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3 years ago

Stuck? helppp aspappp

Mathematics

2 answers:

levacccp [35]3 years ago

5 0

Here’s what I found, good luck bella.

Keith_Richards [23]3 years ago

3 0

The answer to 3) is: x = ±7

Break down the problem into these 2 equations:

x/7 = 1 and -x/7 = 1

Then solve both equation and collect all solutions.

Let me know if this helped you! Good luck Bella4992!

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Also need help on this

Ira Lisetskai [31]

Answer is 0.5
-2 divided by -4 equals 0.5
and same thing for the rest of the equation.

8 0

3 years ago

Read 2 more answers

Plz help<br> Will mark brainliest

Alexandra [31]

Answer:

(0,-1)

Step-by-step explanation:

(0,-1) will work because when you plug them is it is equivalent to 4

6 0

3 years ago

Whats 70-3x6+9 I need the answer plz

kumpel [21]

70-3·6+9

Remember PEMDAS.

You must multiply the 3 and 6 together, and you would get 18.

Then, you subtract 18 from 70, which would give you 52.

And finally, you would add 9 to 52, getting 61 as your final answer.

7 0

3 years ago

How can I write 625 in as many ways as I can in exponents

kirill [66]

Well, 625 is 5 x 5 x 5 x 5 .

So that would be:

(5)⁴

5 x (5)³

(5)² x (5)²

And you could even go the other way if you want to get silly:

(5)⁵ x (5)⁻¹

(5)⁶ x (5)⁻²
.
.
etc.

4 0

3 years ago

Match solutions and differential equations. Note: Each equation may have more than one solution. Select all that apply. (a) 7y''

Nonamiya [84]

Answer:

a- $e^x,e^{-x}$

b- $x^{-2}$

c- $x^3$

Step-by-step explanation:

a.7 y''-7 y =0

Auxillary equation

$D^2-1=0$

$(D-1)(D+1)=0$

D=1,-1

Then , the solution of given differential equation

$y=e^x,y=e^{-x}$

2. $7x^2y''+14xy'-14 y=0$

Y= $y=x^3$

$y=3x^2$

$y''=6x$

Substitute in the given differential equation

$42x^3+42x^3-14x^3\neq 0$

Hence, $x^3$ is not a solution of given differential equation

$e^x,e^{-x}$ are also not a solution of given differential equation.

y= $x^{-2}$

$y'=-2x^{-3}$

$y''=6x^{-4}$

Substitute the values in the differential equation

$7x^2(6x^{-4})+14x(-2x^{-3})-14 x^{-2}$

= $42x^{-2}-28x^{-2}-14x^{-2}=0$

Hence, $x^{-2}$ is a solution of given differential equation.

c. $7x^2y''-42y=0$

$y=x^3$

$y'=3x^2$

$y''=6x$

Substitute the values in the differential equation

$42x^3-42x^3=0$

Hence, $x^3$ is a solution of given differential equation.

a- $e^x,e^{-x}$

b- $x^{-2}$

c- $x^3$

6 0

3 years ago

Read 2 more answers