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morpeh [17]
3 years ago
12

A map scale reads: 2 inches = 25 miles. Which distance does 6 inches representa

Mathematics
2 answers:
xz_007 [3.2K]3 years ago
7 0
150 miles for six miles
Sergio [31]3 years ago
3 0
The answer is supposed to be 75 miles...
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Use the sequence of operations to answer the question.
velikii [3]

Answer:

(8-(3+b))3

Step-by-step explanation:

first of all, we have to variables, 3 and b that we are told to add

that is, 3+b

we are now told to subtract the answer from 8

that is 8 minis the answer

= 8- (3+b)

lastly we triple the answer to give

(8-(3+b)3

or

3(8-(3+b)

6 0
2 years ago
Find the Area And Perimeter
ahrayia [7]

Answer:

Ste-by-step explanation:bf=7x-10 bc=4x-29

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3 years ago
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What is the simplified value of the exponential expression 27 1/3?<br> 1/3<br> 1/9<br> 3<br> 9
madreJ [45]

Answer:

Step-by-step explanation:

27^1/3

27 is cube root of 3

so it can also be written as (3)^3

∴ {(3)^3}^1/3

3 and 3 will get cancelled

so it will be 3^1

= 3

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3 years ago
What is the area of the parallelogram?
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Step-by-step explanation:

formula for area of parallelogram

= A=bh

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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
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