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2 years ago

Which is a part of a line with one endpoint?

Mathematics

2 answers:

valkas [14]2 years ago

4 0

your answer is a ray line.

-Dominant- [34]2 years ago

4 0

The ray is a part of a line with one end point and the other end continues.

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How would you lable a math bar model

Anastasy [175]

Well,. you could say bees in hive + bees in garden = bees in total.

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2 years ago

Factor completely: 3x2 + 2x - 1

Alex777 [14]

$3x^{2}+2x-1=3x^{2}+3x-x-1= \\ \\ 3x(x+1)-(x+1)=(3x-1)(x+1)$

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4 0

2 years ago

Read 2 more answers

For brainiest:):):):)::):):):)

scZoUnD [109]

Answer:

The unit rate is accurately $\frac{1}{4}$ (km/minute)

Step-by-step explanation:

(1,1/4) means that the vehicle traveled 1/4 km within 1 minute.

The unit rate for this problem is also known as the speed of the vehicle.

So to find the speed, you need to use distance over time.

---> Unit rate: $\frac{1}{4}\\$ / 1 = $\frac{1}{4}$ (km/minute)

(2,1/2) means that the vehicle traveled 1/2 km within 2 minutes.

Same process:

---> Unit rate: $\frac{1}{2}$ / 2 =

Because of this, the unit rate is accurately $\frac{1}{4}$ (km/minute)

8 0

2 years ago

Indigo Depot is having a clearance sale of their summer items. A swimsuit that originally cost $90 before any discount is on sal

nekit [7.7K]

Answer:

$43.20

Step-by-step explanation:

A swimsuit that originally cost $90 before any discount is on sale for 20% off.

X = price of swimsuit

X = 90

20% off = 90 * 0.8 = 72

The swimsuit is then put on clearance for an additional 40 % off the sale price.

72 * 0.6 = 43.2

what is the final price?

$43.20

6 0

2 years ago

Read 2 more answers

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