Answer for 20 points

Alexus [3.1K]

(a) It is leaking 18 liters per minute.

Explanation : From 15 minutes to 20 minutes, your starting point at 15 is 590 liters and at 20 you have exactly 500. Next you have 25 which is now at 410. So it's leaking 90 liters every five minutes, so of course dividing the two (90 ÷ 5) you get 18.

(a)² it increases for about the same time that it would decrease of course.

(b) The vat started at 860 liters before they started draining it.

Explanation : Well, all it takes is just working backwards, add 90 liters to the liters at 15 minutes and deduct five minutes from 15. Do this process until you arrive at 0. It's simple

7 0

3 years ago

A circle has an area of 3.14 cm^2 . What is its radius?

Anni [7]

Answer:

Cant be determined

Step-by-step explanation:

You would need the diameter to figure this out

6 0

3 years ago

Read 2 more answers

Can anybody help with this question

noname [10]

You answer will be either 15% or 13%

8 0

3 years ago

Read 2 more answers

If A cylinder has a radius of 3in and a height of 5in, what would the answer be?

tino4ka555 [31]

Answer:

141.37

Step-by-step explanation:

3 0

2 years ago

Solve the equation by graphing. If exact roots cannot be found, state the consecutive integers between which the roots are locat

eimsori [14]

Answer:

There are NO real roots for this equation. The only roots have imaginary parts and therefore cannot be represented on the real x-axis.

Step-by-step explanation:

We notice that the expression on the left of the equation is a quadratic with leading term $2x^2$ , which means that its graph is that of a parabola with branches going up.

Therefore, there can be three different situations:

1) if its vertex is ON the x axis, there would be one unique real solution (root) to the equation.

2) if its vertex is below the x-axis, the parabola's branches are forced to cross it at two locations, giving then two real solutions (roots) to the equation.

3) if its vertex is above the x-axis, it will have NO real solutions (roots) but only non-real ones.

So we proceed to examine the vertex's location, which is also a great way to decide on which set of points to use in order to plot its graph efficiently.

We recall that the x-position of the vertex for a quadratic function of the form $f(x)=ax^2+bx+c$ is given by the expression:

$x_v=\frac{-b}{2a}$

Since in our case $a=2$ and $b=-3$ , we get that the x-position of the vertex is:

$x_v=\frac{-b}{2a}\\x_v=\frac{-(-3)}{2(2)}\\x_v=\frac{3}{4}$

Now we can find the y-value of the vertex by evaluating this quadratic expression for x = 3/4:

$y_v=f(\frac{3}{4})=2( \frac{3}{4})^2-3(\frac{3}{4})+4\\f(\frac{3}{4})=2( \frac{9}{16})-\frac{9}{4}+4\\f(\frac{3}{4})=\frac{9}{8}-\frac{9}{4}+4\\f(\frac{3}{4})=\frac{9}{8}-\frac{18}{8}+\frac{32}{8}\\f(\frac{3}{4})=\frac{23}{8}$

This is a positive value for y, therefore we are in the situation where there is NO x-axis crossing of the parabola's graph, and therefore no real roots.

We can though estimate a few more points of the parabola's graph in order to complete the graph as requested in the problem. For such we select a couple of x-values to the right of the vertex, and a couple to the right so we can draw the branches. For example: x = 1, and x = 2 to the right; and x = 0 and x = -1 to the left of the vertex:

$f(-1) = 2(-1)^2-3(-1)+4= 2+3+4=9\\f(0)=2(0)^2-3(0)+4=0+0+4=4\\f(1)=2(1)^2-3(-1)+4=2-3+4=3\\f(2)=2(2)^2-3(2)+4=8-6+4=6$

See the graph produced in the attached image.

4 0

3 years ago