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2 years ago

What is the exponent for the expression 6x6x6x6x6x6x6?

Mathematics

2 answers:

Artist 52 [7]2 years ago

6 0

Answer:7 is the exponent

Step-by-step explanation:

netineya [11]2 years ago

6 0

Answer:

$6^7$

Step-by-step explanation:

Exponents can be expressed through repeated multiplication, meaning that the amount of times one number is being repeated, would be the same amount as put in an exponent. Therefore :

6 * 6 * 6 * 6 * 6 * 6 * 6

6 is being repeated, therefore it's the constant.

6 is repeated 7 times, therefore the exponent is 7.

$6^7$

You might be interested in

(X^2+y^2+x)dx+xydy=0 Solve for general solution

aksik [14]

Check if the equation is exact, which happens for ODEs of the form

$M(x,y)\,\mathrm dx+N(x,y)\,\mathrm dy=0$

if $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$ .

We have

$M(x,y)=x^2+y^2+x\implies\dfrac{\partial M}{\partial y}=2y$

$N(x,y)=xy\implies\dfrac{\partial N}{\partial x}=y$

so the ODE is not quite exact, but we can find an integrating factor $\mu(x,y)$ so that

$\mu(x,y)M(x,y)\,\mathrm dx+\mu(x,y)N(x,y)\,\mathrm dy=0$

is exact, which would require

$\dfrac{\partial(\mu M)}{\partial y}=\dfrac{\partial(\mu N)}{\partial x}\implies \dfrac{\partial\mu}{\partial y}M+\mu\dfrac{\partial M}{\partial y}=\dfrac{\partial\mu}{\partial x}N+\mu\dfrac{\partial N}{\partial x}$

$\implies\mu\left(\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}\right)=M\dfrac{\partial\mu}{\partial y}-N\dfrac{\partial\mu}{\partial x}$

Notice that

$\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}=y-2y=-y$

is independent of x, and dividing this by $N(x,y)=xy$ gives an expression independent of y. If we assume $\mu=\mu(x)$ is a function of x alone, then $\frac{\partial\mu}{\partial y}=0$ , and the partial differential equation above gives

$-\mu y=-xy\dfrac{\mathrm d\mu}{\mathrm dx}$

which is separable and we can solve for $\mu$ easily.

$-\mu=-x\dfrac{\mathrm d\mu}{\mathrm dx}$

$\dfrac{\mathrm d\mu}\mu=\dfrac{\mathrm dx}x$

$\ln|\mu|=\ln|x|$

$\implies \mu=x$

So, multiply the original ODE by x on both sides:

$(x^3+xy^2+x^2)\,\mathrm dx+x^2y\,\mathrm dy=0$

Now

$\dfrac{\partial(x^3+xy^2+x^2)}{\partial y}=2xy$

$\dfrac{\partial(x^2y)}{\partial x}=2xy$

so the modified ODE is exact.

Now we look for a solution of the form $F(x,y)=C$ , with differential

$\mathrm dF=\dfrac{\partial F}{\partial x}\,\mathrm dx+\dfrac{\partial F}{\partial y}\,\mathrm dy=0$

The solution F satisfies

$\dfrac{\partial F}{\partial x}=x^3+xy^2+x^2$

$\dfrac{\partial F}{\partial y}=x^2y$

Integrating both sides of the first equation with respect to x gives

$F(x,y)=\dfrac{x^4}4+\dfrac{x^2y^2}2+\dfrac{x^3}3+f(y)$

Differentiating both sides with respect to y gives

$\dfrac{\partial F}{\partial y}=x^2y+\dfrac{\mathrm df}{\mathrm dy}=x^2y$

$\implies\dfrac{\mathrm df}{\mathrm dy}=0\implies f(y)=C$

So the solution to the ODE is

$F(x,y)=C\iff \dfrac{x^4}4+\dfrac{x^2y^2}2+\dfrac{x^3}3+C=C$

$\implies\boxed{\dfrac{x^4}4+\dfrac{x^2y^2}2+\dfrac{x^3}3=C}$

5 0

3 years ago

Pls help me with these math problems

user100 [1]

2. W<5
4. D<6 (or equal to)

3 0

3 years ago

Read 2 more answers

-6-5

tankabanditka [31]

The equation of the line that is parallel to the given line and passes through the point (-2,2) is:

$y = \frac{x}{5} + \frac{12}{5}$

<h3>What is a linear function?</h3>

A linear function is modeled by:

y = mx + b

In which:

m is the slope, which is the rate of change, that is, by how much y changes when x changes by 1.

b is the y-intercept, which is the value of y when x = 0, and can also be interpreted as the initial value of the function.

