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mixer [17]
3 years ago
13

Which equation is an identity?

Mathematics
1 answer:
Sav [38]3 years ago
4 0

Answer:

Equation 3

Step-by-step explanation:

An identity is, simply put, an equation that is always true. 1 = 1, 2 = 2, and x = x are all examples of identities, as there's no case in which 1 ≠ 1, 2 ≠ 2, and x ≠ x. Essentially, if we can manipulate and equation so that we end up with the same value on either side, we've found an identity. Let's run through and try to solve each of these equations to see which one fulfills that condition:

8 - (6v + 7) = -6v - 1

8 - 6v - 7 = -6v - 1

1 - 6v = -6v - 1

1 = -1

This is clearly untrue. Moving on to the next equation:

5y + 5 = 5y - 6

5 = -6

Untrue again. Solving the third:

3w + 8 - w = 4w - 2(w - 4)

2w + 8 = 4w - 2w + 8

2w + 8 = 2w + 8

If we created a new variable z = 2w + 8, we could rewrite this equation as

z = z, <em>which is always true</em>. We can stop here, as we've now found that equation 3 is an identity.

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myrzilka [38]
First step: you have to plug in 6y+3 into the equation and solve
Example: 6y+3+2y = 5
3 0
3 years ago
A tractor was purchased for $30,000 six years ago. what is the value of the tractor today if the depreciation rate was 2.5%?
MrRissso [65]
Initial cost = $30,000
Depreciation rate = 2.5%

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In six years,
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6 0
3 years ago
The diameter of a spherical balloon that is being filled with air is increasing at the rate of 3 inches more than the time, t. W
Harrizon [31]

Answer:

V=36 \pi cubic inches

Step-by-step explanation:

Diameter is increasing at 3 inches more than time, When t = 3

Diameter (d) = 3+3 = 6

d = 6

Radius is HALF of diameter, so

r = 6/2 = 3

Now, the volume of a sphere is given by the formula:

V=\frac{4}{3}\pi r^3

Putting r = 3, we get:

V=\frac{4}{3}\pi r^3\\V=\frac{4}{3}\pi (3)^3\\V=36 \pi

The volume would be 36π

5 0
3 years ago
107,609 Divided By 72
Veseljchak [2.6K]
The answer is 1494.569444
8 0
3 years ago
Read 2 more answers
If x&gt;2,then x2-x-6/x2-4=
statuscvo [17]
\frac{x^2-x-6}{x^2-4}=(*)\\------------\\x^2-x-6=0\\a=1;\ b=-1;\ c=-6\\\Delta=b^2-4ac;\ \Delta=(-1)^2-4\cdot1\cdot(-6)=1+24=25\\\\x_1=\frac{-b-\sqrt\Delta}{2a};\ x_2=\frac{-b+\sqrt\Delta}{2a}\\\\x_1=\frac{1-\sqrt{25}}{2\cdot1}=\frac{1-5}{2}=\frac{-4}{2}=-2\\\\x_2=\frac{1+\sqrt{25}}{2\cdot1}=\frac{1+5}{2}=\frac{6}{2}=3\\\\x^2-x-6=(x+2)(x-3)
---------------------\\x^2-4=x^2-2^2=(x-2)(x+2)\\-------------------\\\\(*)=\frac{(x+2)(x-3)}{(x-2)(x+2)}=\frac{x-3}{x-2}


6 0
3 years ago
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