $p$ and any integer $n$ , we have

$n^p\equiv n\pmod p$

From this we immediately know that

$a^{561}\equiv a^{3\times11\times17}\equiv\begin{cases}(a^{11\times17})^3\pmod3\\(a^{3\times17})^{11}\pmod{11}\\(a^{3\times11})^{17}\pmod{17}\end{cases}\equiv\begin{cases}a^{11\times17}\pmod3\\a^{3\times17}\pmod{11}\\a^{3\times11}\pmod{17}\end{cases}$

Now we apply the Euclidean algorithm. Outlining one step at a time, we have in the first case $11\times17=187=62\times3+1$ , so

$a^{11\times17}\equiv a^{62\times3+1}\equiv (a^{62})^3\times a\stackrel{\mathrm{LFT}}\equiv a^{62}\times a\equiv a^{63}\pmod3$

Next, $63=21\times3$ , so

$a^{63}\equiv a^{21\times3}=(a^{21})^3\stackrel{\mathrm{LFT}}\equiv a^{21}\pmod3$

Next, $21=7\times3$ , so

$a^{21}\equiv a^{7\times3}\equiv(a^7)^3\stackrel{\mathrm{LFT}}\equiv a^7\pmod3$

Finally, $7=2\times3+1$ , so

$a^7\equiv a^{2\times3+1}\equiv (a^2)^3\times a\stackrel{\mathrm{LFT}}\equiv a^2\times a\equiv a^3\stackrel{\mathrm{LFT}}\equiv a\pmod3$

We do the same thing for the remaining two cases:

$3\times17=51=4\times11+7\implies a^{51}\equiv a^{4+7}\equiv a\pmod{11}$

$3\times11=33=1\times17+16\implies a^{33}\equiv a^{1+16}\equiv a\pmod{17}$

Now recall the Chinese remainder theorem, which says if $x\equiv a\pmod n$ and $x\equiv b\pmod m$ , with $m,n$ relatively prime, then $x\equiv b{m_n}^{-1}m+a{n_m}^{-1}n\pmod{mn}$ , where ${m_n}^{-1}$ denotes $m^{-1}\pmod n$ .

For this problem, the CRT is saying that, since $a^{561}\equiv a\pmod3$ and $a^{561}\equiv a\pmod{11}$ , it follows that

$a^{561}\equiv a\times{11_3}^{-1}\times11+a\times{3_{11}}^{-1}\times3\pmod{3\times11}$
$\implies a^{561}\equiv a\times2\times11+a\times4\times3\pmod{33}$
$\implies a^{561}\equiv 34a\equiv a\pmod{33}$

And since $a^{561}\equiv a\pmod{17}$ , we also have

$a^{561}\equiv a\times{17_{33}}^{-1}\times17+a\times{33_{17}}^{-1}\times33\pmod{17\times33}$
$\implies a^{561}\equiv a\times2\times17+a\times16\times33\pmod{561}$
$\implies a^{561}\equiv562a\equiv a\pmod{561}$

6 0

4 years ago

I really need help solving this problem from my trigonometry prepbook

padilas [110]

The terminal ray of 145° lies in II Quadrant.

The terminal ray of -83° lies in IV Quadrant.

The terminal ray of -636 lies in I Quadrant.

The terminal ray of 442 lies in I Quadrant.