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miskamm [114]
3 years ago
5

A straight line has an equation y= 4x=5 Write down the equation of the straight line that is parallel to y=4x-5 and passes throu

gh the point (0,3).
Mathematics
2 answers:
Eduardwww [97]3 years ago
7 0


Parallel lines have same slope but different y-intercept. So they gave the equation y=4x-5 and the (0,3) where 3 is the y-intercept.


So the equation would be y=4x+3. HOPE THIS HELPS!!!!!!!!

ratelena [41]3 years ago
6 0
If it is parallel then it has the gradient which is 4 so the equation will be
y = 4x + ?

now to find the y intercept, using the coordinates, x is already 0 so the y intercept is 3

meaning the equation is 
y = 4x + 3
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Answer:

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Step-by-step explanation:

<u>Given functions</u>:

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Solve for p(x) = r(x):

\begin{aligned}p(x) & = r(x)\\\implies 2x-4 & = \dfrac{6x-1}{9x+1}\\(2x-4)(9x+1)&=6x-1\\18x^2+2x-36x-4&=6x-1\\18x^2-40x-3&=0\end{aligned}

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x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0

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a=18, \quad b=-40, \quad c=-3

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\begin{aligned}\implies x & =\dfrac{-(-40) \pm \sqrt{(-40)^2-4(18)(-3)} }{2(18)}\\\\& =\dfrac{40 \pm \sqrt{1816}}{36}\\\\& =\dfrac{40 \pm \sqrt{4 \cdot 454}}{36}\\\\& =\dfrac{40 \pm \sqrt{4}\sqrt{454}}{36}\\\\& =\dfrac{40 \pm2\sqrt{454}}{36}\\\\& =\dfrac{20 \pm\sqrt{454}}{18}\end{aligned}

Therefore, the solutions are:

x =\dfrac{20 +\sqrt{454}}{18}, \quad \dfrac{20 -\sqrt{454}}{18}

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6 0
2 years ago
7. If F(x) = f(g(x)), where f(-2) = 8, f'(-2) = 4, f'(5) = 3, g(5)=-2, gʻ(5) = 6.. Find F'(5) <br>​
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Answer:

I'm going to lay this out in a chart so it's a little easier to see:

F(x) = f(g(x))

x | f (x) | f ' (x) | g (x) | g ' (x)

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

-2 | 8 | 4 |

5 | | 3 | -2 | 6

Remember the chain rule, which says

(f (g (x))) ' = g ' (x) f ' (g (x))

When they ask for F ' (5), they are asking for (f (g (x))) ' when x = 5.

Using the chain rule, that's

F ' (5) = g ' (5) f ' (g (5))

We can simplify using the numbers provided.

F ' (5) = (6) f ' (-2)

F ' (5) = (6) (4)

F ' (5) = 24

I hope that helps!

by jannat <33

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3 years ago
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Option D is correct.

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From the diagram attached, we can see that the triangles on the graph had similar ratios, as such their vertical heights(y-coordinates) to their horizontal (x-coordinates) are equivalent.

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