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4 years ago

Which expression is equivalentHELP ME PLEASE

Mathematics

1 answer:

aalyn [17]4 years ago

7 0

Answer:

the 4th option is correct

Step-by-step explanation:

$xy^{2/9}$ Original expression

$x(y^{2})^{1/9}$ Split up the exponent

$x(\sqrt[9]{y^{2} } )$ The 1/9 of an expression is the same thing is the 9th root of the expression.

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I dont get how to do this please help me solve for the absolute value equation show your work

elena-14-01-66 [18.8K]

Step-by-step explanation:

$| \frac{x}{3} +{ \frac{2}{5} } | = 2$

$\frac{x}{3} = \frac{6}{5}$

$5x = 18$

$x = \frac{18}{5}$

and

x/3=-12/5

x=-36/5

7 0

3 years ago

2/7, solve for n, explain too please

kirill [66]

Answer:

n = 19.2

Step-by-step explanation:

12 / 5 = n / 8

n = 12 * 8 / 5

n = 96 / 5

n = 19.2

3 0

3 years ago

Read 2 more answers

A sports store sells three different packs of tennis balls. Brand A costs $2.50 for two balls, brand B sells three balls for $3.

meriva

Answer:

Step-by-step explanation:

sports store sells three different packs of tennis balls. Brand A costs $2.50 for two balls, brand B sells three balls for $3.90, and brand C has a sale of four balls for $5.12. Which inequality about the unit price per ball is correct

Brand A:

$2.50 for 2

Brand B :

$3.90 for 3

Brand C:

$5.12 for 4

We calculate the unit price :

Brand A:

$2.50 / 2 = $1.25

Brand B:

$3.90 / 3 = $1.30

Brand C:

$5.12 / 4 = $1.2

4 0

3 years ago

Ben and Josh went to the roof of their 40-foot tall high school to throw their math books offthe edge.The initial velocity of Be

Taya2010 [7]

Answer

Josh's textbook reached the ground first

Josh's textbook reached the ground first by a difference of $t=0.6482$

Step-by-step explanation:

Before we proceed let us re write correctly the height equation which in correct form reads:

$h(t)=-16t^2 +v_{o}t+s$ Eqn(1).

Where:

$h(t)$ : is the height range as a function of time

$v_{o}$ : is the initial velocity

$s$ : is the initial heightin feet and is given as 40 feet, thus Eqn(1). becomes:

$h(t)=-16t^2 + v_{o}t + 40$ Eqn(2).

Now let us use the given information and set up our equations for Ben and Josh.

Ben:

We know that $v_{o}=60ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+60t+40$ Eqn.(3)

Josh:

We know that $v_{o}=48ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+48t+40$ Eqn. (4).

Now since we want to find whose textbook reaches the ground first and by how many seconds we need to solve each equation (i.e. Eqns. (3) and (4)) at $h(t)=0$ . Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one.

Thus we have for Ben, Eqn. (3) gives:

$h(t)=0-16t^2+60t+40=0$

Using the quadratic expression to find the roots of the quadratic we have:

$t_{1,2}=\frac{-b+/-\sqrt{b^2-4ac} }{2a} \\t_{1,2}=\frac{-60+/-\sqrt{60^2-4(-16)(40)} }{2(-16)} \\t_{1,2}=\frac{-60+/-\sqrt{6160} }{-32} \\t_{1,2}=\frac{15+/-\sqrt{385} }{8}\\\\t_{1}=4.3276 sec\\t_{2}=-0.5776 sec$