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astraxan [27]
2 years ago
9

What is the distance between parallel lines y=x+3 and y=x+1, rounded to the nearest hundredth?

Mathematics
1 answer:
riadik2000 [5.3K]2 years ago
3 0

Answer:

The distance is 1.41 to nearest hundredth.

Step-by-step explanation:

The line y = x + 3 passes through the point (0, 3) on the yaxis.

Find the perpendicular line passing through this point:

y  - 3 = -1(x + 0)

y = -x + 3

Now find the point where this line intersects the line y = x + 1:

y = -x + 3

y = x + 1

-x + 3 = x  + 1

3 -1 = x + x

2x = 2

x = 1

and y = 2

So this point is (1, 2)

We require the distance between this point and (0, 3)

This = √((3-2)^2 + (0-1)^2) = √2

= 1.4142

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3 years ago
Read 2 more answers
I need help 6x-2y=10 x-2y=-5 solve by elimination
dem82 [27]
<h3><u>Explanation</u></h3>
  • Given the system of equations.

\begin{cases} 6x - 2y = 10 \\ x - 2y =  - 5 \end{cases}

  • Solve the system of equations by eliminating either x-term or y-term. We will eliminate the y-term as it is faster to solve the equation.

To eliminate the y-term, we have to multiply the negative in either the first or second equation so we can get rid of the y-term. I will multiply negative in the second equation.

\begin{cases} 6x - 2y = 10 \\  - x  +  2y =  5 \end{cases}

There as we can get rid of the y-term by adding both equations.

(6x - x) + ( - 2y + 2y) = 10 + 5 \\ 5x + 0 = 15 \\ 5x = 15 \\ x =  \frac{15}{5}  \longrightarrow  \frac{ \cancel{15}}{ \cancel{5}}  =  \frac{3}{1}  \\ x = 3

Hence, the value of x is 3. But we are not finished yet because we need to find the value of y as well. Therefore, we substitute the value of x in any given equations. I will substitute the value of x in the second equation.

x - 2y =  - 5 \\ 3 - 2y =  - 5 \\ 3 + 5 = 2y \\ 8 = 2y \\  \frac{8}{2}  = y \\ y =  \frac{8}{2} \longrightarrow  \frac{ \cancel{8}}{ \cancel{2}}  =  \frac{4}{1}  \\ y = 4

Hence, the value of y is 4. Therefore, we can say that when x = 3, y = 4.

  • Answer Check by substituting both x and y values in both equations.

<u>First</u><u> </u><u>Equation</u>

6x - 2y = 10 \\ 6(3) - 2(4) = 10 \\ 18 - 8 = 10 \\ 10  = 10 \longrightarrow \sf{true} \:  \green{ \checkmark}

<u>Second</u><u> </u><u>Equation</u>

x - 2y =  - 5 \\ 3 - 2(4) =  - 5 \\ 3 - 8 =  - 5 \\  - 5 =  - 5 \longrightarrow  \sf{true} \:  \green{ \checkmark}

Hence, both equations are true for x = 3 and y = 4. Therefore, the solution is (3,4)

<h3><u>Answer</u></h3>

\begin{cases} x = 3 \\ y = 4 \end{cases} \\  \sf \underline{Coordinate \:  \: Form} \\ (3,4)

8 0
3 years ago
Evaluate the expression for x=5 and y= 3
kenny6666 [7]

I will use the positive / negative property of the absolute value to split the equation into two cases, and I will use the fact that the "minus" sign in the negative case indicates "the opposite sign", not "a negative number".

For example, if I have x = –6, then "–x " indicates "the opposite of x" or, in this case, –(–6) = +6, a positive number. The "minus" sign in "–x" just indicates that I am changing the sign on x. It does not indicate a negative number. This distinction is crucial!

Whatever the value of x might be, taking the absolute value of x makes it positive. Since x might originally have been positive and might originally have been negative, I must acknowledge this fact when I remove the absolute-value bars. I do this by splitting the equation into two cases. For this exercise, these cases are as follows:

a. If the value of x was non-negative (that is, if it was positive or zero) to start with, then I can bring that value out of the absolute-value bars without changing its sign, giving me the equation x = 3.

b. If the value of x was negative to start with, then I can bring that value out of the absolute-value bars by changing the sign on x, giving me the equation –x = 3, which solves as x = –3.

The theory can be encapsulated in a main equation called the standard model Lagrangian (named after the 18th-century French mathematician and astronomer Joseph Louis Lagrange), which was chosen by theoretical physicist Lance Dixon of the SLAC National Accelerator Laboratory in California as his favorite formula.

"It has successfully described all elementary particles and forces that we've observed in the laboratory to date — except gravity," Dixon told LiveScience. "That includes, of course, the recently discovered Higgs(like) boson, phi in the formula. It is fully self-consistent with quantum mechanics and special relativity."

The first line is traditionally called forces. That Tensor quantity indicates electromagnetism and both nuclear forces. Importantly, that term does this in a way that preserves certain symmetries. They are called gauge symmetries, and even though they are hard to explain, they are very important in physics. You’ll hear more about them later.

The second line is what I might call motion. It is the reason why electrons and quarks can move. It is also what allows the forces to interact with matter.

The third line is called the Yukawa coupling. It gives mass to things like quarks and electrons. It actually used to look different. That ϕ didn’t used to be there. It was added because it was the only way to have mass while preserving gauge symmetry.

The last line is called the Higgs sector. If you’ve heard of the Higgs boson, it’s the same guy. This term is responsible for ‘spontaneous symmetry breaking’. This means that even though the universe does indeed have gauge symmetry, that symmetry can often be obscured at low temperatures. This explains the masses we observe for particles.


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Step-by-step explanation:

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