Hey i need help with question 3 i put 81.5% but i’ve been told it’s been wrong ?

How do I solve this question?

Sati [7]

Answer:

m=2 and n=3

Step-by-step explanation:

Step :-

Given $[ 2 x^{n}y^{2} ]^m = 4 x^6 y^4$

using algebraic formula $(a^m)^n= a^{mn}$

now

$2^{m} x^{mn} y^{2m} = 4x^{6} y^{4}$

now equating 'x' powers, we get

$x^{mn} = x^{6}$

$mn=6$ ....(1)

now

$y^{2m} = y^{4}$

Equating 'y' powers ,we get

2 m=4

m=2

substitute m= 2 in equation (1)

we get

2 n=6

n=3

verification:-

substitute m=2 and n=3 , we get

$[ 2 x^{n}y^{2} ]^m = 4 x^6 y^4$

$(2 x^{3} y^2)^2 = 4 x^6 y ^4\\$

$4 x^6 y^4 = 4 x^6 y^4$

both are equating so m= 2 and n=3

4 0

3 years ago

paul earns 25.50 for every lawn he mows. this week he moved L lawns. he spends 3.50 per lawn on gasoline. write an expression to

notsponge [240]

If Paul mows the lawn once, he earns 1*25.50. If he mows it twice, he earns 2*25.50, and so on. So, he earns L*25.50. But, he spends 3.50. So the equation is:

$L*25.50-3.50$

3 0

3 years ago

If x=10, write an expression in terms of x for the number 5,364

LenKa [72]

Answer:

(5,354 + x)

or

536.4*x

Step-by-step explanation:

We know that x = 10.

Now we want to write an expression (in terms of x) for the number 5,364.

This could be really trivial, remember that x = 10.

Then: (x - 10) = 0

And if we add zero to a number, the result is the same number, then if we add this to 5,364 the number does not change.

5,364 = 5,364 + (x - 10) = 5,364 + x - 10

5,364 = 5,354 + x

So (5,354 + x) is a expression for the number 5,364 in terms of x.

Of course, this is a really simple example, we could do a more complex case if we know that:

x/10 = 1

And the product between any real number and 1 is the same number.

Then:

(5,364)*(x/10) = 5,364

(5,364/10)*x = 5,364

536.4*x = 5,364

So we just found another expression for the number 5,364 in terms of x.

5 0

3 years ago

What are the solution(s) of the quadratic equation 98 – x2 = 0

DanielleElmas [232]

To solve first move the 98 to the other side by subtracting it from both sides. Then divide by - 1 on both sides to that the x^2 is no longer a negative. You are left with:

$x^{2} = 98$

Now square root both sides to get

$x = +/- \sqrt{98}$

The two factors that make up 98 are 49 and 2. 49 can be square rooted. So we are left with these answers:

$x = - 7 \sqrt{2}$ and $x = 7 \sqrt{2}$

4 0

3 years ago

Read 2 more answers

PLEASE HELP MEEEE HURRRY!!! :)

sammy [17]

Answer:

Option D

Step-by-step explanation:

We are given the following equations -

$\begin{bmatrix}-5x-12y-43z=-136\\ -4x-14y-52z=-146\\ 21x+72y+267z=756\end{bmatrix}$

It would be best to solve this equation in matrix form. Write down the coefficients of each terms, and reduce to " row echelon form " -

$\begin{bmatrix}-5&-12&-43&-136\\ -4&-14&-52&-146\\ 21&72&267&756\end{bmatrix}$ First, I swapped the first and third rows.

$\begin{bmatrix}21&72&267&756\\ -4&-14&-52&-146\\ -5&-12&-43&-136\end{bmatrix}$ Leading coefficient of row 2 canceled.

$\begin{bmatrix}21&72&267&756\\ 0&-\frac{2}{7}&-\frac{8}{7}&-2\\ -5&-12&-43&-136\end{bmatrix}$ The start value of row 3 was canceled.

$\begin{bmatrix}21&72&267&756\\ 0&-\frac{2}{7}&-\frac{8}{7}&-2\\ 0&\frac{36}{7}&\frac{144}{7}&44\end{bmatrix}$ Matrix rows 2 and 3 were swapped.

$\begin{bmatrix}21&72&267&756\\ 0&\frac{36}{7}&\frac{144}{7}&44\\ 0&-\frac{2}{7}&-\frac{8}{7}&-2\end{bmatrix}$ Leading coefficient in row 3 was canceled.

$\begin{bmatrix}21&72&267&756\\ 0&\frac{36}{7}&\frac{144}{7}&44\\ 0&0&0&\frac{4}{9}\end{bmatrix}$

And at this point, I came to the conclusion that this system of equations had no solutions, considering it reduced to this -

$\begin{bmatrix}1&0&-1&0\\ 0&1&4&0\\ 0&0&0&1\end{bmatrix}$

The positioning of the zeros indicated that there was no solution!

Hope that helps!

6 0

3 years ago