How do we use the Pythagoras theorem in "everyday life".

1 answer:

5 0

I use it all the time, but as my profession is engineering, mathematics is integral in my everyday life. I usually do it when evaluating for the magnitude of vector force components. Usually in 2D space, so Pythagorean theorem is quite useful.

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Choose the correct simplification of x to the 12th power times z to the 11th power all over x to the 2nd power times z to the 4t

Umnica [9.8K]

Answer:

<h3>The option A) $x^{10}z^7$ is correct answer.</h3><h3>The correct simplification for the given expression $\frac{x^{12}\times z^{11}}{x^2\times z^4}$ is $x^{10}z^7$ </h3>

Step-by-step explanation:

Given expression is x to the 12th power times z to the 11th power all over x to the 2nd power times z to the 4th power.

The given expression can be written as $\frac{x^{12}\times z^{11}}{x^2\times z^4}$

<h3>To choose the correct simplification of the given expression :</h3>

Now we have to simplify the given expression as below

$\frac{x^{12}\times z^{11}}{x^2\times z^4}$

$=x^{12}\times z^{11}(x^{-2}\times z^{-4})$ ( by using the identity $\frac{1}{a^m}=a^{-m}$ )

$=(x^{12}.x^{-2})\times (z^{11}.z^{-4})$

$=x^{12-2}\times z^{11-4}$ ( by using the identity $a^m.a^n=a^{m+n}$ )

$=x^{10}\times z^7$

$=x^{10}z^7$

∴ $\frac{x^{12}\times z^{11}}{x^2\times z^4}=x^{10}z^7$

<h3>The correct simplification for the given expression $\frac{x^{12}\times z^{11}}{x^2\times z^4}$ is $x^{10}z^7$ </h3><h3>Hence option A) $x^{10}z^7$ is correct answer.</h3>

4 0

4 years ago

Can you help thank youu!! will mark the first answer!!

ELEN [110]

Answer:

0.3× $10^{5}$

Step-by-step explanation:

0.3× $10^{5}$ =30000

8 0

3 years ago

Read 2 more answers

Type the correct answer in the box. Write the standard form of the quadratic equation modeled by the points shown in the table b

White raven [17]

Y =ax² + bx +c

1) Point (0,7)

7 = a*0² +b*0 +c
c = 7
y=ax² + bx + 7

2) Point (1,4)
4=a*1² + b*1 + 7, ----> 4 = a +b + 7, ------> a+b= - 3
3) Point (2, 5)
5=a*2² + b*2 + 7, ----> 5=4a+2b +7,---> -2=4a+2b, ----> -1=2a + b

4) a+b= - 3, ----> b= -3 - a (substitute in the second equation)
2a+b= -1
2a - 3 - a = -1, ----> a - 3 = -1, a =2

5) a+b= - 3
2 + b = -3
b = -5

y=2x² - 5x + 7