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Bogdan [553]
3 years ago
10

What’s the equation for the two points (19, 3) and (20,3)

Mathematics
1 answer:
Orlov [11]3 years ago
7 0

Answer:

Equation of line passing through point A = ( 19 , 3 ) , and point B = ( 20 , 3 ) is

y−3=0

OR

Equation of line passing through point A = ( 19 , 3 ) , and point B = ( 29 , 3 ) is

y=0x+3

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Solve for x 11x+4<15 OR 12x−7>−25
never [62]

Isolate the x in both cases. What you do to one side, you do to the other.


11x + 4 < 15


Subtract 4 from both sides


11x + 4 (-4) < 15 (-4)


11x < 11


isolate the x, divide 11 from both sides


11x/11 < 11/11


x< 1


x < 1 is your answer for the first one

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


Again, isolate the x.


12x - 7 > -25


Add 7 to both sides


12x - 7 (+7) > -25 (+7)


12x > -18


Isolate the x, divide 12 from both sides


12x/12 > -18/12


x > -1.5 is your answer for the second one.


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



hope this helps

3 0
3 years ago
Read 2 more answers
A polynomial p has zeros when x=5, x=-1 and x=-1/4
kiruha [24]

Answer:

p(x) = (5x - 1) (x + 4) (x - 2)

4 0
3 years ago
Read 2 more answers
Using subsitution, which value of x makes this equation true?
julia-pushkina [17]

Answer:

x = 3/5

Step-by-step explanation:

After substituting each option into the equation, we find that the only option that makes the equation true is x=3/5, since 4/5 - 3/5 = 1/5.

8 0
2 years ago
Read 2 more answers
The width of a rectangle is 6 in. less than its length. The perimeter is 68 in.
erma4kov [3.2K]
2L + 2(L-6) = 68

2L + 2L -12 = 68

4L = 80

L = 20

The length is 20 in. and the width is 14 in.
8 0
4 years ago
Consider the initial value problem y′′+36y=2cos(6t),y(0)=0,y′(0)=0. y″+36y=2cos⁡(6t),y(0)=0,y′(0)=0. Take the Laplace transform
Bingel [31]

Recall the Laplace transform of a second-order derivative,

L(y''(t)) = s^2Y(s)-sy(0)-y'(0)

and the transform of cosine,

L(\cos(at))=\dfrac s{a^2+s^2}

Here, both y(0)=y'(0)=0, so taking the transform of both sides of

y''(t)+36y(t)=2\cos(6t)

gives

s^2Y(s)+36Y(s)=\dfrac{2s}{36+s^2}

\implies Y(s)=\dfrac{2s}{(s^2+36)^2}

4 0
3 years ago
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