All categories

2 years ago

Solve foR a12-3(a-1)=11+a

Mathematics

2 answers:

9966 [12]2 years ago

8 0

Answer:

Solve for a

a12−3(a−1)=11+a

Write the problem as a mathematical expression.

a12−3(a−1)=11+a

Simplify each term.

a12−3a+3=11+a

Graph each side of the equation. The solution is the x-value of the point of intersection.

a≈−1.1114621,1.23792371

Step-by-step explanation:

sergeinik [125]2 years ago

7 0

Answer:

The answer is...

a=-2

You might be interested in

Help help help help help help

maw [93]

Answer:

B) d=4t

Step-by-step explanation:

let's look at two points

1. t=0, d = 0

2. t = 1, d = 4

slope = change in vertical axis (d)/ change in horizontal axis (t)

change in d = 4 = d₂-d₁

change in t = 1 = t₂-t₁

slope = 4/1 = 4

vertical variable = slope * horizontal variable + horizontal axis intercept

d = 4 * t + horizontal axis intercept

horizontal axis intercept is the vertical axis variable (d) when the horizontal axis variable (t) is at 0. at t=0, d=0

thus, d = 4 * t + 0

d = 4t (B)

4 0

1 year ago

Pre algebra help ( will give brainliest ) sorry if it's hard to see

vladimir2022 [97]

Its eq 4 its 41 and all the other are 21 and lower.

3 0

2 years ago

Read 2 more answers

WHAT IS THE ANWSER TO<br> 6(G+9)=

k0ka [10]

6(G+9)

you have to use the distributive property.

6*G=6G

6*9=54

answer is 6G+54

6 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]

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

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

A line represented by y = 2x − 3 and a line perpendicular to it intersect at R(2, 1). What is the equation of the perpendicular

neonofarm [45]

Answer:

y = -1/2x + 2

Step-by-step explanation:

y = 2x - 3. The slope here is 2. A perpendicular line will have a negative reciprocal slope. To find the negative reciprocal, just flip the slope and change the sign.

slope = 2 or 2/1.....flip it.....1/2....change the sign...-1/2. So our perpendicular line will have a slope of -1/2.

y = mx + b

slope(m) = - 1/2

(2,1)....x = 2 and y = 1

now we sub and find b, the y intercept

1 = -1/2(2) + b

1 = -1 + b

1 + 1 = b

2 = b

so ur equation is : y = -1/2x + 2 <===

and u can check it with ur points....(2,1)

y = 2x - 3

1 = 2(2) - 3

1 = 4 - 3

1 = 1 (yep...it checks out)

8 0

2 years ago