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3 years ago

Is (4, 4) a solution of the graphed system of inequalities?answers are A) Yes B) No

Mathematics

1 answer:

Softa [21]3 years ago

8 0

Answer: A) Yes

If (4, 4) lies within the double-shaded area that covers the solution set of both inequalities, then it's a solution.

It does, therefore, it's a solution to the system of inequalities.

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What is your sister’s total cost under each of the two plans?2. Suppose your sister doubles her monthly usage to 3,500 minutes a

chubhunter [2.5K]

Answer:

(1) The total cost under Plan A is $92 and the total cost under Plan B is $515.

(2) The total cost under Plan A is $92 and the total cost under Plan B is $1,030.

Step-by-step explanation:

The question is incomplete, so the complete question is below:

$A\ cell \ phone \ company\ offers\ two \ different \ plans.$

$Plan\ A\ costs\ \$92\ per\ month\ for\ unlimited\ talk\ and\ text.\ Plan\ B\ costs \ $0.20\ per\ minute\ plus\ \$0.10\ per\ text\ message\ sent.$ $You\ need \ to\ purchase \ a \ plan \ for\ your \ 14-year-old \ sister$ $.Your\ sister\ currently \ uses\ 1,750 \ minutes\ and \ sends\ 1,650 \ texts\ each \ month.$ $(1) What\ is\ your \ sister's\ total\ cost\ under \ each\ of\ the\ two\ plans?$

$(2) Suppose\ your\ sister\ doubles\ her\ monthly\ usage \ to\ 3,500\ minutes \ and\ sends\ 3,300 \ texts.$ $What \ is \ your \ sister's \ total \ cost \ under \ each \ of \ the \ two \ plans?$

Now, to find (1) total cost for each of the two plans. (2) Total cost under each of the two plans, if sister doubles her monthly usage to 3,500 minutes and sends 3,300 texts.

As, the Plan A is for unlimited talk and text

Cost of the Plan A = $92.

Now, to find the cost under Plan B:

According to question:

Rate of call per minute is $0.20 and per text is $0.10.

Calls she uses currently is 1,750 minutes and text 1,650.

So, to get the cost of Plan B:

$0.20\times 1750+0.10\times 1650$

$=350+165$

$=515.$

Thus, Plan B costs $515.

Now, to get the total cost as, sister doubles her monthly usage to 3,500 minutes and sends 3,300 texts.

As, the Plan A is same in both cases and is for unlimited text and calls.

So, cost of Plan A = $92.

As, the monthly usage is double.

Calls in minutes are 3,500.

Texts are 3,300.

Now, to get the total cost under Plan B:

$0.20\times 3500+0.10\times 3300$

$=700+330$

$=1030.$

Hence, the cost of Plan B = $1,030.

Therefore, (1) The total cost under Plan A is $92 and the total cost under Plan B is $515.

(2) The total cost under Plan A is $92 and the total cost under Plan B is $1,030.

7 0

3 years ago

The notation Ak meas the matrix A Multiplied with itself k times (a) For the 2×2 identity matrix I, show that I2 =I (b)For the n

djverab [1.8K]

Answer:

Entries of I^k are are also identity elements.

Step-by-step explanation:

a) For the 2×2 identity matrix I, show that I² =I

$I^{2}=\left[\begin{array}{cc}1&0\\0&1\end{array}\right] \times \left[\begin{array}{cc}1&0\\0&1\end{array}\right] \\\\=\left[\begin{array}{cc}1\times 1+0\times 0&1\times 0+0\times 1\\0\times 1+1\times 0&0\times 0+1\times1\end{array}\right] \\\\=\left[\begin{array}{cc}1&0\\0&1\end{array}\right]$

Hence proved I² =I

b) For the n×n identity matrix I, show that I² =I

n×n identity matrix is as shown in figure

Elements of identity matrix are

$\delta I_{ij}=1\quad if\quad i=j\\\delta I_{ij}=0\quad if\quad i\ne j\\$

As square of 1 is equal to 1 so for n×n identity matrix I, I² =I

As mentioned above

$\delta I_{ij}=1\quad if\quad i=j\\\delta I_{ij}=0\quad if\quad i\ne j\\$

Any power of 1 is equal to 1 so kth power of 1 is also 1. According to this Ik=I

6 0

2 years ago

Help pls see picture

Nady [450]

We can first add all them up:
$P_t = 2x+3+5x+3+5x-2 = 12x+4$

Since it's for a square, we have to divide by 4:
$P_s = \frac{12x+4}{4} = 3x + 1$

So, the expression for a square with the exact same perimeter is 3x + 1

8 0

3 years ago

Find an integer x such that 0<=x<527 and x^37===3 mod 527

Greeley [361]

Since $527=17\times31$ , we have that

$x^{37}\equiv3\mod{527}\implies\begin{cases}x^{37}\equiv3\mod{17}\\x^{37}\equiv3\mod{31}\end{cases}$

