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3 years ago

Which of these is irrational?Picture below :)Please help I’m so confused!

Mathematics

1 answer:

Kamila [148]3 years ago

6 0

I believe that the answer is D because when you go through

$\sqrt{5 } = \: 2.23606797749979$

and an irrational number is a number that does not end and can not be put in a fraction

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A right triangle is labeled with the hypotenuse 12m and an acute angle 23 degree

IgorLugansk [536]

Answer:

sin23= x/12

sin23(12)=x

0.391(12)=x

x=4.7 or 5

3 0

3 years ago

What is the m∠2 if m∠4 = 81° and m∠3 = 40°? a. 40° b. 59° c. 41° d. 60°

velikii [3]

Answer:

C. 41

Step-by-step explanation:

180 - 81 = 99

99+40 = 139

180 -139 = 41

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3 years ago

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The base of a pyramid is a square whose perimeter is 936936 m. what is the length of each side of this square?

djverab [1.8K]

936936 is the perimeter and perimeter of a square is 4 times the length of one of the sides. therefore, you divide 936936 by 4 and get your answer, which is 234234.

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4 years ago

A 100 gallon tank initially contains 100 gallons of sugar water at a concentration of 0.25 pounds of sugar per gallon suppose th

Vsevolod [243]

At the start, the tank contains

(0.25 lb/gal) * (100 gal) = 25 lb

of sugar. Let $S(t)$ be the amount of sugar in the tank at time $t$ . Then $S(0)=25$ .

Sugar is added to the tank at a rate of P lb/min, and removed at a rate of

$\left(1\frac{\rm gal}{\rm min}\right)\left(\dfrac{S(t)}{100}\dfrac{\rm lb}{\rm gal}\right)=\dfrac{S(t)}{100}\dfrac{\rm lb}{\rm min}$

and so the amount of sugar in the tank changes at a net rate according to the separable differential equation,

$\dfrac{\mathrm dS}{\mathrm dt}=P-\dfrac S{100}$

Separate variables, integrate, and solve for S.

$\dfrac{\mathrm dS}{P-\frac S{100}}=\mathrm dt$

$\displaystyle\int\dfrac{\mathrm dS}{P-\frac S{100}}=\int\mathrm dt$

$-100\ln\left|P-\dfrac S{100}\right|=t+C$

$\ln\left|P-\dfrac S{100}\right|=-100t-100C=C-100t$

$P-\dfrac S{100}=e^{C-100t}=e^Ce^{-100t}=Ce^{-100t}$

$\dfrac S{100}=P-Ce^{-100t}$

$S(t)=100P-100Ce^{-100t}=100P-Ce^{-100t}$

Use the initial value to solve for C :

$S(0)=25\implies 25=100P-C\implies C=100P-25$

$\implies S(t)=100P-(100P-25)e^{-100t}$

The solution is being drained at a constant rate of 1 gal/min; there will be 5 gal of solution remaining after time

$1000\,\mathrm{gal}+\left(-1\dfrac{\rm gal}{\rm min}\right)t=5\,\mathrm{gal}\implies t=995\,\mathrm{min}$

has passed. At this time, we want the tank to contain

(0.5 lb/gal) * (5 gal) = 2.5 lb

of sugar, so we pick P such that

$S(995)=100P-(100P-25)e^{-99,500}=2.5\implies\boxed{P\approx0.025}$

5 0

3 years ago

Choose the kind(s) of symmetry. (Select all that apply.) the letter (D)

k0ka [10]

Answer:Point and line

Step-by-step explanation:

5 0

3 years ago

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