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2 years ago

8

Helpppppp fastttt ASAP

1 answer:

Xelga [282]2 years ago

6 0

Answer:

It is an octogan and it isn't regular, you are correct.

Step-by-step explanation:

Normal octagons look like a stop sign. This one looks like a star.

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Find the slope of the line with the given equation. Show your work.<br> 6x – 2y = 18

Vera_Pavlovna [14]

Currently the equation 6x - 2y = 18 is in standard form. Convert the standard form equation into a slope-intercept form and we can find the slope easily.

Solve for y.

6x - 2y = 18
-2y = 18 - 6x <-- Subtract 6x from each side. This is to isolate the 2y term
2y / -2 = $\frac{18 - 6x}{2}$ <-- Divide each side by 2. This it to
get rid of the 2 coefficient.
y = -9 + 3x

Rearrange the right-hand side a bit.

y = -9 + 3x becomes y = 3x - 9

Now it is in slope-intercept form.
The slope is the coefficient of the x variable.

So, 3 is the slope.

7 0

3 years ago

The sum of two consecutive odd integers is greater than 36 (solve with an inequality and show work please)

mixas84 [53]

O+o+2>36
2o>34
o>17
o1>17
o2>19
o is equal to odd number

4 0

3 years ago

Olivia wants to cut 3 3/4 inches from a piece of string.She has already cut off 2 9/16 inches from the piece of string.How much

Roman55 [17]

Answer: she still has to cut off 19/16 inches.

Step-by-step explanation:

Olivia wants to cut 3 3/4 inches from a piece of string. Converting 3 3/4 inches to improper fraction, it becomes 15/4 inches.

She has already cut off 2 9/16 inches from the piece of string. Converting 2 9/16 inches to improper fraction, it becomes 41/16 inches.

Therefore, the length of string that is left for her to cut off would be

15/4 - 41/16 = (60 - 41)/16

= 19/16 inches

6 0

3 years ago

Chuck built a garden in the shape of a rectangle in his backyard. The width of the garden is 7 feet and the length of the garden

Nastasia [14]

The answer is 25 because of the Pythagorean theorem. 49+576=625 625 squared=25

5 0

3 years ago

A tank contains 1600 L of pure water. Solution that contains 0.04 kg of sugar per liter enters the tank at the rate 2 L/min, and

goldfiish [28.3K]

Let S(t) denote the amount of sugar in the tank at time t. Sugar flows in at a rate of

(0.04 kg/L) * (2 L/min) = 0.08 kg/min = 8/100 kg/min

and flows out at a rate of

(S(t)/1600 kg/L) * (2 L/min) = S(t)/800 kg/min

Then the net flow rate is governed by the differential equation

$\dfrac{\mathrm dS(t)}{\mathrm dt}=\dfrac8{100}-\dfrac{S(t)}{800}$

Solve for S(t):

$\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{S(t)}{800}=\dfrac8{100}$

$e^{t/800}\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{e^{t/800}}{800}S(t)=\dfrac8{100}e^{t/800}$

The left side is the derivative of a product:

$\dfrac{\mathrm d}{\mathrm dt}\left[e^{t/800}S(t)\right]=\dfrac8{100}e^{t/800}$

Integrate both sides:

$e^{t/800}S(t)=\displaystyle\frac8{100}\int e^{t/800}\,\mathrm dt$

$e^{t/800}S(t)=64e^{t/800}+C$

$S(t)=64+Ce^{-t/800}$

There's no sugar in the water at the start, so (a) S(0) = 0, which gives

$0=64+C\impleis C=-64$

and so (b) the amount of sugar in the tank at time t is

$S(t)=64\left(1-e^{-t/800}\right)$

As $t\to\infty$ , the exponential term vanishes and (c) the tank will eventually contain 64 kg of sugar.

7 0

3 years ago