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Gekata [30.6K]
3 years ago
15

Y - 1 > -7 A.y ≥ -6 B.y > -6 C.y -8

Mathematics
1 answer:
Radda [10]3 years ago
5 0
Answer: B

Step by Step:
y-1>-7
y>-7+1
y>-6
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A Ferris wheel rotates 9π/8 radians prior to making a stop. The total height of the Ferris wheel is 246 ft. How far around did t
WARRIOR [948]

Answer:

The ferris wheel travelled 434.717 ft.

Step-by-step explanation:

Assuming that the ferris wheel is a perfect circle, then it's height is the same as the diameter of the circle, therefore it's radius is:

radius = height/2 = 246/2 = 123 ft

The distance travelled by the ferris wheel is the same as the length of the arc created by a angle of (9pi/8). The length of an arc is given by:

length = r*(angle in radians)

length = 123*(9pi/8)

length = 434.717 ft

The ferris wheel travelled 434.717 ft.

5 0
3 years ago
What is -2 = 5 / (2x+3)
LUCKY_DIMON [66]

Answer:

x = -11/4

Step-by-step explanation:

Given that:

-2 = 5 / (2x+3)

Multiplying both sides by (2x + 3)

-2(2x + 3) = 5

"-" sign will alter the inner signs

-4x -6 = 5

Adding 6 on both sides:

-4x -6 +6 = 5 + 6

-4x  = 11

Dividing both sides by -4

-4x/-4 = 11/-4

x = -11/4

i hope it will help you!

3 0
3 years ago
....................................Graph y=−x−2.
pychu [463]
See attached photo for the graph

4 0
2 years ago
A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 30 ft/s. Its height
Crank

Answer:

a) h = 0.1: \bar v = -11\,\frac{ft}{s}, h = 0.01: \bar v = -10.1\,\frac{ft}{s}, h = 0.001: \bar v = -10\,\frac{ft}{s}, b) The instantaneous velocity of the ball when t = 2\,s is -10 feet per second.

Step-by-step explanation:

a) We know that y = 30\cdot t -10\cdot t^{2} describes the position of the ball, measured in feet, in time, measured in seconds, and the average velocity (\bar v), measured in feet per second, can be done by means of the following definition:

\bar v = \frac{y(2+h)-y(2)}{h}

Where:

y(2) - Position of the ball evaluated at t = 2\,s, measured in feet.

y(2+h) - Position of the ball evaluated at t =(2+h)\,s, measured in feet.

h - Change interval, measured in seconds.

Now, we obtained different average velocities by means of different change intervals:

h = 0.1\,s

y(2) = 30\cdot (2) - 10\cdot (2)^{2}

y (2) = 20\,ft

y(2.1) = 30\cdot (2.1)-10\cdot (2.1)^{2}

y(2.1) = 18.9\,ft

\bar v = \frac{18.9\,ft-20\,ft}{0.1\,s}

\bar v = -11\,\frac{ft}{s}

h = 0.01\,s

y(2) = 30\cdot (2) - 10\cdot (2)^{2}

y (2) = 20\,ft

y(2.01) = 30\cdot (2.01)-10\cdot (2.01)^{2}

y(2.01) = 19.899\,ft

\bar v = \frac{19.899\,ft-20\,ft}{0.01\,s}

\bar v = -10.1\,\frac{ft}{s}

h = 0.001\,s

y(2) = 30\cdot (2) - 10\cdot (2)^{2}

y (2) = 20\,ft

y(2.001) = 30\cdot (2.001)-10\cdot (2.001)^{2}

y(2.001) = 19.99\,ft

\bar v = \frac{19.99\,ft-20\,ft}{0.001\,s}

\bar v = -10\,\frac{ft}{s}

b) The instantaneous velocity when t = 2\,s can be obtained by using the following limit:

v(t) = \lim_{h \to 0} \frac{x(t+h)-x(t)}{h}

v(t) =  \lim_{h \to 0} \frac{30\cdot (t+h)-10\cdot (t+h)^{2}-30\cdot t +10\cdot t^{2}}{h}

v(t) =  \lim_{h \to 0} \frac{30\cdot t +30\cdot h -10\cdot (t^{2}+2\cdot t\cdot h +h^{2})-30\cdot t +10\cdot t^{2}}{h}

v(t) =  \lim_{h \to 0} \frac{30\cdot t +30\cdot h-10\cdot t^{2}-20\cdot t \cdot h-10\cdot h^{2}-30\cdot t +10\cdot t^{2}}{h}

v(t) =  \lim_{h \to 0} \frac{30\cdot h-20\cdot t\cdot h-10\cdot h^{2}}{h}

v(t) =  \lim_{h \to 0} 30-20\cdot t-10\cdot h

v(t) = 30\cdot  \lim_{h \to 0} 1 - 20\cdot t \cdot  \lim_{h \to 0} 1 - 10\cdot  \lim_{h \to 0} h

v(t) = 30-20\cdot t

And we finally evaluate the instantaneous velocity at t = 2\,s:

v(2) = 30-20\cdot (2)

v(2) = -10\,\frac{ft}{s}

The instantaneous velocity of the ball when t = 2\,s is -10 feet per second.

8 0
3 years ago
Evaluate the numerical expression 2+65 + 7*3=
riadik2000 [5.3K]
2 + 65 + 7 × 3
= 2 + 65 + 21
= 67 + 21
= 88

answer is 88
7 0
3 years ago
Read 2 more answers
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