What will be the local time in Nigeria 15oE when the time is 7.00 p.m

Complete the equivalent equation for -7x-60 =x^2 +10x

alex41 [277]

Answer:

(x+12)(x+5)

Step-by-step explanation:

Formula use: a²+bx+c

Make one side equal to zero:

Original:

-7x-60 =x² +10x

New:

x² + 17x + 60

Multiply A with C

New:

(1)x x 60 = 60

Find factors of 60 that when added, equal to 17.

New:

10 × 6, 60 × 1, 20 × 3, <u>5 × 12</u>, 4 × 15

5 times 12 equal 60, but when added equal to 17.

Replace the 17 with 5 and 12

New:

x² + 5x + 12x + 60

Break them off into two equations

New:

x² + 5x l 12x + 60

Divide each equation into it's simpilest form. Make sure the numbers in the ( ) are the same.

New:

x(x + 5) l +12(x+5)

Answer: (x+12)(x+5)

6 0

3 years ago

FeCl3 + O2  Fe2O3 + Cl2 BALANCE the equation

Tomtit [17]

4 miles of FeCl3 x (6 Cl2 / 4 FeCl3)=6 moles Cl2 ( Yahoo )

3 0

2 years ago

Do all perpendicular lines have negative reciprocal slopes?

inna [77]

Not necessarily, the more correct definition is opposite reciprocal slopes.

The example used is how horizontal and vertical lines are parallel. Horizontal lines have a slope of 0, also written as 0/1. However, vertical lines have an undefined slope, which isn't necessarily negative. It has a slope of 1/0, which is undefined. In this case, the reciprocal isn't negative.

In all other cases (1 and -1, 2 and -1/2, etc.) yes, the perpendicular pairs are negative and reciprocal.

6 0

2 years ago

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

2 years ago

I need to be able to find the value of x

drek231 [11]

If a 4 sided figure has 360°...for each side we add, we add 180°.

you have a 7 sided figure (360° + 3(180°) = 900°)

adding up your angles (781) and subtract it from 900 =

x = 119°

8 0

3 years ago

What will be the local time in Nigeria 15oE when the time is 7.00 p.m​

What will be the local time in Nigeria 15oE when the time is 7.00 p.m