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Andrew [12]
3 years ago
13

Simply and find the value of the following

Mathematics
2 answers:
Digiron [165]3 years ago
5 0

Answer:

this is the answer of your question

anyanavicka [17]3 years ago
4 0

Step-by-step explanation:

i) \:  {(729)}^{ \frac{1}{6} }

=  {(3 \times 3 \times 3 \times 3 \times 3 \times 3)}^{ \frac{1}{6} }

=    \sqrt[6]{(3 \times 3 \times 3 \times 3 \times 3 \times 3)}

= <em><u>3 (Ans)</u></em>

ii) \:  {(64)}^{ \frac{2}{3} }

=  \sqrt[3]{ {64}^{2} }

=  \sqrt[3]{(64 \times 64)}

=  \sqrt[3]{ {2}^{12} }

=  {(2)}^{ \frac{12}{3} }

=  {2}^{4}

= <em><u>16 (Ans)</u></em>

iii) \:  {(243)}^{ \frac{6}{5} }

= {( \sqrt[5]{ {243} })}^{6}

=  {3}^{6}

= <em><u>729 (Ans)</u></em>

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Evaluate the expression using x = -6 and y= 3
FinnZ [79.3K]

Answer:

-45

Step-by-step explanation:

put x = -6 and y=3 in this expression : y(2x - y)

3(2(-6)-3) = 3(-12 -3)= 3(-15) = -45

6 0
3 years ago
Read 2 more answers
Help what are the answers im so stuck!
AnnZ [28]

Answer:

  see below

Step-by-step explanation:

A percentage change is computed as follows:

  percent change = ((new value) -(original value))/(original value) × 100%

For your first one, this looks like ...

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__

When there are a lot of similar calculations to do, I like to use a spreadsheet. See below for the other percentage changes (rightmost column).

4 0
3 years ago
The product of two number is 1575 and their quotient is 9/2 find the number
Otrada [13]
(1) p x q = 1575
(2) p / q = 9 / 2
divide 1 by 2 to eliminate p:
q^2 = 1575 / (9/2)
q^2 = 350
q = sqrt(350)

Substitue q in either 1 or 2 to find p.
5 0
4 years ago
Which of the following functions is graphed below?
Vladimir [108]

Answer:

Step-by-step explanation:

The answer is D because the parabola is an empty circle and is positive which would mean its greater than 1 and the line has a solid point and is negative which means less than or equal to.

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3 years ago
On the moon, the height of a golf ball hit with an initial velocity of 96 ft/sec is given by the equation h= -2.7t^2+96t, where
nikdorinn [45]
So hmmm notice the picture below

"x" being how many seconds the object was going

and when y = 0, the object hits the ground, either on the moon or on earth

so, if we set y = 0 on each equation, we'll know when that happen, how long of a "x-value" or seconds it took

\bf \begin{array}{llll}&#10;\textit{on the moon}\\\\&#10;h=-2.7t^2+96t\implies 0=t(96-2.7t)&#10;\end{array} \quad &#10;\begin{cases}&#10;0=t\\&#10;-------\\&#10;0=96-2.7t\\&#10;2.7t=96\\&#10;t=\frac{96}{2.7}\\&#10;t\approx 35.\overline{55}&#10;\end{cases}\\\\&#10;-------------------------------\\\\&#10;\begin{array}{llll}&#10;\textit{on earth}\\\\&#10;h=-16t^2+96t\implies 0=t(96-16t)&#10;\end{array} \quad &#10;\begin{cases}&#10;0=t\\&#10;------\\&#10;0=96-16t\\&#10;16t=96\\&#10;t=\frac{96}{16}\\&#10;t=6&#10;\end{cases}

4 0
3 years ago
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