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2 years ago

6

2x+5=12 i don´t know what the value of x is

1 answer:

Ilya [14]2 years ago

3 0

Answer: X = 7/2 or 3.5

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Helpppppp<br> I NEED THIS<br> HELP

ohaa [14]

Answer:

A

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=if%20%5C%3A%20%28%20%5Cfrac%7B3%7D%7B4%7D%20%29%5E%7B6%7D%20%20%5Ctimes%20%28%20%5Cfrac%7B16%7

zhenek [66]

Answer:

2

Step-by-step explanation:

$(\frac{3}{4})^6 \times (\frac{16}{9})^5=(\frac{4}{3})^{x+2}$

$\frac{3^6}{4^6} \cdot \frac{16^5}{9^5}=\frac{4^{x+2}}{3^{x+2}}$

$\frac{3^6}{4^6} \cdot \frac{(4^2)^5}{(3^2)^5}=\frac{4^{x+2}}{3^{x+2}}$

$\frac{3^6}{4^6} \cdot \frac{4^{10}}{3^{10}}=\frac{4^{x+2}}{3^{x+2}}$

$\frac{3^6}{3^{10}} \cdot \frac{4^{10}}{4^6}=\frac{4^{x+2}}{3^{x+2}}$

$3^{-4} \cdot 4^{4}=4^{x+2}3^{-(x+2)}$

This implies

x+2=4

and

-(x+2)=-4.

x+2=4 implies x=2 since subtract 2 on both sides gives us x=2.

Solving -(x+2)=-4 should give us the same value.

Multiply both sides by -1:

x+2=4

It is the same equation as the other.

You will get x=2 either way.

Let's check:

$(\frac{3}{4})^6 \times (\frac{16}{9})^5=(\frac{4}{3})^{2+2}$

$(\frac{3}{4})^6 \times (\frac{16}{9})^5=(\frac{4}{3})^{4}$

Put both sides into your calculator and see if you get the same thing on both sides:

Left hand side gives 256/81.

Right hand side gives 256/81.

Both side are indeed the same for x=2.

4 0

2 years ago

Which expression is equivalent to r^9/r^3

Deffense [45]

$\cfrac{r^9}{r^3} =r^{9-3}=r^6$

4 0

3 years ago

Read 2 more answers

Given the system of linear equations:

Vanyuwa [196]

Answer:

Step-by-step explanation:

hello :

Part A : x+6y =6 means : 6y = - x+6

so : y = (-1/6)x+1 an equation for the line (D)

y = (1/3)x -2 is the line (D')

PartB : solution of the system : y = (-1/6)x+1 ....(1) color red

y = (1/3)x -2 ....(2) color bleu

is the intersection point : (6 ; 0)

PartC : Algebraically by (1) and (2) : (-1/6)x+1 = (1/3)x -2

(-1/6)x - (1/3)x = -2-1

(-x-2x)/6= -3

-3x = -18

so : x = 6 put this value in (1) or (2) : y = (-1/6)(6)+1 =0 the solution is : (6 ;0)

3 0

3 years ago

For each of the following vector fields

olga nikolaevna [1]

(A)

$\dfrac{\partial f}{\partial x}=-16x+2y$

$\implies f(x,y)=-8x^2+2xy+g(y)$

$\implies\dfrac{\partial f}{\partial y}=2x+\dfrac{\mathrm dg}{\mathrm dy}=2x+10y$

$\implies\dfrac{\mathrm dg}{\mathrm dy}=10y$

$\implies g(y)=5y^2+C$

$\implies f(x,y)=\boxed{-8x^2+2xy+5y^2+C}$

(B)

$\dfrac{\partial f}{\partial x}=-8y$

$\implies f(x,y)=-8xy+g(y)$

$\implies\dfrac{\partial f}{\partial y}=-8x+\dfrac{\mathrm dg}{\mathrm dy}=-7x$

$\implies \dfrac{\mathrm dg}{\mathrm dy}=x$

But we assume $g(y)$ is a function of $y$ alone, so there is not potential function here.

(C)

$\dfrac{\partial f}{\partial x}=-8\sin y$

$\implies f(x,y)=-8x\sin y+g(x,y)$

$\implies\dfrac{\partial f}{\partial y}=-8x\cos y+\dfrac{\mathrm dg}{\mathrm dy}=4y-8x\cos y$

$\implies\dfrac{\mathrm dg}{\mathrm dy}=4y$

$\implies g(y)=2y^2+C$

$\implies f(x,y)=\boxed{-8x\sin y+2y^2+C}$

For (A) and (C), we have $f(0,0)=0$ , which makes $C=0$ for both.

4 0

2 years ago