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2 years ago

9

the graph of y=e^tanx-2 crosses the x axis at one point in the interval [0,1]. What is the slope of the graph at this point? I g

et 3.08267 but it's not the right answer

1 answer:

Svetlanka [38]2 years ago

3 0

Y = e^tanx - 2

To find at which point it crosses x axis we state that y= 0

e^tanx - 2 = 0
e^tanx = 2
tanx = ln 2
tanx = 0.69314
x = 0.6061

to find slope at that point first we need to find first derivative of funtion y.

y' = (e^tanx)*1/cos^2(x)

now we express x = 0.6061 in y' and we get:
y' = k = 2,9599

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I’ll use 1/2 for each Just multiply the 1/2 by the 8

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3 years ago

Pls help, and explain. I will give brainliest

-BARSIC- [3]

Answer:

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3 years ago

A cyclist bikes at a constant speed for 22 miles. He then returns home at the same speed but takes a different route. His return

brilliants [131]

How concerned are you with how to get the answer?
Because I know the answer is 5 miles per hour
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Alright this is how you solve it
Distance over time equals speed
So 22/y=x and 27/y+1=x
Start by solving for time by having these two equations equal each other
22/y=27/y+1
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3 years ago

Х- а<br>x-b<br>If f(x) = b.x-a÷b-a + a.x-b÷a - b<br>Prove that: f (a) + f(b) = f (a + b)

GenaCL600 [577]

Given:

Consider the given function:

$f(x)=\dfrac{b\cdot(x-a)}{b-a}+\dfrac{a\cdot(x-b)}{a-b}$

To prove:

$f(a)+f(b)=f(a+b)$

Solution:

We have,

$f(x)=\dfrac{b\cdot(x-a)}{b-a}+\dfrac{a\cdot (x-b)}{a-b}$

Substituting $x=a$ , we get

$f(a)=\dfrac{b\cdot(a-a)}{b-a}+\dfrac{a\cdot (a-b)}{a-b}$

$f(a)=\dfrac{b\cdot 0}{b-a}+\dfrac{a}{1}$

$f(a)=0+a$

$f(a)=a$

Substituting $x=b$ , we get

$f(b)=\dfrac{b\cdot(b-a)}{b-a}+\dfrac{a\cdot (b-b)}{a-b}$

$f(b)=\dfrac{b}{1}+\dfrac{a\cdot 0}{a-b}$

$f(b)=b+0$

$f(b)=b$

Substituting $x=a+b$ , we get

$f(a+b)=\dfrac{b\cdot(a+b-a)}{b-a}+\dfrac{a\cdot (a+b-b)}{a-b}$

$f(a+b)=\dfrac{b\cdot (b)}{b-a}+\dfrac{a\cdot (a)}{-(b-a)}$

$f(a+b)=\dfrac{b^2}{b-a}-\dfrac{a^2}{b-a}$

$f(a+b)=\dfrac{b^2-a^2}{b-a}$

Using the algebraic formula, we get

$f(a+b)=\dfrac{(b-a)(b+a)}{b-a}$ $[\because b^2-a^2=(b-a)(b+a)]$

$f(a+b)=b+a$

$f(a+b)=a+b$ [Commutative property of addition]

Now,

$LHS=f(a)+f(b)$

$LHS=a+b$

$LHS=f(a+b)$

$LHS=RHS$

Hence proved.

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2 years ago

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netineya [11]

29.9999 would be the answer here's how i worked it out

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