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4 years ago

14

Find the length of the segment with endpoints of (3,2) and (-3,-6).

1 answer:

MakcuM [25]4 years ago

5 0

The formula of the length of the segment AB:
$A(x_A;\ y_A);\ B(x_B;\ y_B)\\\\|AB|=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}$
We have:
$A(3;\ 2)\to x_A=3;\ y_A=2\\\\B(-3;\ -6)\to x_B=-3;\ y_B=-6$
Substitute:
$|AB|=\sqrt{(-3-3)^2+(-6-2)^2}=\sqrt{(-6)^2+(-8)^2}\\\\=\sqrt{36+64}=\sqrt{100}=10$
Answer: B) 10.

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Two cities are 252 miles apart. How far are the two cities from one another on a map with a scale of 1/4 inches: 18 miles?

tekilochka [14]

Answer:

3.5 inches

Step-by-step explanation:

create a proportion:

inches / miles = inches / miles

let 'x' = distance in inches

<u>1/4</u> = <u> x </u>

18 252

18x = 252(1/4)

18x = 63

x = 63 / 18

x = 3.5

5 0

3 years ago

Write a fraction that is less than 5/6 and has a denominator of 8

KatRina [158]

You'll want to make a common denominator with 6 and 8.
That denominator would be 24.
24/6=4 so you would have to multiply 5/6 by 4/4 to get 20/24.
Next, 24/8=3 so 1/8 could be multiplied by 3/3 to get 3/24.
Since 3 is less than 20, 1/8 is smaller than 5/6.

If you want the same numerator, 5/8 = 15/24. This would make 5/8 smaller than 5/6 as well.
The highest eighth you can go is 6/8 which is 18/24.
So you can use any numerator between 1 and 6 with a denominator of 8 to get a fraction smaller than 5/6.

3 0

3 years ago

Simplify 2⁰+5¹+4³/7

Alekssandra [29.7K]

Your question has been heard loud and clear.

2^0= 1

5^1= 5

4^3=64

2^0+5^1+4^3/7= 1+5+64/7 = 6+64/7= 15.14

So , 2^0+5^1+4^3/7 = 15.14

Thank you

8 0

4 years ago

Help me I’m. It big brain

umka21 [38]

Answer:

A 35

Step-by-step explanation:

75-40=35

4 0

3 years ago

Read 2 more answers

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

tino4ka555 [31]

a.

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

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.

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

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

b.

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

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

3 years ago