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3 years ago

(8a^3÷27x^-3)1÷3 simplify

Mathematics

2 answers:

Damm [24]3 years ago

8 0

Answer: (8a^3x^3)/81

Step-by-step explanation: Simplifying the expression :)

NeTakaya3 years ago

4 0

Answer:

$\frac{8a^{3}x^{3} }{81}$ is the simplified version of (8a^3÷27x^-3)1÷3.

Step-by-step explanation:

$27x^{-3}$ would be put on the top of the equation, since it was in the denominator when dividing the $8ax^{3}$ .

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The marked price of an article is 30% above it's selling price and the cost price is 35%less than it's marked price. find the di

SashulF [63]

Answer:

Let C be the cost price of the article. Then:

C(1+30%)=marked price

So:

C(1+30%)=C(1.3)

=1.3C

If the article is then marked down 35%, then its’ price would be:

1.3C(1–35%)=1.3C(.65)

=0.845 or 84.5% of its’ cost

Then:

0.845–1=-0.155

or, a 15.5% loss on the item

8 0

3 years ago

Show work and explain with formulas:

Marina86 [1]

17 Answer: 205

Step-by-step explanation:

$\{1\dfrac{2}{3}+1\dfrac{5}{6}+2+...+8\dfrac{1}{3}\}\implies a_1=1\dfrac{2}{3},\ d=\dfrac{1}{6}\\\\\\a_n=a_1+d(n-1)\qquad solve\ for\ n\\\\8\dfrac{1}{3}=1\dfrac{2}{3}+\dfrac{1}{6}(n-1)\\\\\\\dfrac{25}{3}=\dfrac{5}{3}+\dfrac{1}{6}n-\dfrac{1}{6}\\\\\\\dfrac{50}{6}=\dfrac{10}{6}+\dfrac{1}{6}n-\dfrac{1}{6}\\\\\\\dfrac{41}{6}=\dfrac{1}{6}n\\\\\\\dfrac{41}{6}\cdot 6=n\\\\41=n$

$\text{Now use the sum formula:}\\\\S_n=\dfrac{a_1+a_n}{2}\cdot n\\\\\\S_{41}=\dfrac{1\dfrac{2}{3}+8\dfrac{1}{3}}{2}\cdot 41\\\\\\.\quad =\dfrac{10}{2}\cdot 41\\\\\\.\quad =5\cdot 41\\\\.\quad =\large\boxed{205}$

18 Answer: 1968

Step-by-step explanation:

$a_1=-6,\ d=2,\ n=48, \quad \text{solve for }a_{48}\\\\a_{n}=a_1+d(n-1)\\\\a_{48}=-6+2(48-1)\\\\.\quad =-6+2(47)\\\\.\quad =-6+94\\\\.\quad =88$

$\text{Now use the sum formula:}\\\\S_n=\dfrac{a_1+a_n}{2}\cdot n\\\\\\S_{48}=\dfrac{-6+88}{2}\cdot 48\\\\\\.\quad =\dfrac{82}{2}\cdot 48\\\\\\.\quad =41\cdot 48\\\\.\quad =\large\boxed{1968}$

19 Answer: -116

Step-by-step explanation:

$\{3, -2, -7, ...\}\\a_1=3,\ d=-5,\ n=8, \quad \text{solve for }a_{8}\\\\a_{n}=a_1+d(n-1)\\\\a_{8}=3-5(8-1)\\\\.\quad =3-5(7)\\\\.\quad =3-35\\\\.\quad =-32\\\\\text{Now use the sum formula:}\\\\S_8=\dfrac{a_1+a_8}{2}\cdot 8\\\\\\S_{8}=\dfrac{3-32}{2}\cdot 8\\\\\\.\quad =\dfrac{-29}{2}\cdot 8\\\\\\.\quad =-29\cdot 4\\\\.\quad =\large\boxed{-116}$

6 0

3 years ago

Solve using quadratic formula 6x^2-x=2

Nadya [2.5K]

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$x = \cfrac{ -(-1) \pm \sqrt { (-1)^2 -4(6)(-2)}}{2(6)}\implies x = \cfrac{1\pm\sqrt{1+48}}{12} \\\\\\ x = \cfrac{1\pm\sqrt{49}}{12}\implies x = \cfrac{1\pm 7}{12}\implies x = \begin{cases} \frac{8}{12}\to &\frac{2}{3}\\[1em] -\frac{6}{12}\to &-\frac{1}{2} \end{cases}$

7 0

3 years ago

If 1/2 of 5 is 3, then what is 1/3 of 10?

Dahasolnce [82]

3.3 repeated hope this helps

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4 years ago

Use the elimination method to solve 4x+8y=16 4x-8y=0

Rina8888 [55]

Answer:

x=2, y=1

Step-by-step explanation:

4x+8y=16

4x-8y=0

Add the two equations together:

8x=16

Divide both sides by 8:

x=2

Plug this back into one of the original equations to find y:

4(2)-8y=0

8-8y=0

y=1

Hope this helps!

3 0

3 years ago

Read 2 more answers

(8a^3÷27x^-3)1÷3 simplify​

(8a^3÷27x^-3)1÷3 simplify