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2 years ago

Please Please please

Mathematics

1 answer:

Nadusha1986 [10]2 years ago

4 0

Answer:

is the name of your account from little mermaid? :DDD

Step-by-step explanation:

Use SOH CAH TOA to recall how the trig functions fit on a triangle

SOH: Sin(Ф)= Opp / Hyp

CAH: Cos(Ф)= Adj / Hyp

TOA: Tan(Ф) = Opp / Adj

now just refer to the above ( also copy and paste the above on your computer some where so you have it always)

sin( D )= 28 / 53

sin( E ) = 45 / 53

cos(D) = 45 / 53

cos(E) = 28 / 53

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What number or fractions are between 2/3 and 1 1/3

olganol [36]

To find the fraction inbetween,
1. Add up the 2 fractions
2 Divide it by 2

Step (!) Add them up:
2/3 + 1 1/3 = 2

Step (2) Divide the answer by 2:
2÷ 2 = 1

Answer: The number between 2/3 and 1 1/3 is 1.

5 0

2 years ago

Please help me. I don’t understand at all.

erastovalidia [21]

Answer:

Option 2

Step-by-step explanation:

Evaluating the options :

Option 1

2 |x - 5| - 4 < -8
2 |x - 5| < -4
|x - 5| < -2
Empty set as modulus cannot be less than 0

Option 2

|2x - 1| - 7 < -6
|2x - 1| < 1
x < 1
There is a solution set other than empty set

Option 2 is the right answer.

7 0

1 year ago

Help me!! Plz....(ᗒᗣᗕ)՞

maksim [4K]

Answer:

the base is 7 units and the height is 3 units

5 0

2 years ago

Solve the following equations: (a) x^11=13 mod 35 (b) x^5=3 mod 64

tino4ka555 [31]

$x^{11}=13\pmod{35}\implies\begin{cases}x^{11}\equiv13\equiv3\pmod5\\x^{11}\equiv13\equiv6\pmod7\end{cases}$

By Fermat's little theorem, we have

$x^{11}\equiv (x^5)^2x\equiv x^3\equiv3\pmod5$

$x^{11}\equiv x^7x^4\equiv x^5\equiv6\pmod 7$

5 and 7 are both prime, so $\varphi(5)=4$ and $\varphi(7)=6$ . By Euler's theorem, we get

$x^4\equiv1\pmod5\implies x\equiv3^{-1}\equiv2\pmod5$

$x^6\equiv1\pmod7\impleis x\equiv6^{-1}\equiv6\pmod7$

Now we can use the Chinese remainder theorem to solve for $x$ . Start with

$x=2\cdot7+5\cdot6$

Taken mod 5, the second term vanishes and $14\equiv4\pmod5$ . Multiply by the inverse of 4 mod 5 (4), then by 2.

$x=2\cdot7\cdot4\cdot2+5\cdot6$

Taken mod 7, the first term vanishes and $30\equiv2\pmod7$ . Multiply by the inverse of 2 mod 7 (4), then by 6.

$x=2\cdot7\cdot4\cdot2+5\cdot6\cdot4\cdot6$

$\implies x\equiv832\pmod{5\cdot7}\implies\boxed{x\equiv27\pmod{35}}$

$x^5\equiv3\pmod{64}$

We have $\varphi(64)=32$ , so by Euler's theorem,

$x^{32}\equiv1\pmod{64}$

Now, raising both sides of the original congruence to the power of 6 gives

$x^{30}\equiv3^6\equiv729\equiv25\pmod{64}$

Then multiplying both sides by $x^2$ gives

$x^{32}\equiv25x^2\equiv1\pmod{64}$

so that $x^2$ is the inverse of 25 mod 64. To find this inverse, solve for $y$ in $25y\equiv1\pmod{64}$ . Using the Euclidean algorithm, we have

64 = 2*25 + 14

25 = 1*14 + 11

14 = 1*11 + 3

11 = 3*3 + 2

3 = 1*2 + 1

=> 1 = 9*64 - 23*25

so that $(-23)\cdot25\equiv1\pmod{64}\implies y=25^{-1}\equiv-23\equiv41\pmod{64}$ .

So we know

$25x^2\equiv1\pmod{64}\implies x^2\equiv41\pmod{64}$

Squaring both sides of this gives

$x^4\equiv1681\equiv17\pmod{64}$

and multiplying both sides by $x$ tells us

$x^5\equiv17x\equiv3\pmod{64}$

Use the Euclidean algorithm to solve for $x$ .

64 = 3*17 + 13

17 = 1*13 + 4

13 = 3*4 + 1

=> 1 = 4*64 - 15*17

so that $(-15)\cdot17\equiv1\pmod{64}\implies17^{-1}\equiv-15\equiv49\pmod{64}$ , and so $x\equiv147\pmod{64}\implies\boxed{x\equiv19\pmod{64}}$

5 0

2 years ago

Someone help pleaseeee

qaws [65]

Answer:

Part A = D

Part B = 5 and 50

Step-by-step explanation:

Part A:

area = 5*4x = 20x

so, $100 \leq 20x \leq 1000$

Part B:

to leave it only by x, you divide both sides by 20

100/20 = 5, 1000/20 = 50

which $5\leq x\leq 50$

7 0

3 years ago