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prisoha [69]
3 years ago
10

What is the difference between (5,3) and (-2,9)

Mathematics
1 answer:
Sidana [21]3 years ago
7 0

Answer:

(7,-7) that's it according to me

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Chris is working on math homework. He solves the equation m/6=48 and says that the solution is m=8. Do you agree or disagree wit
vredina [299]

Chris is incorrect. To check the answer, replace m with 8 and simplify the left side. So m/6 turns into 8/6 which converts to the decimal value 1.333 (use a calculator)

Therefore the original equation m/6 = 48 becomes 1.333 = 48 after you plug in m = 8 and simplify fully. The two sides of 1.333 = 48 are not the same, so m = 8 is not the solution.

--------------------

The solution is actually 288 and we can prove it so like this

m/6 = 48

288/6 = 48 .... replace m with 288

48 = 48 .... use a calculator to compute 288/6

So the solution m = 288 is confirmed

------------------

How did I get this answer? By multiplying both sides by 6

m/6 = 48

6*(m/6) = 6*48 ..... multiply both sides by 6

m = 288

So Chris mistakenly divided both sides by 6, which explains how he got 48/6 = 8 as the solution. Instead he should have multiplied both sides by 6. This is to undo the operation "divide by 6" that is being applied to m in the original equation.

7 0
3 years ago
Solve the following 3 × 3 system. Enter the coordinates of the solution below.
ICE Princess25 [194]

Step-by-step explanation:

2x - 3y - 2z = 4

[2] x + 3y + 2z = -7

[3] -4x - 4y - 2z = 10

Solve by Substitution :

// Solve equation [2] for the variable x

[2] x = -3y - 2z - 7

// Plug this in for variable x in equation [1]

[1] 2•(-3y-2z-7) - 3y - 2z = 4

[1] - 9y - 6z = 18

// Plug this in for variable x in equation [3]

[3] -4•(-3y-2z-7) - 4y - 2z = 10

[3] 8y + 6z = -18

// Solve equation [3] for the variable z

[3] 6z = -8y - 18

[3] z = -4y/3 - 3

// Plug this in for variable z in equation [1]

[1] - 9y - 6•(-4y/3-3) = 18

[1] - y = 0

// Solve equation [1] for the variable y

[1] y = 0

// By now we know this much :

x = -3y-2z-7

y = 0

z = -4y/3-3

// Use the y value to solve for z

z = -(4/3)(0)-3 = -3

// Use the y and z values to solve for x

x = -3(0)-2(-3)-7 = -1

Solution :

{x,y,z} = {-1,0,-3}

6 0
3 years ago
What is the quadratic. Formula?
balu736 [363]

Answer:

x =   \frac{ - b   ± \sqrt{ {b }^{2}  -  4ac } }{2a}

Step-by-step explanation:

For example, we'll use this quadratic equation.

{x}^{2}  + 5x + 6

To understand how to plug it into the formula we need to know what each term represents.

a {x}^{2}  + bx + c

So the equation above would be put into the formula like this.

x =  \frac{ - 5± \sqrt{ {5}^{2}  -  4(1)(6) } }{2(1)}

Then we would solve

\frac{ - 5± \sqrt{25 - 24} }{2}  \\ \\  =  \frac{ -5±1}{2}

Now, the equation will branch off into one that solves when addition and one when subtraction.

\frac{ - 5 + 1}{2}  =  \frac{ - 4}{2}  =  - 2 \\  \\   \frac{ - 5 - 1}{2}  =  \frac{ - 6}{2}  =  - 3

So x={-3, -2} (-3 and -2)

6 0
3 years ago
What is the slope of the line segment?<br><br> 5<br> 1/5<br> -1/5<br> -5
Elanso [62]

Answer:

5

Step-by-step explanation:

Calculate slope by choosing two points. (1, 5) and (2, 10)

slope = (10 - 5) / (2 - 1) = 5/1 = 5

6 0
3 years ago
For f(x)=√(2x+1) , find the following:
Wittaler [7]

Part a.

The domain is the set of x values such that x \ge -\frac{1}{2}, basically x can be equal to -1/2 or it can be larger than -1/2. To get this answer, you solve 2x+1 \ge 0 for x (subtract 1 from both sides; then divide both sides by 2). I set 2x+1 larger or equal to 0 because we want to avoid the stuff under the square root to be negative.

If you want the domain in interval notation, then it would be \Big[ -\frac{1}{2} , \infty \Big) which means the interval starts at -1/2 (including -1/2) and then it stops at infinity. So technically it never stops and goes on forever to the right.

-----------------------

Part b.

I'm going to use "sqrt" as shorthand for "square root"

f(x) = sqrt(2x+1)

f(10) = sqrt(2*10+1) ... every x replaced by 10

f(10) = sqrt(20+1)

f(10) = sqrt(21)

f(10) = 4.58257569 which is approximate

-----------------------

Part c.

f(x) = sqrt(2x+1)

f(x) = sqrt(2(x)+1)

f(x+2a) = sqrt(2(x+2a)+1) ... every x replaced by (x+2a)

f(x+2a) = sqrt(2x+4a+1) .... distribute

we can't simplify any further

6 0
3 years ago
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