When two lines are parallel, they have the same slope. In this problem, the given line passes through (-5,-4) and (0,-3), hence the slope is:

m = (-3 - (-4))/(0 - (-5)) = 1/5.

Hence the equation is:

$y = \frac{x}{5} + b$

When x = -2, y = 2, then:

$y = \frac{x}{5} + b$

$2 = \frac{-2}{5} + b$

$b = \frac{12}{5}$

Hence:

$y = \frac{x}{5} + \frac{12}{5}$

More can be learned about linear equations at brainly.com/question/24808124

#SPJ1

7 0

1 year ago

What's the length of AB?

Nastasia [14]

Use the pythagorean theorem to solve this: a^2 + b^2 = c^2.
You already have a and b which are 3 and 4. Square them up and you will get 9 + 16 = c^2.
Add.
25 = c^2
Lastly, square root both sides.
5 = c
The length of AB is 5.

8 0

2 years ago

Show work please \sqrt(x+12)-\sqrt(2x+1)=1

Nesterboy [21]

Answer:

$x=4$

Step-by-step explanation:

Given $\displaystyle\\\sqrt{x+12}-\sqrt{2x+1}=1$ , start by squaring both sides to work towards isolating $x$ :

$\displaystyle\\\left(\sqrt{x+12}-\sqrt{2x+1}\right)^2=\left(1\right)^2$

Recall $(a-b)^2=a^2-2ab+b^2$ and $\sqrt{a}\cdot \sqrt{b}=\sqrt{a\cdot b}$ :

$\displaystyle\\\left(\sqrt{x+12}-\sqrt{2x+1}\right)^2=\left(1\right)^2\\\implies x+12-2\sqrt{(x+12)(2x+1)}+2x+1=1$

Isolate the radical:

$\displaystyle\\x+12-2\sqrt{(x+12)(2x+1)}+2x+1=1\\\implies -2\sqrt{(x+12)(2x+1)}=-3x-12\\\implies \sqrt{(x+12)(2x+1)}=\frac{-3x-12}{-2}$

Square both sides:

$\displaystyle\\(x+12)(2x+1)=\left(\frac{-3x-12}{-2}\right)^2$

Expand using FOIL and $(a+b)^2=a^2+2ab+b^2$ :

$\displaystyle\\2x^2+25x+12=\frac{9}{4}x^2+18x+36$

Move everything to one side to get a quadratic:

$\displaystyle-\frac{1}{4}x^2+7x-24=0$

Solving using the quadratic formula:

A quadratic in $ax^2+bx+c$ has real solutions $\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ . In $\displaystyle-\frac{1}{4}x^2+7x-24$ , assign values:

$\displaystyle \\a=-\frac{1}{4}\\b=7\\c=-24$

Solving yields:

$\displaystyle\\x=\frac{-7\pm \sqrt{7^2-4\left(-\frac{1}{4}\right)\left(-24\right)}}{2\left(-\frac{1}{4}\right)}\\\\x=\frac{-7\pm \sqrt{25}}{-\frac{1}{2}}\\\\\begin{cases}x=\frac{-7+5}{-0.5}=\frac{-2}{-0.5}=\boxed{4}\\x=\frac{-7-5}{-0.5}=\frac{-12}{-0.5}=24 \:(\text{Extraneous})\end{cases}$

Only $x=4$ works when plugged in the original equation. Therefore, $x=24$ is extraneous and the only solution is $\boxed{x=4}$

4 0

2 years ago