Find the distance between the pair of parallel lines, y = x-11 & y = x-7

Answers

Answer 1

Answer: $k:\ y = x-11\ \ \ \Leftrightarrow\ \ \ x-y-11=0\n and\n l:\ y = x-7\ \ \ \Leftrightarrow\ \ \ x-y-7=0\n\nthe\ distance:\n\n d(k;l)= \frac{\big{|-11-(-7)|}}{\big{ √(1^2+1^2) }} =\frac{\big{|-11+7|}}{\big{ √(2) }} =\frac{\big{|-4|}}{\big{ √(2) }} =\frac{\big{4\cdot √(2) }}{\big{ √(2)\cdot √(2) }} =\frac{\big{4 √(2) }}{\big{2 }} =2 √(2)$

Answer 2

Answer: $Given \ the \ equations \ of \ two \ non-vertical \ parallel \ lines:\n\ny = mx+b_1\ny = mx+b_2\n\nthe \ distance \ between \ them \ can \ be \ expressed \ as : \n\nd= (|b_(1)-b_(2)|)/( √( m^2+1) )$

$y = x-11 \n y = x-7 \n\n\nd= (| -11- (-7)|)/( √( 1^2+1) ) =(| -11+7|)/( √( 1+1) ) = (|-4|)/( √(2) ) = (4)/( √(2) )\cdot (√(2))/(√(2))=(4√(2))/(2)=2√(2)$

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√21 + √13 + √7 + √4 =
The product of 9 and a number is 12 less than three times that number.What is the number? Giving 20 Points....
Help ASAP Brian ran 4 1/4 miles in 3/4 of an hour. How fast was brian running.
A projectile is fired straight upward with an initial velocity of 400 feet per second. The height of the projectile, h(t), if represented by the function h(t)=-16t^2+400t, where t is the time in seconds. How long does it take the projectile to reach the maximum height?

3a² . 2a³ find the product

Answers

Answer:

$6a^(5)$ .

Step-by-step explanation:

To multiply exponents, you will need to begin by multiplying the coefficients as normal. In this instance:

3 · 2 = 6.

When multiplying exponents, however, remember that you must ADD the exponents. Therefore:

3a² · 2a³ = $6a^(5)$

Final answer:

The product of 3a² and 2a³ is found by multiplying the coefficients (the numbers in front of 'a') and adding the exponents for 'a'. The result is 6a⁵.

Explanation:

To find the product of 3a² and 2a³, first you multiply the numerical coefficients and then you add the exponents of 'a'. The numerical coefficient for the first expression is 3 and for the second expression it's 2, multiplying these together we get 6. For the powers of 'a', since both 'a' terms in the two expressions are raised to powers (2 and 3), you add these powers together when multiplying, thus a² * a³ = a⁵. So the final answer is 6a⁵.

Learn more about Exponent rules here:

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Find the inverse of the following f(x)=-3x^2+3 X>0

Answers

Answer

f^-1(x) =√[(3-x)÷3]

Step-by-step explanation:

y=-3x^2+3

interchange 'y' with 'x'

x=-3y^2+3

make y the subject

3y^2=3-x

divide by '3' both sides

y^2=(3-x)÷3

apply square root both sides

√(y^2)=√[(3-x)÷3]

y=√[(3-x)÷3]

f^-1(x) =√[(3-x)÷3]

the inverse of the given function is

f^-1(x)=√[(3-x)÷3]

The equation x2 + y2 − 2x + 2y − 1 = 0 is the general form of the equation of a circle. What is the standard form of the equation?

Answers

Answer:

$(x-1)^2+ (y+1)^2=3$

Step-by-step explanation:

$x^2 + y^2 - 2x + 2y - 1 = 0$

Standard form of the equation is $(x-h)^2 + (y-k)^2= r^2$

To get standard form we apply completing the square method

$x^2-2x+ y^2+ 2y - 1 = 0$

Take coefficient of x and y . Divide it by 2 and then square it

$(2)/(2) =1$ and 1^2=1

Add and subtract 1

$(x^2-2x)+(y^2+ 2y) - 1 = 0$

$(x^2-2x+1-1)+(y^2+ 2y+1-1) - 1 = 0$

$(x^2-2x+1)+(y^2+ 2y+1)-1-1- 1 = 0$

$(x^2-2x+1)+(y^2+ 2y+1)-3= 0$

Now write the parenthesis in square form

$(x-1)(x-1)+ (y+1)(y+1)-3= 0$

$(x-1)^2+ (y+1)^2-3= 0$ , add 3 on both sides

$(x-1)^2+ (y+1)^2=3$ is the standard form

x^2 + y^2 - 2x + 2y - 1 = 0

(x^2 - 2x) + (y^2 + 2y) - 1 = 0

(x^2 - 2x + 1) + (y^2 + 2y + 1) - 1 - 1 - 1 = 0

(x - 1)^2 + (y + 1)^2 - 3 = 0

(x - 1)^2 + (y + 1)^2 = 3

The volume of a rectangular box is (x3 - 7x2 - 9x + 63) cubic units. Determine the dimensions of the rectangular box by factoring the volume expression completely.

Answers

Hello,
well done.
As x^3-7x²-9x+63=(x-3)(x+3)(x-7) we must have x>3 and x>-3 and x>7 thus x>7

Explanations:
x^3-7x²-9x+63=x^3-9x+(-7x²+63)=x(x²-9)-7(x²-9)=(x²-9)(x-7)=(x+3)(x-3)(x-7)

3n-5=-8(6+5n) SOLVE

Answers

It's gonna be a long problem: Remember PEMDAS
3n-5=8(6+5n)
3n-5= 48 + 40n Distribute 8 into parenthesis
3n-3n-5=48+40n-3n Put variable on one side.
-5-48=48-48+37n Isolate the variable
-43/37=37n/37 Isolate variable more
-1.162=n Simplify
The answer is repeating; that is a shortened version.

t's gonna be a long problem: Remember PEMDAS

3n-5=8(6+5n)

3n-5= 48 + 40n Distribute 8 into parenthesis

3n-3n-5=48+40n-3n Put variable on one side.

-5-48=48-48+37n Isolate the variable

-43/37=37n/37 Isolate variable more

-1.162=n Simplify

The answer is repeating; that is a shortened version.

Three of four numbers have a sum of 22. If the average of the four numbers is 8, what is the fourth number?

Answers

$a+b+c=22\n \n (a+b+c+x)/(4)=8\n \n a+b+c+x=32\n \n 22+x=32\n \n \boxed{x=10}$

Answer:

Step-by-step explanation:

Let's call the four numbers x1, x2, x3, and x4. Their average is (x1 + x2 + x3 + x4)/4 = 8, so x1 + x2 + x3 + x4 = 4 × 8 = 32. And we know that three of the numbers have a sum of 22, so x1 + x2 + x3 = 22.

Subtracting this last equation from the first equation, we have:

$\sf:\implies{x4 = 32 - (x1 + x2 + x3) = }$

$\sf:\implies{32 - 22 = 10}$

So the fourth number is 10.

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