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Lapatulllka [165]

3 years ago

8

Samantha works at a nursery where she pots plants. It takes her 1 1/2 hours to pot 12 plants. How many plants can Samantha pot i

n 1 hour? (Please give me the equation you used.) Please help fast!

1 answer:

ahrayia [7]3 years ago

3 0

She can plot 9 in 1 hour

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(2pm^-1q^0)^-4 • 2m ^-1 p^3 / 2pq^2

Montano1993 [528]

Answer:

$\dfrac{m^3}{16p^2q^2}$

Step-by-step explanation:

Given:

$(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}$

1.

$m^{-1}=\dfrac{1}{m}$

2.

$q^0=1$

3.

$2pm^{-1}q^0=2p\cdot \dfrac{1}{m}\cdot 1=\dfrac{2p}{m}$

4.

$(2pm^{-1}q^0)^{-4}=\left(\dfrac{2p}{m}\right)^{-4}=\left(\dfrac{m}{2p}\right)^4=\dfrac{m^4}{(2p)^4}=\dfrac{m^4}{16p^4}$

5.

$m^{-1}=\dfrac{1}{m}$

6.

$2m^{-1} p^3=2\cdot \dfrac{1}{m}\cdot p^3=\dfrac{2p^3}{m}$

7.

$\dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{\frac{2p^3}{m}}{2pq^2}=\dfrac{2p^3}{m}\cdot \dfrac{1}{2pq^2}=\dfrac{p^2}{mq^2}$

8.

$(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{m^4}{16p^4}\cdot \dfrac{p^2}{mq^2}=\dfrac{m^3}{16p^2q^2}$

8 0

3 years ago

Suppose you deposit $600 in a bank account with a simple interest rate of 2.5%. You want to keep your deposit in the bank long

8090 [49]

Answer:

120/2.5%

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

In the diagram, the radius of the outer circle is

JulijaS [17]

the value of x is 8 cm.

<u>Step-by-step explanation:</u>

Correct Question : In the diagram, the radius of the outer circle is 2x cm and the radius of the inside circle is 6 cm. The area of the shaded region is 220π cm2. What is the value of x? Enter your answer in the box.

We have ,

the area of a circle = πr²

the outer circle area = $\pi(2x)^2 =4\pi x^2$

the inside circle area = $\pi (6)^2= 36\pi$

According to Question,

the outer circle area - the inside circle area = he shaded region

⇒ $\pi (4x^2)-36\pi =220\pi$

⇒ $x^2-9 =55$

⇒ $x^2=64$

⇒ $x =\sqrt{64}=8$

Therefore , the value of x is 8 cm.

5 0

3 years ago

Solve:<br><br> (15/x)+(9x-7)/(x+2)=9

hram777 [196]

Answer:x=3

Step-by-step explanation:

D( x )

x+2 = 0

x = 0

x+2 = 0

x+2 = 0

x+2 = 0 // - 2

x = -2

x = 0

x = 0

x in (-oo:-2) U (-2:0) U (0:+oo)

(9*x-7)/(x+2)+15/x = 9 // - 9

(9*x-7)/(x+2)+15/x-9 = 0

(x*(9*x-7))/(x*(x+2))+(15*(x+2))/(x*(x+2))+(-9*x*(x+2))/(x*(x+2)) = 0

x*(9*x-7)+15*(x+2)-9*x*(x+2) = 0

9*x^2-9*x^2+8*x-18*x+30 = 0

30-10*x = 0

(30-10*x)/(x*(x+2)) = 0

(30-10*x)/(x*(x+2)) = 0 // * x*(x+2)

30-10*x = 0

30-10*x = 0 // - 30

-10*x = -30 // : -10

x = -30/(-10)

x = 3

x = 3

3 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%5Ctan%28x%20%29%20%20%3D%20%20%5Cfrac%7B10%7D%7B60%7D%20" id="TexFormula1" title=" \tan(x

chubhunter [2.5K]

tan(x) = 10/60

tan(x) = 1/6

tan(x) = a -> x = arctan(a) + πn

x = arctan(1/6) + πn

x = 0.165 + πn

Best of Luck!

7 0

3 years ago