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3 years ago

I will give 50 pts, please answer, im so sad and need help

Mathematics

2 answers:

arlik [135]3 years ago

8 0

Answer:

b. (-4/3)

Step-by-step explanation:

the average rate of change in the interval is the slope

(-1,4) (2,0)

slope = (y2-y1)/(x2-x1)

slope = (0-4)/(2--1)

slope = (-4)/(2+1)

slope = (-4/3

Afina-wow [57]3 years ago

5 0

Answer:

First, you want to solve for the equation for this problem, would be:

0.05N + 0.10D = 20.50

While N = The amount of nickels, and D = The amount of dimes.

Since N = 164, and D = 123. It would add up to 287 coins. 164 + 123 = 287.

Now that you have the number for those two variables, solve the equation for when N = 164, and D = 123.

0.05N + 0.10D = 20.50

0.05(164) + 0.10(123) = 20.50

8.2 + 12.3 = 20.50

20.50 = 20.50

So, there is 164 nickels, and 123 dimes for your answer.

I hope this helps!

Step-by-step explanation:

You might be interested in

what is the slope that passed through the points (-3, -9) and (22, -4)? Write your answer in a simplest form

andrew11 [14]

Answer:

साहस को सलाम पाठ/अरुणिमा सिन्हा कहां बैठी थी

Step-by-step explanation:

I hope you understand

8 0

3 years ago

Will mark brainly if u show work !

mihalych1998 [28]

Answer:

Step-by-step explanation:

The 6 comes from the 24(6*4=24)

The 3 comes from the 12(3*4=12)

The 4 is common

7 0

3 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

3 years ago

The area of a rectangular of length L width W is given by the formula A =LW. Ann's backyard is rectangular with a length of 27.6

marta [7]

300.84m^2

First, set up the equation

A=27.6•10.9

Next, multiply(use a calculator)

And you get the answer.

^2 means squared.

3 0

3 years ago

The temperature is 45°F. The temperature will decrease by 2°F each hour. Let h be the number of hours.

Varvara68 [4.7K]

45 - 2t < 32 ( where t is time and you can replace it with x or h if you prefer)
Solving for t you get:
45 - 2t - 32 < 0
45 - 32 < 2t
13 < 2t
13/2 < t
Or the temperature will drop below 32F after 6 and a half hours

8 0

3 years ago