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3 years ago

Which of the following does not have a scale factor of 1/60? (Hint: Use Unit Conversion).

1 answer:

4 0

1in : 5ft
1in : (5 x 12) in
1in : 60 in

1ft : 20yd
1ft : (20 x 3) ft
1ft : 60 ft

1 cm : 60cm

1in : 60ft
1in : (60 x 12)in
1in : 720 in
Hence, 1in : 60ft does not have a scale factor of 1/60

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Ella has a points card for a movie theater.

MAXImum [283]

Answer:

175<=12.5x+40

Step-by-step explanation:

She starts with 40 points already so you add in those 40 then since she gets 12.5 points per visit that is your x, so you set that as greater than or equal to 175

7 0

3 years ago

Please help!! Due really soon!

Alisiya [41]

9514 1404 393

Answer:

f(x) = 6x +1

Step-by-step explanation:

Differences in x-values (first row) are 1, 1, 1.

Differences in y-values (second row) are 6, 6, 6.

The constant ratio of differences (6/1) tells you the function is linear, and has a slope of m = 6/1 = 6.

Using the first point in the form ...

y = mx + b

we have ...

y = 6x + b

7 = 6·1 + b . . . . (x, y) = (1, 7)

1 = b . . . . . . subtract 6

Then the equation can be written ...

y = 6x +1

In functional form, this is ...

f(x) = 6x +1

4 0

3 years ago

In the diagram, AABC ~ ADEF. Find the value of x.

Andrew [12]

Answer:

Step-by-step explanation:

bdeehvdvjwjkwkwwjehheeh

If you dont know the anwser its always B

3 0

3 years ago

Find the volume of the wedge with vertices at points (0,0,0), (1,0,0), (0,1,0), (0,0,1) by integrating the area of cross-section

Angelina_Jolie [31]

Answer:

V = 1/6 cubic units

Step-by-step explanation:

Applying the concept of integrals for volume calculation:

$V = \int\limits^b_a {S(x)} \, dx$ (1)

V = volume of the solid bounded by x = a and x = b

S(x) = cross section area of the solid, perpendicular to the x axis

From the figure we have that S is the area of a triangle that has base Z and height Y

Area of the triangle = $S(x)=\frac{y(x)*z(x)}{2}$ (2)

Calculation of y(x) and z(x)

We apply the equation of the point-slope line (plane xy):

Slope = $m = \frac{y_{2} - y_{1} }{x_{2} - x_{1}}$ (3)

Equation of the line = $y - y_{1} =m(x-x_{1} )$ (4)

Replacing the points (1,0) and (0,1) in (3):

$m=\frac{1-0}{0-1} =-1$

Replacing the point (1,0) and m = -1 in (4):

$y-0=(-1)(x-1)$

y(x) = -x + 1 (Line A-B) (5)

We apply the equation of the point-slope line (plane xz):

Slope = $m = \frac{z_{2} - z_{1} }{x_{2} - x_{1}}$ (6)

Equation of the line = $z - z_{1} =m(x-x_{1} )$ (7)

Replacing the points (1,0) and (0,1) in (6):

$m=\frac{1-0}{0-1} =-1$

Replacing the point (1,0) and m = -1 in (7):

$z-0=(-1)(x-1)$

z(x) = -x + 1 (Line A-C) (8)

Replacing (5) and (8) in (2)

$S(x) = \frac{(-x + 1) * (-x + 1)}{2} =\frac{(-x + 1)^{2} }{2}$ (9)

Replacing (9) in (1) and knowing that a = 0 and b = 1:

$V = \int\limits^1_0 {\frac{(-x + 1)^{2} }{2}} \, dx = \int\limits^1_0 {\frac{x^{2}-2x+1 }{2}} \, dx$

$V =\frac{1}{2} (\frac{x^{3} }{3} -2\frac{x^{2} }{2} +x)$ evaluated from x=0 to x=1

$V= \frac{1}{2} (\frac{1}{3} -1 +1) = \frac{1}{6}$

3 0

4 years ago

an inverted conical water tank with a height of 20 ft and a radius of 8 ft is drained through a hole in the vertex (bottom) at a

viktelen [127]

Answer:

the rate of change of the water depth when the water depth is 10 ft is; $\mathbf{\dfrac{dh}{dt} = \dfrac{-25}{100 \pi} \ \ ft/s}$

Step-by-step explanation:

Given that:

the inverted conical water tank with a height of 20 ft and a radius of 8 ft is drained through a hole in the vertex (bottom) at a rate of 4 ft^3/sec.

We are meant to find the rate of change of the water depth when the water depth is 10 ft.

The diagrammatic expression below clearly interprets the question.

From the image below, assuming h = the depth of the tank at a time t and r = radius of the cone shaped at a time t

Then the similar triangles ΔOCD and ΔOAB is as follows:

$\dfrac{h}{r}= \dfrac{20}{8}$ ( similar triangle property)

$\dfrac{h}{r}= \dfrac{5}{2}$

$\dfrac{h}{r}= 2.5$

h = 2.5r

$r = \dfrac{h}{2.5}$

The volume of the water in the tank is represented by the equation:

$V = \dfrac{1}{3} \pi r^2 h$

$V = \dfrac{1}{3} \pi (\dfrac{h^2}{6.25}) h$

$V = \dfrac{1}{18.75} \pi \ h^3$

The rate of change of the water depth is :

$\dfrac{dv}{dt}= \dfrac{\pi r^2}{6.25}\ \dfrac{dh}{dt}$

Since the water is drained through a hole in the vertex (bottom) at a rate of 4 ft^3/sec

Then,

$\dfrac{dv}{dt}= - 4 \ ft^3/sec$

Therefore,

$-4 = \dfrac{\pi r^2}{6.25}\ \dfrac{dh}{dt}$

the rate of change of the water at depth h = 10 ft is:

$-4 = \dfrac{ 100 \ \pi }{6.25}\ \dfrac{dh}{dt}$

$100 \pi \dfrac{dh}{dt} = -4 \times 6.25$

$100 \pi \dfrac{dh}{dt} = -25$

$\dfrac{dh}{dt} = \dfrac{-25}{100 \pi}$

Thus, the rate of change of the water depth when the water depth is 10 ft is; $\mathtt{\dfrac{dh}{dt} = \dfrac{-25}{100 \pi} \ \ ft/s}$

4 0

4 years ago