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3 years ago

11

Which algebraic expression represents "the difference of 54 and a number"?

2 answers:

atroni [7]3 years ago

5 0

$\huge{\boxed{54-x}}$

The difference is the result of a subtraction problem.

We are given two values that are being subtracted: 54 and a number, represented by $x$

So, represent this mathematically with $\boxed{54-x}$ .

Musya8 [376]3 years ago

4 0

54-x is the correct answer, the difference between number is x-54 or 54-x and the number of algebraic expressions subtracted or number symbol like x.

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adelina 88 [10]

Answer:

b ≈ 9.5, c ≈ 14.7

Step-by-step explanation:

Using the Sine rule in Δ ABC, that is

$\frac{a}{sinA}$ = $\frac{b}{sinB}$ , substitute values

$\frac{7}{sin23}$ = $\frac{b}{sin32}$ ( cross- multiply )

b × sin23° = 7 × sin32° ( divide both sides by sin23° )

b = $\frac{7sin32}{sin23}$ ≈ 9.5 ( to the nearest tenth )

Also

$\frac{a}{sinA}$ = $\frac{c}{sinC}$

$\frac{7}{sin23}$ = $\frac{c}{sin125}$ ( cross- multiply )

c × sin23° = 7 × sin125° ( divide both sides by sin23° )

c = $\frac{7sin125}{sin23}$ ≈ 14.7 ( to the nearest tenth )

3 0

3 years ago

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]

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Answer:

$Rate = 0.935042^\circ /cm$

Step-by-step explanation:

Given

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

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

$r = 1m$

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

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

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:

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

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

$\triangle T = 0.935042$

Hence, the rate is:

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

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