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3 years ago

15

Solve the following equation:x^2=10

2 answers:

torisob [31]3 years ago

8 0

Answer:

x=- (square root sign) 10 x=(square root sign) 10

Step-by-step explanation:

kow [346]3 years ago

5 0

Question

Solve the following equation:x^2=10

Answer:

<h2>x = √10</h2>

or

<h2>x = ± 3.16</h2>

Step-by-step explanation:

x² = 10

√x² = √10

x = √10 or ± 3.16

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which digit has the same value in both numbers? what is the value of that digit 123456789. 987654321

r-ruslan [8.4K]

The digit 5 does because it is in the same place value.

8 0

3 years ago

Subtract the fractions. Write each difference in simplest form. <br><br><br><br> 8/9 - 5/6 =

Stolb23 [73]

Answer:

1/18

Step-by-step explanation:

Change the denominators to 18 to make like denonminators (make sure to multiply the top)

(16/18 - 15/18)

1/18

5 0

2 years ago

Read 2 more answers

Morgan writes the expression

DerKrebs [107]

He didn’t follow PEMDAS which is the an acronym for the order of operations. according to pemdas the first step is parenthesis (which is the p in pemdas) there fore the first step is to add the 1 and the 2
then since there are no exponents you multiply so you multiple 5(3) which is 15
then you can add all the number left to right
so 15+2+3= 20
so the answer is 20
hope this helped
let me know if you have any questions

5 0

3 years ago

Consider the following differential equation to be solved by undetermined coefficients. y(4) − 2y''' + y'' = ex + 1 Write the gi

kompoz [17]

Answer:

The general solution is

$y= (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x +\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

+ $\frac{x^2}{2}$

Step-by-step explanation:

Step :1:-

Given differential equation y(4) − 2y''' + y'' = e^x + 1

The differential operator form of the given differential equation

$(D^4 -2D^3+D^2)y = e^x+1$

comparing f(D)y = e^ x+1

The auxiliary equation (A.E) f(m) = 0

$m^4 -2m^3+m^2 = 0$

$m^2(m^2 -2m+1) = 0$

$(m^2 -2m+1) this is the expansion of (a-b)^2$

$m^2 =0 and (m-1)^2 =0$

The roots are m=0,0 and m =1,1

complementary function is $y_{c} = (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x$

<u>Step 2</u>:-

The particular equation is $\frac{1}{f(D)} Q$

P.I = $\frac{1}{D^2(D-1)^2} e^x+1$

P.I = $\frac{1}{D^2(D-1)^2} e^x+\frac{1}{D^2(D-1)^2}e^{0x}$

P.I = $I_{1} +I_{2}$

$\frac{1}{D^2} (\frac{x^2}{2!} )e^x + \frac{1}{D^{2} } e^{0x}$

$\frac{1}{D} means integration$

$\frac{1}{D^2} (\frac{x^2}{2!} )e^x = \frac{1}{2D} \int\limits {x^2e^x} \, dx$

applying in integration u v formula

$\int\limits {uv} \, dx = u\int\limits {v} \, dx - \int\limits ({u^{l}\int\limits{v} \, dx } )\, dx$

$I_{1}$ = $\frac{1}{D^2(D-1)^2} e^x$

$\frac{1}{2D} (e^x(x^2)-e^x(2x)+e^x(2))$

$\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

$I_{2}$ $= \frac{1}{D^2(D-1)^2}e^{0x}$

$\frac{1}{D} \int\limits {1} \, dx= \frac{1}{D} x$

again integration $\frac{1}{D} x = \frac{x^2}{2!}$

The general solution is $y = y_{C} +y_{P}$

$y= (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x +\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

+ $\frac{x^2}{2!}$

3 0

3 years ago

Identify the inverse of f(x)=x^2-9

Reika [66]

A
x=y^2-9
x+9=y^2
square root both sides
+-sqrt(x+9)

7 0

2 years ago