Ask question

Ask question

All categories

uranmaximum [27]

2 years ago

10

Sophia’s vegetable garden is divided into 12 equal sections. She plants carrots in 8 of the sections. Write two fractions that a

re equivalent to the part of Sophia’s garden that is planted with carrots.

1 answer:

givi [52]2 years ago

4 0

Answer:

Sophia's Vegetable garden - 1 thats divided into 12 sections

the carrots are planted in 8 of the sections making

8/12 or 2/3

You might be interested in

Answers: <br> 3x + 26<br> 5x + 15<br> x + 30<br> 10x - 7<br> please answer fast!

pentagon [3]

Answer: I think it's A. 3x + 26 but im not completely sure

Step-by-step explanation:

5 0

3 years ago

Read this excerpt from “Incidents in the Life of a Slave Girl.” Again and again I had [traveled] those dreary twelve miles to an

Kamila [148]

Answer:

i think its D but im not entirely sure, im sincerly sry if im wrong

Step-by-step explanation:

3 0

2 years ago

Solve the equation<br> 3x/ 8 - 4x/3 =4

Virty [35]

Answer:

x = -96/23

Step-by-step explanation

$\frac{3x}{8}$ - $\frac{4x}{3}$ = 4

multiply each side of equation by 24 to eliminate denominators

(24) $\frac{3x}{8}$ - $\frac{4x}{3}$ = 4 (24)

3(3x) - 8(4x) = 96

9x - 32x = 96

-23x = 96

x = -96/23

6 0

2 years ago

Sam decides to buy 100 shares of DEF company. They cost $16.60 a share. The commission is 6% of the total price. What is Sam's t

Serga [27]

100×16.60=1,660

Sam cost
1,660+1,660×0.06=1,759.6

5 0

2 years ago

Read 2 more answers

Factor completely. <br> <img src="https://tex.z-dn.net/?f=x%5E%7B8%7D-%5Cfrac%7B1%7D%7B81%7D" id="TexFormula1" title="x^{8}-\fra

Eduardwww [97]

We have 3⁴ = 81, so we can factorize this as a difference of squares twice:

$x^8 - \dfrac1{81} = \left(x^2\right)^4 - \left(\dfrac13\right)^4 \\\\ x^8 - \dfrac1{81} = \left(\left(x^2\right)^2 - \left(\dfrac13\right)^2\right) \left(\left(x^2\right)^2 + \left(\dfrac13\right)^2\right) \\\\ x^8 - \dfrac1{81} = \left(x^2 - \dfrac13\right) \left(x^2 + \dfrac13\right) \left(\left(x^2\right)^2 + \left(\dfrac13\right)^2\right) \\\\ x^8 - \dfrac1{81} = \left(x^2 - \dfrac13\right) \left(x^2 + \dfrac13\right) \left(x^4 + \dfrac19\right)$

Depending on the precise definition of "completely" in this context, you can go a bit further and factorize $x^2-\frac13$ as yet another difference of squares:

$x^2 - \dfrac13 = x^2 - \left(\dfrac1{\sqrt3}\right)^2 = \left(x-\dfrac1{\sqrt3}\right)\left(x+\dfrac1{\sqrt3}\right)$

And if you're working over the field of complex numbers, you can go even further. For instance,

$x^4 + \dfrac19 = \left(x^2\right)^2 - \left(i\dfrac13\right)^2 = \left(x^2 - i\dfrac13\right) \left(x^2 + i\dfrac13\right)$

But I think you'd be fine stopping at the first result,

$x^8 - \dfrac1{81} = \boxed{\left(x^2 - \dfrac13\right) \left(x^2 + \dfrac13\right) \left(x^4 + \dfrac19\right)}$

6 0

2 years ago