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Simora [160]
3 years ago
11

Solve the question. ​

Mathematics
2 answers:
Rama09 [41]3 years ago
5 0

Answer: 1

(\frac{x^{a+b} }{x^{c} }) ^{a-b}.(\frac{x^{b+c} }{x^{a} }) ^{b-c} .(\frac{x^{c+a} }{x^{b} })^{c-a}\\\\= \frac{x^{(a+b)(a-b)} }{x^{c(a-b)} }.\frac{x^{(b+c)(b-c)} }{x^{a(b-c)} }.\frac{x^{(c+a)(c-a)} }{x^{b(c-a)} }\\\\=\frac{x^{a^{2}-b^{2}  } }{x^{ac-bc} }.\frac{x^{b^{2}-c^{2}  } }{x^{ab-ac} }.\frac{x^{c^{2}-a^{2}  } }{x^{bc-ab} }      \\\\=\frac{x^{a^{2}-b^{2}+b^{2}-c^{2}+c^{2}-a^{2}   } }{x^{ac-bc+ab-ac+bc-ab} }\\\\=\frac{x^{0} }{x^{0} }=1

Step-by-step explanation:

Serggg [28]3 years ago
5 0
Wow the person who answered above me is smart
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The quadratic mean of two real numbers x and y equals p (x 2 y 2)/2. By computing the arithmetic and quadratic means of differen
BARSIC [14]

Answer:

The quadratic mean of 2 real positive numbers is greater than or equal to the arithmetic mean.

Step-by-step explanation:

x and y      Quadratic Mean       Arithmetic mean

3 and 3                    3                                    3

2 and 3                   2.55                               2.5

3 and  6                  4.74                                4.5

2 and 5                   3.8                                  3.5

2 and 17                 12.1                                   9.5

18 and 28              23.5                                  23

10 and  48             34.7                                  29

The quadratic mean is always greater than the arithmetic mean except when  x and y are the same.

When the difference between the pairs is  small the difference in the means is also small. As that difference increases the difference in the means also increases.

So we conjecture that the quadratic mean is always greater than or equal to the arithmetic mean.

Proof.

Suppose it is true then:

√(x^2 + y^2) / 2) ≥ (x + y)/2       Squaring  both sides:

(x ^2 + y^2) / 2 ≥ (x + y)^2 / 4    Multiply through by 4:

2x^2 +2y^2  ≥  (x + y)^2

2x^2 +2y^2 >=  x^2 + 2xy + y^2

x^2 + y^2 >= 2xy.

x^2 - 2xy + y^2 ≥ 0

(x - y)^2 ≥ 0

This is true  because the square of any real number is positive so the original inequality must also be true.

6 0
3 years ago
Which transformations would result in the image shown?
Paul [167]

Answer:

a.ABCD is reflected over both axes.

Step-by-step explanation:

Reflection over x-axis transforms point (x, y) into (x, -y)

Reflection over y-axis transforms point (x, y) into (-x, y)

Then, reflection over both axes transforms point (x, y) into (-x, -y). Applying this rule to figure ABCD:

A (-2, 2) -> (2, -2) which corresponds to point A'

B (-1, 2) -> (1, -2) which corresponds to point B'

C (-1, 1) -> (1, -1) which corresponds to point C'

D (-3, 1) -> (3, -1) which corresponds to point D'

6 0
3 years ago
Can someone answer this question please answer it correctly pleas help me if it’s correct I will mark you brainliest
elena-14-01-66 [18.8K]

Answer:

5/39

Step-by-step explanation:

<u>Step 1:  Find the probability</u>

Cross country / total

10 / 78

(10/2) / (78/2)

5 / 39

Answer:  5/39

5 0
3 years ago
HELP MEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE please
VMariaS [17]

Answer:

50.5

Step-by-step explanation:

Sooo first we calculate edf. 11/2= 5.5, 5.5x5=27.5

And then we do the other part.9/5=4, and 4x11.5=46, and we need to cut that in half cuz it’s a triangle 46/2=23. We add up the rectangle and the triangle: 27.5+23=50.5

3 0
3 years ago
Write an equation of a line parallel to line GH below in slope-intercept form that passes through the point (−5, 6).
icang [17]
The slope of GH is 1, so any line parallel to GH will have the same slope.

Since this other line passes through (-5,6), you can use the point-slope formula:

y-6=1(x-(-5))\implies y-6=x+5\implies y=x+11

so you are correct.
6 0
3 years ago
Read 2 more answers
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