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3 years ago

Can someone answer this please

Mathematics

2 answers:

love history [14]3 years ago

7 0

Answer:

7^4

Step-by-step explanation:

7^12 / 7^8

We know that a^b / a^c = a^(b-c)

7^(12-8)

7^4

suter [353]3 years ago

6 0

Answer:

7^4

Step-by-step explanation:

7^(12 - 8)

7^4

Bye!

Have a nice day! :-)

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HELP PLEASE <br><br>i dont understand at all...

bija089 [108]

An exponent signifies repeated multiplication.

$x\cdot x\cdot x=x^{3}$ the factor x is repeated 3 times

Exponents can be added and subtracted to express the effects of multiplication and division.

$\dfrac{x\cdot x\cdot x\cdot x\cdot x}{x\cdot x\cdot x}=\dfrac{x^{5}}{x^{3}}\\\\=\dfrac{x\cdot x\cdot x}{x\cdot x\cdot x}\cdot x\cdot x=x\cdot x\\\\=x^{(5-3)}=x^{2}$

The addition and subtraction of exponents works the same even when there are more denominator factors than numerator factors.

$\dfrac{x\cdot x\cdot x}{x\cdot x\cdot x\cdot x\cdot x}=\dfrac{x^{3}}{x^{5}}=\dfrac{1}{x^{2}}\\\\=x^{(3-5)}=x^{-2}$

That is, a negative numerator exponent is the same as a positive denominator exponent and vice versa. You can move a factor with an exponent from denominator to numerator and change the sign of the exponent, and vice versa.

Your expression has 3 in the denominator with a negative exponent. It can be moved to the numerator and the exponent changed to positive:

$\dfrac{1}{3^{-2}}=3^{2}\\\\=3\cdot 3=\bf{9}$

8 0

3 years ago

division equation where 13 is quotient with the dividend a whole number and a unit fraction is the divisor

lys-0071 [83]

The division equation that has a quotient of 13 is 1/1/13 = 13

<h3>How to determine the equation?</h3>

The given parameters are:

Quotient = 13
Dividend = Whole number
Divisor = Unit fraction i.e. 1/n where n is an integer.

A division equation is represented as:

Dividend/Divisor = Quotient

Substitute 13 for the Quotient

Dividend/Divisor = 13

Recall that:

Unit fraction = 1/n

So, we have:

Dividend/1/n = 13

Let n = 13.

So, we have:

Dividend/1/13 = 13

This gives

13 * Dividend = 13

Divide both sides by 13

Dividend = 1

So, we have:

1/1/13 = 13

Hence, the division equation is 1/1/13 = 13

Read more about division equations at:

brainly.com/question/1622425

#SPJ1

3 0

2 years ago

3y''-6y'+6y=e*x sexcx

Simora [160]

From the homogeneous part of the ODE, we can get two fundamental solutions. The characteristic equation is

$3r^2-6r+6=0\iff r^2-2r+2=0$

which has roots at $r=1\pm i$ . This admits the two fundamental solutions

$y_1=e^x\cos x$
$y_2=e^x\sin x$

The particular solution is easiest to obtain via variation of parameters. We're looking for a solution of the form

$y_p=u_1y_1+u_2y_2$

where

$u_1=-\displaystyle\frac13\int\frac{y_2e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$
$u_2=\displaystyle\frac13\int\frac{y_1e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$

and $W(y_1,y_2)$ is the Wronskian of the fundamental solutions. We have

$W(e^x\cos x,e^x\sin x)=\begin{vmatrix}e^x\cos x&e^x\sin x\\e^x(\cos x-\sin x)&e^x(\cos x+\sin x)\end{vmatrix}=e^{2x}$

and so

$u_1=-\displaystyle\frac13\int\frac{e^{2x}\sin x\sec x}{e^{2x}}\,\mathrm dx=-\int\tan x\,\mathrm dx$
$u_1=\dfrac13\ln|\cos x|$

$u_2=\displaystyle\frac13\int\frac{e^{2x}\cos x\sec x}{e^{2x}}\,\mathrm dx=\int\mathrm dx$
$u_2=\dfrac13x$

Therefore the particular solution is

$y_p=\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

so that the general solution to the ODE is

$y=C_1e^x\cos x+C_2e^x\sin x+\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

7 0

3 years ago

Luke delivered 225 circulars on traft street 134 on 87th street 218 on roosevelt street and 229 on jones circle. how many did he

11111nata11111 [884]

225+134+218+229=806 circulars

6 0

3 years ago

Read 2 more answers

What is 4 + 7 x 2 – 8

Inessa [10]

Answer:

The answer is 10.

Step-by-step explanation:

4 + 7 x 2 - 8.

7 x 2 = 14

4 + 14 = 18

18 - 8 = 10

So, the answer is 10

Hope this helps! :)

8 0

2 years ago