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2 years ago

How many yards long is a piece of fabric witha width of 5/8 yrd nd a area of 2 3/4 square yards?

Mathematics

1 answer:

Dmitry_Shevchenko [17]2 years ago

5 0

Answer:

D. $4\frac{2}{5}$

Step-by-step explanation:

Area of fabrics = 2¾ square yards

Width of fabrics = ⅝ yard

Length of fabrics = x

Area = length * width

Therefore:

2¾ = ⅝*x

Change the mixed fraction to improper fraction

$\frac{11}{4} = \frac{5x}{8}$

Cross multiply

$4*5x = 11*8$

$20x = 88$

Divide both sides by 20

$\frac{20x}{20} = \frac{88}{20}$

$x = \frac{22}{5}$

$x = 4\frac{2}{5}$

$length = 4\frac{2}{5}$

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Suppose that the Celsius temperature at the point (x, y) in the xy-plane is T(x, y) = x sin 2y and that distance in the xy-plane

liraira [26]

Missing information:

How fast is the temperature experienced by the particle changing in degrees Celsius per meter at the point

$P = (\frac{1}{2}, \frac{\sqrt 3}{2})$

Answer:

$Rate = 0.935042^\circ /cm$

Step-by-step explanation:

Given

$P = (\frac{1}{2}, \frac{\sqrt 3}{2})$

$T(x,y) =x\sin2y$

$r = 1m$

$v = 2m/s$

Express the given point P as a unit tangent vector:

$P = (\frac{1}{2}, \frac{\sqrt 3}{2})$

$u = \frac{\sqrt 3}{2}i - \frac{1}{2}j$

Next, find the gradient of P and T using: $\triangle T = \nabla T * u$

Where

$\nabla T|_{(\frac{1}{2}, \frac{\sqrt 3}{2})} = (sin \sqrt 3)i + (cos \sqrt 3)j$

So: the gradient becomes:

$\triangle T = \nabla T * u$

$\triangle T = [(sin \sqrt 3)i + (cos \sqrt 3)j] * [\frac{\sqrt 3}{2}i - \frac{1}{2}j]$

By vector multiplication, we have:

$\triangle T = (sin \sqrt 3)* \frac{\sqrt 3}{2} - (cos \sqrt 3) \frac{1}{2}$

$\triangle T = 0.9870 * 0.8660 - (-0.1606 * 0.5)$

$\triangle T = 0.9870 * 0.8660 +0.1606 * 0.5$

$\triangle T = 0.935042$

Hence, the rate is:

$Rate = \triangle T = 0.935042^\circ /cm$

3 0

2 years ago

The average daily temperature, t, in degrees fahrenheit for a city as a function of the month of the year, m, can be modeled by

Vikentia [17]

The answer will be t = $sin((m.\frac{\pi}{6})+55)$ average daily temperature, t, in degrees Fahrenheit for a city as a function of the month of the year.

<h3>What is temperature?</h3>

Temperature is the degree or intensity of heat present in a substance or object, especially as expressed according to a comparative scale and shown by a thermometer or perceived by touch.

<h3>TO SOLVE:</h3>

$35cos(\frac{\pi}{6(m+3)}+55)\\\\35 cos(\frac{\pi m}{6} + 55 + \frac{\pi}{2})\\$

suppose $x = m.\frac{\pi}{6}+55$ and $\frac{\pi}{2} = 90$ degree

We know, cos(x+90°) = - sin(X)

⇒ $cos((m.\frac{\pi}{6})+55+90degree)$ degree = - $sin((m.\frac{\pi}{6})+55)$

⇒ t = $sin((m.\frac{\pi}{6})+55)$

Hence the answer will be t = $sin((m.\frac{\pi}{6})+55)$ average daily temperature, t, in degrees Fahrenheit for a city as a function of the month of the year.

To learn more visit:

brainly.com/question/27441712

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