At what value of x does the graph of the following function f(x) have a vertical asymptope? f(x)=5x/2x-6

Harry drove 2 hours on the freeway then decreased his speed by 20mph and drove 6 more hours. total trip was 352 what was his spe

poizon [28]

Recall your d = rt, distance = rate * time

we know the total trip was 352miles, so, let's say he was going for the first 2hrs at "r" speed, he went "d" miles, then he slowed down to "r - 20" and went 6 more hours, since the total trip is 352 miles, the 6hrs take the slack from 352 - d, and that's how many miles he covered on those 6hours

$\bf \begin{array}{lccclll}
&distance&rate&time\\
&-----&-----&-----\\
\textit{before slowing}&d&r&2\\
\textit{after slowing}&352-d&r-20&6
\end{array}
\\\\\\

\begin{cases}
\boxed{d}=2r\\
352-d=(r-20)6\\
----------\\
352-\boxed{2r}=(r-20)6
\end{cases}$

solve for "r".

7 0

4 years ago

20 POINTS ANSWER PLEASE LIKE NOWWW

spayn [35]

I don't see the options
5.708×10^−8

What is the number in standard form?

5.708×10^−8 = 0.00000005708

hope it helps

5 0

3 years ago

Read 2 more answers

A slide 15 feet long makes an angle of 330 with its vertical ladder. To the

Shkiper50 [21]

Answer:

13ft

Step-by-step explanation:

Kindly find attached a rough draft of the situation.

Step one:

Given data

The length of the slide represents the

Hypotenuse of the situation on the rough sketch

Angle =33°

Required

The height of the ladder which is the adjacent of the rough sketch represented by x

Step two:

Applying SOH CAH TOA

Cos θ= adj/hyp

Cos 33=x/15

0.84=x/15

Cross multiplying

x=0.84*15

x=12.6

To the nearest foot the ladder is 13ft tall

6 0

3 years ago

Solve for x in the equation x squared 11 x startfraction 121 over 4 endfraction = startfraction 125 over 4 endfraction.

Mrrafil [7]

The solutions are { -1,-11}

Lets solve the given equation:

x^2 + 11x + 121/4 = 125/4

Subtracting 125/4 from both sides:

=>x^2 + 11x + 121/4-125/4= 125/4 -125/4

=> x^2 + 11x - (-4/4) = 0

=> x^2 +11x -(-1) = 0

=>x^2 + 11 x + 1 = 0

This is a quadratic equation so we will use the determinant (b^2-4ac)

a = 1

b = 11

c = 1

b^2-4ac = 11^2-4*1*1 = 117

So this equation has two solutions:

x = (-b -/+ √(b^2-4ac) ) / 2a

x = (-11 -/+ √(117) ) / 2

x = (-11 -/+ 3√(13))/ 2

x = -0.91 or x = -10.9

x = -1 or x = -11 ( rounding off)

hence the solutions are { -1,-11}.

Learn more about Quadratic Equation on:

brainly.com/question/1214333

#SPJ4

6 0

2 years ago

I need help with this math work

Akimi4 [234]

9514 1404 393

Answer:

see attached

Step-by-step explanation:

If you plot the points on the graph, you notice that the run (change in x) is 3 for each "rise" (change in y) of -1, so the slope is ...

m = rise/run = -1/3

The middle point is on the y-axis, so is the y-intercept.

In slope-intercept form, the equation is ...

y = mx + b . . . . . . where m is the slope (-1/3) and b is the y-intercept (4)

y = -1/3x +4 . . . . equation

6 0

3 years ago