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3 years ago

What is the reciprocal of 159?

Mathematics

1 answer:

wariber [46]3 years ago

5 0

<h2>Hey there! </h2>

<h2>The Reciprocal of 159 is:</h2>

$\frac{1}{159}$

${where \: 1 \: is \: hidden}$

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Help plz:)))I’ll mark u Brainliest

sergey [27]

Answer:

10 ft away

Step-by-step explanation:

do pythagorean theorum

24^2 + b^2 = 26^2

576 + b^2 = 676

*subtract 576 from both sides*

b^2 = 100

*find square root of 100*

b = 10

4 0

2 years ago

An auto mechanic converts a metric part to 0.492 in. The part comes only in fractional sizes given to the nearest 64th of an inc

tatuchka [14]

Answer:

the closest part the mechanic can chose is 0.5 inch

Step-by-step explanation:

fractional sizes given to the nearest 64th of an inch = 1/64 = 0.0156

the closest size to 0.492in = 0.492 + 0.0156 = 0.5076 inches

approximately, the closest part the mechanic can chose is 0.5 inch

4 0

3 years ago

Ethan purchased a container of sports drink that contains 300 calories. The label says that 80% of the calories are from carbohy

MA_775_DIABLO [31]

240 each 10 percent is 30 calories multiply them by ten gets you 300

7 0

3 years ago

A 75-gallon tank is filled with brine (water nearly saturated with salt; used as a preservative) holding 11 pounds of salt in so

Debora [2.8K]

Let $A(t)$ = amount of salt (in pounds) in the tank at time $t$ (in minutes). Then $A(0) = 11$ .

Salt flows in at a rate

$\left(0.6\dfrac{\rm lb}{\rm gal}\right) \left(3\dfrac{\rm gal}{\rm min}\right) = \dfrac95 \dfrac{\rm lb}{\rm min}$

and flows out at a rate

$\left(\dfrac{A(t)\,\rm lb}{75\,\rm gal + \left(3\frac{\rm gal}{\rm min} - 3.25\frac{\rm gal}{\rm min}\right)t}\right) \left(3.25\dfrac{\rm gal}{\rm min}\right) = \dfrac{13A(t)}{300-t} \dfrac{\rm lb}{\rm min}$

where 4 quarts = 1 gallon so 13 quarts = 3.25 gallon.

Then the net rate of salt flow is given by the differential equation

$\dfrac{dA}{dt} = \dfrac95 - \dfrac{13A}{300-t}$

which I'll solve with the integrating factor method.

$\dfrac{dA}{dt} + \dfrac{13}{300-t} A = \dfrac95$

$-\dfrac1{(300-t)^{13}} \dfrac{dA}{dt} - \dfrac{13}{(300-t)^{14}} A = -\dfrac9{5(300-t)^{13}}$

$\dfrac d{dt} \left(-\dfrac1{(300-t)^{13}} A\right) = -\dfrac9{5(300-t)^{13}}$

Integrate both sides. By the fundamental theorem of calculus,

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac1{(300-t)^{13}} A\bigg|_{t=0} - \frac95 \int_0^t \frac{du}{(300-u)^{13}}$

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac{11}{300^{13}} - \frac95 \times \dfrac1{12} \left(\frac1{(300-t)^{12}} - \frac1{300^{12}}\right)$

$\displaystyle -\dfrac1{(300-t)^{13}} A = \dfrac{34}{300^{13}} - \frac3{20}\frac1{(300-t)^{12}}$

$\displaystyle A = \frac3{20} (300-t) - \dfrac{34}{300^{13}}(300-t)^{13}$

$\displaystyle A = 45 \left(1 - \frac t{300}\right) - 34 \left(1 - \frac t{300}\right)^{13}$

After 1 hour = 60 minutes, the tank will contain

$A(60) = 45 \left(1 - \dfrac {60}{300}\right) - 34 \left(1 - \dfrac {60}{300}\right)^{13} = 45\left(\dfrac45\right) - 34 \left(\dfrac45\right)^{13} \approx 34.131$

pounds of salt.

7 0

1 year ago

What is the distance between points M and N?

Free_Kalibri [48]

By using <em>triangle</em> properties and the law of the cosine twice, we find that the distance between points M and N is approximately 9.8 meters.

<h3>How to determine the distance between two points</h3>

In this problem we must determine the distance between two points that are part of a triangle and we can take advantage of properties of triangles to find it. First, we determine the measure of angle L by the law of the cosine:

$\cos L = \frac{(19.6\,m)^{2}-(14.8\,m)^{2}-(21.4\,m)^{2}}{-2\cdot (14.8\,m)\cdot (21.4\,m)}$

L ≈ 62.464°

Then, we get the distance between points M and N by the law of the cosine once again:

$MN = \sqrt{(7.4\,m)^{2}+(10.7\,m)^{2}-2\cdot (7.4\,m)\cdot (10.7\,m)\cdot \cos 62.464^{\circ}}$

MN ≈ 9.8 m

By using <em>triangle</em> properties and the law of the cosine twice, we find that the distance between points M and N is approximately 9.8 meters.

To learn more on triangles: brainly.com/question/2773823

#SPJ1

6 0

1 year ago