All categories

2 years ago

A=2x+6xz solve for x

Mathematics

2 answers:

rodikova [14]2 years ago

6 0

Answer:

The value of the provide equation for x is $x=\frac{A}{2+6z}$ .

Step-by-step explanation:

Consider the provided equation.

$A=2x+6xz$

We need to solve the equation for x.

Take x common from right side.

$A=x(2+6z)$

Divide both the sides by 2+6z.

$\frac{A}{2+6z}=\frac{x(2+6z)}{2+6z}$

$x=\frac{A}{2+6z}$

Hence, the value of the provide equation for x is $x=\frac{A}{2+6z}$ .

Nitella [24]2 years ago

5 0

To answer your question the answer will be and or equal to 8z that is the answer.

You might be interested in

Find the volume of this triangular prism.

Ivanshal [37]

Answer:

45m^3

Step-by-step explanation:

Given that the area of thee triangle is 3m^2 and the volume formula is b x h x w. You have b x h of the triangle so you just need to multiply by the width of the prism which is 15. 3x15=45

7 0

2 years ago

A chocolate chip cookie recipe requires one and three quarters cups of flour to one cup of chocolate chips. If four and two thir

nikitadnepr [17]

1 cup of chocolate chips for every 1and 3/4 cup of flour

so divide 4 and 2/3 x 1 and 3/4

turn them both into fractions

4 2/3 = 14/3

1 3/4 = 7/4

14/3 / 7/4 = 14/3 x 4/7 = 56/21 = 2 and 2/3 cups of chocolate chips are needed

8 0

2 years ago

The point (1, −1) is on the terminal side of angle θ, in standard position. What are the values of sine, cosine, and tangent of

mixer [17]

Notice the picture below

x = 1 and y = -1

now... recall your SOH CAH TOA $\bf sin(\theta)=\cfrac{y}{r}
\qquad 
% cosine
cos(\theta)=\cfrac{x}{r}
\qquad 
% tangent
tan(\theta)=\cfrac{y}{x}$

now.. what's the value for "r" or the radius? well, using the pythagorean theorem $\bf c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}\qquad 
\begin{cases}
a=x\\
b=y\\
c=r
\end{cases}$

6 0

2 years ago

Lim x-> vô cùng ((căn bậc ba 3 (3x^3+3x^2+x-1)) -(căn bậc 3 (3x^3-x^2+1)))

NNADVOKAT [17]

I believe the given limit is

$\displaystyle \lim_{x\to\infty} \bigg(\sqrt[3]{3x^3+3x^2+x-1} - \sqrt[3]{3x^3-x^2+1}\bigg)$

Let

$a = 3x^3+3x^2+x-1 \text{ and }b = 3x^3-x^2+1$

Now rewrite the expression as a difference of cubes:

$a^{1/3}-b^{1/3} = \dfrac{\left(a^{1/3}-b^{1/3}\right)\left(a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}\right)}{\left(a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}\right)} \\\\ = \dfrac{a-b}{a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}}$

Then

$a-b = (3x^3+3x^2+x-1) - (3x^3-x^2+1) \\\\ = 4x^2+x-2$

The limit is then equivalent to

$\displaystyle \lim_{x\to\infty} \frac{4x^2+x-2}{a^{2/3}+(ab)^{1/3}+b^{2/3}}$

From each remaining cube root expression, remove the cubic terms:

$a^{2/3} = \left(3x^3+3x^2+x-1\right)^{2/3} \\\\ = \left(x^3\right)^{2/3} \left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} \\\\ = x^2 \left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3}$

$(ab)^{1/3} = \left((3x^3+3x^2+x-1)(3x^3-x^2+1)\right)^{1/3} \\\\ = \left(\left(x^3\right)^{1/3}\right)^2 \left(\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1x\right)\left(3-\dfrac1x+\dfrac1{x^3}\right)\right)^{1/3} \\\\ = x^2 \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3}$

$b^{2/3} = \left(3x^3-x^2+1\right)^{2/3} \\\\ = \left(x^3\right)^{2/3} \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3} \\\\ = x^2 \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}$

Now that we see each term in the denominator has a factor of x ², we can eliminate it :

$\displaystyle \lim_{x\to\infty} \frac{4x^2+x-2}{a^{2/3}+(ab)^{1/3}+b^{2/3}} \\\\ = \lim_{x\to\infty} \frac{4x^2+x-2}{x^2 \left(\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} + \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3} + \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}\right)}$

$=\displaystyle \lim_{x\to\infty} \frac{4+\dfrac1x-\dfrac2{x^2}}{\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} + \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3} + \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}}$

As x goes to infinity, each of the 1/x ⁿ terms converge to 0, leaving us with the overall limit,

$\displaystyle \frac{4+0-0}{(3+0+0-0)^{2/3} + (9+0-0+0+0-0)^{1/3} + (3-0+0)^{2/3}} \\\\ = \frac{4}{3^{2/3}+(3^2)^{1/3}+3^{2/3}} \\\\ = \frac{4}{3\cdot 3^{2/3}} = \boxed{\frac{4}{3^{5/3}}}$

8 0

2 years ago

The table gives some information about the heights of 30 plants.

kati45 [8]

The answer would be c

4 0

2 years ago