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Dafna1 [17]
3 years ago
8

Describe what the inequality x ≥ 3 looks like when graphed

Mathematics
1 answer:
Pachacha [2.7K]3 years ago
4 0

<, > - open circle

≤, ≥ - closed circle

<, ≤ - draw to the left

>, ≥ - draw to the right

--------------------------------------------

We have x ≥ 3

closed circle on number 3 and draw the line to the right

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The size of a small ant is about 0.002 meters, which can be written as 2 x 10"
Nesterboy [21]

Answer:

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Step-by-step explanation:

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6 0
2 years ago
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Jerald jumped from a bungee tower. If the equation that models his height, in feet, is h = –16t2 + 729, where t is the time in s
Snezhnost [94]

Answer. First option: t > 6.25


Solution:

Height (in feet): h=-16t^2+729

For which interval of time is h less than 104 feet above the ground?

h < 104

Replacing h for -16t^2+729

-16t^2+729 < 104

Solving for h: Subtracting 729 both sides of the inequality:

-16t^2+729-729 < 104-729

-16t^2 < -625

Multiplying the inequality by -1:

(-1)(-16t^2 < -625)

16t^2 > 625

Dividing both sides of the inequality by 16:

16t^2/16 > 625/16

t^2 > 39.0625

Replacing t^2 by [ Absolute value (t) ]^2:

[ Absolute value (t) ]^2 > 39.0625

Square root both sides of the inequality:

sqrt { [ Absolute value (t) ]^2 } > sqrt (39.0625)

Absolute value (t) > 6.25

t < -6.25 or t > 6.25, but t can not be negative, then the solution is:

t > 6.25



5 0
3 years ago
Read 2 more answers
A street light is mounted at the top of a 15-ft-tall pole. A man 6 ft tall walks away from the pole with a speed of 4 ft/s along
mina [271]

Answer:

The tip of the man shadow moves at the rate of \frac{20}{3} ft.sec

Step-by-step explanation:

Let's draw a figure that describes the given situation.

Let "x" be the distance between the man and the pole and "y" be distance between the pole and man's shadows tip point.

Here it forms two similar triangles.

Let's find the distance "y" using proportion.

From the figure, we can form a proportion.

\frac{y - x}{y} = \frac{6}{15}

Cross multiplying, we get

15(y -x) = 6y

15y - 15x = 6y

15y - 6y = 15x

9y = 15x

y = \frac{15x}{9\\} y = \frac{5x}{3}

We need to find rate of change of the shadow. So we need to differentiate y with respect to the time (t).

\frac{dy}{t} = \frac{5}{3} \frac{dx}{dt} ----(1)

We are given \frac{dx}{dt} = 4 ft/sec. Plug in the equation (1), we get

\frac{dy}{dt} = \frac{5}{3} *4 ft/sect\\= \frac{20}{3} ft/sec

Here the distance between the man and the pole 45 ft does not need because we asked to find the how fast the shadow of the man moves.

7 0
3 years ago
Which expression is equivalent to 8x-2x+x+x<br><br> a. 4x<br> b. 8x<br> c. 6x-2x<br> d. 10x-2y
Deffense [45]
B - 8x

To find this, combine like terms.

8x - 2x is 6x, then add the two x's on the side to bring it back to 8x.

Hope this helps!
4 0
3 years ago
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hello this is 6th grade hw and can someone help me the area and perimeter for all of them and plz explain how you got them
Scilla [17]
Text me ----- [email protected]
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3 years ago
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