By Fermat's little theorem, and the fact that $37=2(17)+3=1(31)+6$ , we know that

$x^{37}\equiv(x^2)^{17}x^3\equiv x^5\mod{17}$
$x^{37}\equiv(x^1)^{31}x^6\equiv x^7\mod{31}$

so we have

$\begin{cases}x^5\equiv3\mod{17}\\x^7\equiv3\mod{31}\end{cases}$

Consider the first case. By Fermat's little theorem, we know that

$x^{17}\equiv x^{16}x\equiv x\mod{17}$

so if we were to raise $x^5$ to the $n$ th power such that

$(x^5)^n\equiv x^{5n}\equiv x\mod{17}$

we would need to choose $n$ such that $5n\equiv1\mod{16}$ (because $16+1\equiv1\mod{16}$ ). We can find such an $n$ by applying the Euclidean algorithm:

$16=3(5)+1$
$\implies1=16-3(5)$
$\implies16-3(5)\equiv-3(5)\equiv1\mod{16}$

which makes $-3\equiv13\mod{16}$ the inverse of 5 modulo 16, and so $n=13$ .

Now,

$x^5\equiv3\mod{17}$
$\implies (x^5)^{13}\equiv x^{65}\equiv x\equiv3^{13}\equiv(3^4)^2\times3^4\times3^1\mod{17}$

$3^1\equiv3\mod{17}$
$3^4\equiv81\equiv4(17)+13\equiv13\equiv-4\mod{17}$
$3^8\equiv(3^4)^2\equiv(-4)^2\mod{17}$
$\implies3^{13}\equiv(-4)^2\times(-4)\times3\equiv(-1)\times(-4)\times3\equiv12\mod{17}$

Similarly, we can look for $m$ such that $7m\equiv1\mod{30}$ . Apply the Euclidean algorithm:

$30=4(7)+2$
$7=3(2)+1$
$\implies1=7-3(2)=7-3(30-4(7))=13(7)-3(30)$
$\implies13(7)-3(30)\equiv13(7)equiv1\mod{30}$

so that $m=13$ is also the inverse of 7 modulo 30.

And similarly,

$x^7\equiv3\mod{31}[/ex] [tex]\implies (x^7)^{13}\equiv3^{13}\mod{31}$

Decomposing the power of 3 in a similar fashion, we have

$3^{13}\equiv(3^3)^4\times3\mod{31}$

$3\equiv3\mod{31}$
$3^3\equiv27\equiv-4\mod{31}$
$\implies3^{13}\equiv(-4)^4\times3\equiv256\times3\equiv(8(31)+8)\times3\equiv24\mod{31}$

So we have two linear congruences,

$\begin{cases}x\equiv12\mod{17}\\x\equiv24\mod{31}\end{cases}$

and because $\mathrm{gcd}\,(17,31)=1$ , we can use the Chinese remainder theorem to solve for $x$ .

Suppose $x=31+17$ . Then modulo 17, we have

$x\equiv31\equiv14\mod{17}$

but we want to obtain $x\equiv12\mod{17}$ . So let's assume $x=31y+17$ , so that modulo 17 this reduces to

$x\equiv31y+17\equiv14y\equiv1\mod{17}$

Using the Euclidean algorithm:

$17=1(14)+3$
$14=4(3)+2$
$3=1(2)+1$
$\implies1=3-2=5(3)-14=5(17)-6(14)$
$\implies-6(14)\equiv11(14)\equiv1\mod{17}$

we find that $y=11$ is the inverse of 14 modulo 17, and so multiplying by 12, we guarantee that we are left with 12 modulo 17:

$x\equiv31(11)(12)+17\equiv12\mod{17}$

To satisfy the second condition that $x\equiv24\mod{31}$ , taking $x$ modulo 31 gives

$x\equiv31(11)(12)+17\equiv17\mod{31}$

To get this remainder to be 24, we first multiply by the inverse of 17 modulo 31, then multiply by 24. So let's find $z$ such that $17z\equiv1\mod{31}$ . Euclidean algorithm:

$31=1(17)+14$
$17=1(14)+3$

and so on - we've already done this. So $z=11$ is the inverse of 17 modulo 31. Now, we take

$x\equiv31(11)(12)+17(11)(24)\equiv24\mod{31}$

as required. This means the congruence $x^{37}\equiv3\mod{527}$ is satisfied by

$x=31(11)(12)+17(11)(24)=8580$

We want $0\le x$ , so just subtract as many multples of 527 from 8580 until this occurs.

$8580=16(527)+148\implies x=148$

3 0

3 years ago

I think the answer is 12 am i correct?

vichka [17]

How did you get 12? It should be 11 because 9+2=11.

8 0

3 years ago

Read 2 more answers

Is (4, 4) a solution of the graphed system of inequalities?answers are A) Yes B) No​

Is (4, 4) a solution of the graphed system of inequalities?answers are A) Yes B) No