All categories

3 years ago

Need help pls question 3 pls pls pls

Mathematics

1 answer:

MaRussiya [10]3 years ago

4 0

Answer:

maybe 67 ft not sure though

You might be interested in

What is equivalent to (4xy-3z)^2 and<br> what type of special product is it?

sergey [27]

Answer:

the answer on edgenuty is option D

Step-by-step explanation:

i just took the quiz

3 0

3 years ago

while roping a fence post jane catches the post 5/6 of the time. time catches the post 3/4 of the time. who catches the post mor

White raven [17]

5/6 is equivalent to 20/24

3/4 is equivalent to 18/24

so 5/6 is bigger than 3/4 by 2/24

3 0

3 years ago

Find the area between the graph of the function and the x-axis over the given interval, if possible.

vodomira [7]

Answer:

$A= -\frac{5}{-1} - \lim_{x\to\infty} \frac{5}{x-2} = 5-0 = 5$

So then the integral converges and the area below the curve and the x axis would be 5.

Step-by-step explanation:

In order to calculate the area between the function and the x axis we need to solve the following integral:

$A = \int_{-\infty}^1 \frac{5}{(x-2)^2}$

For this case we can use the following substitution $u = x-2$ and we have $dx = du$

$A = \int_{a}^b \frac{5}{u^2} du = 5\int_{a}^b u^{-2}du$

And if we solve the integral we got:

$A= -\frac{5}{u} \Big|_a^b$

And we can rewrite the expression again in terms of x and we got:

$A = -\frac{5}{x-2} \Big|_{-\infty}^1$

And we can solve this using the fundamental theorem of calculus like this:

$A= -\frac{5}{-1} - \lim_{x\to\infty} \frac{5}{x-2} = 5-0 = 5$

So then the integral converges and the area below the curve and the x axis would be 5.

7 0

3 years ago

What are the vertex, axis of symmetry, domain and range of the function:

DiKsa [7]

Y=(-x*x)-4x+3
y=-(x^2+4x-3)
4+y=-(x^2+4x+4)+3
y+4=-(x-2)^2.) +3
-4. -4
y=-((x-2)^2)-1

Answers:
Vertex: (2,-1)
AOS:x = 2
Domain: All Real Numbers
Range:[-1,Infinity)

3 0

3 years ago

What are the discontinuities of the function y= 1500/x

Ratling [72]

The discontinuity occurs at x = 0, since that is the only "problem" place in the graph that makes the function undefined. A vertical asymptote exists there. It is nonremoveable.

4 0

3 years ago