What is 6/7 Answer quickly please!!

Need answer asap evaluate the limit or state that the limit does not exist 7n-8n/2n

bagirrra123 [75]

$\lim_{n \to \ 0 } \frac{7n-8n}{2n} = \lim_{n \to \ 0 } \frac{-n}{2n} 
= \lim_{n \to \ 0 } \frac{-1}{2} = \frac{-1}{2}$

4 0

3 years ago

If a fire is burning a building and it takes 2 hours to put it out using 2000 gal./min., what minimum diameter cylindrical tank

strojnjashka [21]

Answer:

The minimum diameter of the cylindrical tank needed to store the quantity needed to put out the fire is approximately 58.415 feet.

Step-by-step explanation:

A gallon equals 0.134 cubic feet. First, we determine the amount of water ( $Q$ ), measured in cubic feet, needed to put out the fire under the assumption that water is consumed at constant rate:

$Q = \dot Q \cdot \Delta t$ (1)

Where:

$\dot Q$ - Volume rate, measured in feet per minute.

$\Delta t$ - Time, measured in minutes.

If we know that $\dot Q = 2000\,\frac{gal}{min}$ and $\Delta t = 120\,min$ , then the amount of water is:

$Q = \left(2000\,\frac{gal}{min} \right)\cdot (120\,min) \cdot \left(0.134\,\frac{ft^{3}}{gal} \right)$

$Q = 32160\,ft^{3}$

And the diameter of the cylindrical tank based on the capacity found above is determined by volume formula for a cylinder:

$Q = \frac{\pi}{4}\cdot D^{2}\cdot h$ (2)

Where:

$D$ - Diameter, measured in feet.

$h$ - Height, measured in feet.

If we know that $Q = 32160\,ft^{3}$ and $h = 12\,ft$ , then the minimum diameter is:

$D^{2} = \frac{4\cdot Q}{\pi\cdot h}$

$D = 2\cdot \sqrt{\frac{Q}{\pi\cdot h} }$

$D = 2\cdot \sqrt{\frac{32160\,ft^{3}}{\pi\cdot (12\,ft)} }$

$D \approx 58.415\,ft$

The minimum diameter of the cylindrical tank needed to store the quantity needed to put out the fire is approximately 58.415 feet.

8 0

2 years ago

Find the slope given (2,7) and (4,4)

8090 [49]

<h3>Answer:</h3>

m= -3/2

m=y1-y2/x1-x2 = 4-7/4-2 = -3/2

8 0

3 years ago

What is the surface area of the triangular prism?

Elanso [62]

Answer:

135 square feet

Step-by-step explanation:

area of base and top=2(1/2×6×2.5)=15 square feet.

area of three faces=(2.5+6+6.5)×8=15×8=120 ft

total surface area=15+120=135 square feet.

4 0

2 years ago

Read 2 more answers

A certain square is to be drawn on a coordinate plane. One of the vertices must be on the origin, and the square is to have an a

Scrat [10]

Answer:

The answer is (C) 8

Step-by-step explanation:

First, let's calculate the length of the side of the square.

$A_{square}=a^2$ , where $a$ is the length of the side. Now, let's try to build the square. First we need to find a point which distance from (0, 0) is 10. For this, we can use the distance formula in the plane:

$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ which for $x_1=0$ and $y_1 = 0$ transforms as $d=\sqrt{(x_2)^2 + (y_2)^2}$ . The first point we are looking for is connected to the origin and therefore, its components will form a right triangle in which, the Pythagoras theorem holds, see the first attached figure. Then, $x_2$ , $y_2$ and 10 are a Pythagorean triple. From this, $x_2= 6$ or $x_2=8$ while $y_2= 6$ or $y_2=8$ . This leads us with the set of coordinates:

$(\pm 6, \pm 8)$ and $(\pm 8, \pm 6)$ . (A)

The next step is to find the coordinates of points that lie on lines which are perpendicular to the lines that joins the origin of the coordinate system with the set of points given in (A):

Let's do this for the point (6, 8).

The equation of the line that join the point (6, 8) with the origin (0, 0) has the equation $y = mx +n$ , however, we only need to find its slope in order to find a perpendicular line to it. Thus,

$m = \frac{y_2-y_1}{x_2-x_1} \\m = \frac{8-0}{6-0} \\m = 8/6$

Then, a perpendicular line has an slope $m_{\bot} = -\frac{1}{m} = -\frac{6}{8}$ (perpendicularity condition of two lines). With the equation of the slope of the perpendicular line and the given point (6, 8), together with the equation of the distance we can form a system of equations to find the coordinates of two points that lie on this perpendicular line.

$m_{\bot}=\frac{6}{8} = \frac{8-y}{6-x}\\ 6(6-x)+8(8-y)=0$ (1)

$d^2 = \sqrt{(y_o-y)^2+(x_o-x)^2} \\(10)^2=\sqrt{(8-y)^2+(6-x)^2}\\100 = \sqrt{(8-y)^2+(6-x)^2}$ (2)

This system has solutions in the coordinates (-2, 14) and (14, 2). Until here, we have three vertices of the square. Let's now find the fourth one in the same way we found the third one using the point (14,2). A line perpendicular to the line that joins the point (6, 8) and (14, 2) has an slope $m = 8/6$ based on the perpendicularity condition. Thus, we can form the system:

$\frac{8}{6} =\frac{2-y}{14-x} \\8(14-x) - 6(2-y) = 0$ (1)

$100 = \sqrt{(14-x)^2+(2-y)^2}$ (2)

with solution the coordinates (8, -6) and (20, 10). If you draw a line joining the coordinates (0, 0), (6, 8), (14, 2) and (8, -6) you will get one of the squares that fulfill the conditions of the problem. By repeating this process with the coordinates in (A), the following squares are found:

(0, 0), (6, 8), (14, 2), (8, -6)

(0, 0), (8, 6), (14, -2), (6, -8)
(0, 0), (-6, 8), (-14, 2), (-8, -6)
(0, 0), (-8, 6), (-14, -2), (-6, -8)

Now, notice that the equation of distance between the two points separated a distance of 10 has the trivial solution $(\pm10, 0)$ and $(0, \pm10)$ . By combining this points we get the following squares:

(0, 0), (10, 0), (10, 10), (0, 10)
(0, 0), (0, 10), (-10, 10), (-10, 0)
(0, 0), (-10, 0), (-10, -10), (0, -10)
(0, 0), (0, -10), (-10, -10), (10, 0)

See the attached second attached figure. Therefore, 8 squares can be drawn

8 0

3 years ago