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Tom [10]
3 years ago
10

Suppose that when you factor 4t^6-r^10, you get (2t^x+r^y)(2t^x-r^y). determine the sum of t and r.

Mathematics
1 answer:
Alexxandr [17]3 years ago
5 0

Answer:

8

Step-by-step explanation:

Given the expression

4t^6-r^10 we are to factorize in the form;

(2t^x+r^y)(2t^x-r^y)

First we need to factorize (2t^x+r^y)(2t^x-r^y)

=  (2t^x+r^y)(2t^x-r^y)\\= 4t^{2x}-2t^xr^y + 2t^xr^y - r^{2y}\\=  4t^{2x}-r^{2y}

Compare 4t^{2x}-r^{2y} \ with \ 4t^6-r^{10}

4t^{2x} = 4t^6\\t^{2x} = t^6\\equate \ the \ powers\\2x = 6\\x = 3\\

Get y;

r^{2y} = r^{10}\\2y = 10\\y = 5\\Hence \ x + y = 3+5 = 8

Hence the sum of x and y is 8

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Eduardwww [97]

Answer:

2

Step-by-step explanation:

3 0
3 years ago
Which expression represents the distance between point (0, a) and point (a, 0) on a coordinate grid?
o-na [289]

Answer:

\sqrt{2} a

Step-by-step explanation:

The distance between point (x1,y1) and (x2,y2) on a coordinate plane is given by expression \sqrt{(x1-x2)^2+(y1-y2)^2}

In the problem given coordinate points are

(0, a) and (a, 0).

Thus, distance between these points are given by

distance = \sqrt{(x1-x2)^2+(y1-y2)^2}\\=>distance = \sqrt{(0-a)^2+(a-0)^2}\\=>distance = \sqrt{(a)^2+(a)^2}\\\\=>distance = \sqrt{2a^2}\\=>distance = \sqrt{2} a

Thus , expression \sqrt{2} a gives the distance between point (0, a) and point (a, 0) on a coordinate grid.

5 0
3 years ago
Read 2 more answers
Find the domain of the function y = 3 tan(23x)
solmaris [256]

Answer:

\mathbb{R} \backslash \displaystyle \left\lbrace \left. \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right)  \; \right| k \in \mathbb{Z}  \right\rbrace.

In other words, the x in f(x) = 3\, \tan(23\, x) could be any real number as long as x \ne \displaystyle \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right) for all integer k (including negative integers.)

Step-by-step explanation:

The tangent function y = \tan(x) has a real value for real inputs x as long as the input x \ne \displaystyle k\, \pi + \frac{\pi}{2} for all integer k.

Hence, the domain of the original tangent function is \mathbb{R} \backslash \displaystyle \left\lbrace \left. \left(k\, \pi + \frac{\pi}{2}\right)  \; \right| k \in \mathbb{Z}  \right\rbrace.

On the other hand, in the function f(x) = 3\, \tan(23\, x), the input to the tangent function is replaced with (23\, x).

The transformed tangent function \tan(23\, x) would have a real value as long as its input (23\, x) ensures that 23\, x\ne \displaystyle k\, \pi + \frac{\pi}{2} for all integer k.

In other words, \tan(23\, x) would have a real value as long as x\ne \displaystyle \frac{1}{23} \, \left(k\, \pi + \frac{\pi}{2}\right).

Accordingly, the domain of f(x) = 3\, \tan(23\, x) would be \mathbb{R} \backslash \displaystyle \left\lbrace \left. \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right)  \; \right| k \in \mathbb{Z}  \right\rbrace.

4 0
3 years ago
PLEASE ANSWER ASAP!!!
julsineya [31]

Answer:

The answer to your question is A) x = 1

Step-by-step explanation:

We know that f(x) = 2x + 3 and g(x) = -x + 6

But   f(x) = g(x)

Then,                          2x + 3 = - x + 6

                                   2x + x = 6 - 3

                                         3x = 3

                                          x = 3/3

                                          x = 1

The solution is circle in the graph below.

8 0
3 years ago
Which statement best describes the solution to the equation below ?
SIZIF [17.4K]
Im going to say option 1
8 0
3 years ago
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