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3 years ago

What is 2 to the second power

Mathematics

2 answers:

valentinak56 [21]3 years ago

8 0

Answer:

The answer is 4

Step-by-step explanation:

Another term for 2 to the second power is 2 squared.

Any time you square something you multiply that number by itself. So in reality that problem would be 2 times 2.

If it were a bigger number you would multiply the number by itself the about of times you see if anything is to the first power it stays the same so squared is a number times itself, and cubed which is the third power would be:

2 x 2 x 2.

2 x 2 or 2 to the second power or 2 squared equals 4.

Hope this helps.

-Que

kodGreya [7K]3 years ago

6 0

Answer:

Step-by-step explanation:

2 to the second power = 2^2.

2^2 means 2*2 = 4.

The 2 in ^2 is called the 'exponent' .

Note 3^3 = 3*3*3

and 5^5 = 5*5*5*5*5.

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If line segment RU is considered the base of parallelogram RSTU, what is the corresponding height of the parallelogram?

klasskru [66]

Given that line segment RU with vertices R(1, 1) and U(4, 5) is considered the base of parallelogram RSTU.

Then, the line segment ST with vertices S(7, 0) and T(10, 4) is the top of the parallelogram.

The corresponding height of the parallelogram is the length of a line with endponts at RU and ST and perpendicular to both RU and ST.

The equation of the line segment RU is given by
$\frac{y-1}{x-1} = \frac{5-1}{4-1} = \frac{4}{3} \\ \\ 3(y-1)=4(x-1) \\ \\ 3y-3=4x-4 \\ \\ 3y=4x-1 \\ \\ y= \frac{4}{3} x- \frac{1}{3}$

Recall that given that two lines are perpendicular, the product of the slope of the two lines is -1.
Let the slope of the line perpendicular to line RU be m, then
$\frac{4}{3} m=-1 \\ \\ m=- \frac{3}{4}$

Thus, the equation of the line perpendicular to RU passing through point (1, 1) is given by
$y-1=- \frac{3}{4} (x-1) \\ \\ 4(y-1)=-3(x-1) \\ \\ 4y-4=-3x+3 \\ \\ 4y=-3x+7 \\ \\ y=- \frac{3}{4} x+ \frac{7}{4}$

The equation of the line segment ST is given by
$\frac{y-0}{x-7} = \frac{4-0}{10-7} = \frac{4}{3} \\ \\ 3y=4(x-7)=4x-28 \\ \\ y= \frac{4}{3} x- \frac{28}{3}$

The line perpendicular to line segment RU intersected line segment ST at the point given by
$- \frac{3}{4} x+ \frac{7}{4}=\frac{4}{3} x- \frac{28}{3} \\ \\ \frac{4}{3} x+\frac{3}{4} x=\frac{7}{4}+\frac{28}{3} \\ \\ \frac{25}{12} x= \frac{133}{12} \\ \\ x= \frac{133}{25} \\ \\ y=\frac{4}{3} \left(\frac{133}{25}\right)- \frac{28}{3}= -\frac{56}{25}$

Thus, the corresponding height of the parallelogram is the line with endpoints
$(1,1) \ and \ \left(\frac{133}{25},-\frac{56}{25}\right)$

Recall that the length of a line passing through points
$(x_1,y_1) \ and \ (x_2,y_2)$
is given by
$l= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Thus, the length of the line passing through points
$(1,1) \ and \ \left(\frac{133}{25},-\frac{56}{25}\right)$
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$l= \sqrt{\left(\frac{133}{25}-1\right)^2+\left(-\frac{56}{25}-1\right)^2} \\ \\ = \sqrt{\left( \frac{108}{25}\right)^2+\left(- \frac{81}{25} \right)^2}= \sqrt{ \frac{11,664}{625} + \frac{6,561}{625} } \\ \\ = \sqrt{ \frac{729}{25} } = \frac{27}{5} =5.4$

Therefore, the corresponding height of the given parallelogram is 5.4 units

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