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LenaWriter [7]
4 years ago
13

CAN YALL HELP PLZZzz

Mathematics
2 answers:
yulyashka [42]4 years ago
6 0
The answer would be the third one .
Perpendicular would always be the opposite of your original slope.
Hope this helps! (:
notka56 [123]4 years ago
6 0
Y - 3 = (-3/8)(x + 2)


Because,


The gradient has been inverted, showing that the line is perpendicular (indicated by the -1/m).

Additionally, slotting -2 in to that equation gives us 3, which is the y component we need (assuming that y is isolated, and made the subject, otherwise it would equal 0). This shows that this line passes through (-2, 3).
You might be interested in
What is the slope of the line that passes through the points (-2, 7) and (2, -5)?
amm1812

Answer:-3

Step-by-step explanation:

6 0
3 years ago
I really need help some one please
Alika [10]
The factorization of A is y = (x - 8)(x + 7).
The factorization of B is y = (x + 1)(x - 4)(x - 5)

In order to find these, you must first find where each graph crosses the x-axis. In the first problem it does so at 8 and -7. In order to find the correct parenthesis for those, you need to write it out as a statement and then solve for 0. 

x = 8 ---> subtract 8 from both sides
x - 8 = 0
This means we use (x -  8) in our factorization. 

You then need to repeat the process until you have all the pieces. In the second problem, there will be 3 instead of 2 since it crosses the axis 3 times. 
7 0
3 years ago
How do I find the value ?
ipn [44]
Value of what buddy?
5 0
3 years ago
Read 2 more answers
The tangent of the graph of y=<img src="https://tex.z-dn.net/?f=e%5E%7Bax%7D" id="TexFormula1" title="e^{ax}" alt="e^{ax}" align
bearhunter [10]

a) The tangent to y=e^{ax} at x=p has slope

y' = ae^{ax} \implies y'(p) = ae^{ap}

and y=e^{ap} at this point. It passes through the origin, so its equation is

y - 0 = ae^{ap} (x - 0) \implies y = ae^{ap} x

It also passes through the point (p,e^{ap}) on the curve, so

y - e^{ap} = ae^{ap} (x - p) \implies y = e^{ap} + ae^{ap} x - ape^{ap}

By substitution,

ae^{ap} x = e^{ap} + ae^{ap} x - ape^{ap} \implies e^{ap} = ape^{ap} \implies ap=1 \\\\ \implies \boxed{p=\dfrac1a}

b) The normal to y=e^{ax} at x=2p has slope

-\dfrac1{y'(2p)} = -\dfrac1{ae^{2ap}} = -1 \implies ae^{2ap} = 1

It follows that

ae^{2ap} = ae^2 = 1 \implies \boxed{a = \dfrac1{e^2} \text{ and } p = e^2}

c) The tangent line equation is then

y = \dfrac1{e^2} e^1 x \implies \boxed{y = \dfrac xe}

7 0
2 years ago
Suppose that y varies inversely with x, and y = 2 when x = 4. What is an<br>variation?​
Alexeev081 [22]

Answer:

xy=8

Step-by-step explanation:

The equation for inverse variation is

xy = k where k is the constant of variation

4*2 = k

8=k

The equation is

xy=8

3 0
4 years ago
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