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3 years ago

If a = b, then a - c = b - c. This is the _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ property of equality.

Mathematics

2 answers:

inessss [21]3 years ago

4 0

Answer:

The correct answer is Subtraction property of the equality.

Step-by-step explanation:

Let's recall that it's possible to perform the same operation on each side of an equivalent linear equation without changing the equality.

That's why we can add, subtract, multiply (same non-zero number) and divide (same non-zero number) the same number to each side of a linear equation and, that operation,gives us an equivalent equation .

In this particular case, we are subtracting c at both sides of the equation and this a representation of the Subtraction property of the equality.

Whitepunk [10]3 years ago

3 0

Answer:

Associative Property

Step-by-step explanation:

Associative property involves computations that has to do with regrouping of quantities. That is simply the technique employed there.

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3 years ago

What is the value of x in the equation ex- y = 30, when y = 15?

Bumek [7]

Answer:

x = 45

Step-by-step explanation:

Solution:

x - y = 30

x - 15 = 30 ( y = 15 given)

x = 30 + 15

Therefore, x = 45

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3 years ago

A circle is growing so that the radius is increasing at the rate of 3 cm/min. How fast is the area of the circle changing at the

Naya [18.7K]

Answer:

The area is growing at a rate of $\frac{dA}{dt} =226.2 \,\frac{cm^2}{min}$

Step-by-step explanation:

Notice that this problem requires the use of implicit differentiation in related rates (some some calculus concepts to be understood), and not all middle school students cover such.

We identify that the info given on the increasing rate of the circle's radius is 3 $\frac{cm}{min}$ and we identify such as the following differential rate:

$\frac{dr}{dt} = 3\,\frac{cm}{min}$

Our unknown is the rate at which the area (A) of the circle is growing under these circumstances,that is, we need to find $\frac{dA}{dt}$ .

So we look into a formula for the area (A) of a circle in terms of its radius (r), so as to have a way of connecting both quantities (A and r):

$A=\pi\,r^2$

We now apply the derivative operator with respect to time ( $\frac{d}{dt}$ ) to this equation, and use chain rule as we find the quadratic form of the radius:

$\frac{d}{dt} [A=\pi\,r^2]\\\frac{dA}{dt} =\pi\,*2*r*\frac{dr}{dt}$

Now we replace the known values of the rate at which the radius is growing ( $\frac{dr}{dt} = 3\,\frac{cm}{min}$ ), and also the value of the radius (r = 12 cm) at which we need to find he specific rate of change for the area :

$\frac{dA}{dt} =\pi\,*2*r*\frac{dr}{dt}\\\frac{dA}{dt} =\pi\,*2*(12\,cm)*(3\,\frac{cm}{min}) \\\frac{dA}{dt} =226.19467 \,\frac{cm^2}{min}\\$