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2 years ago

If m is 2 times n minus 7 and y is the square of m, then write y as a function of n.

Mathematics

1 answer:

Elenna [48]2 years ago

3 0

Answer:

y = 4n^2

Step-by-step explanation:

Because m = 2n, and y = m^2, then substitute m = 2n into y equation, can get y = (2n)^2 = 4n^2.

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1. Express <img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bx%282x%2B3%29%20%7D" id="TexFormula1" title="\frac{1}{x(2x+3) }" a

katovenus [111]

1. Let a and b be coefficients such that

$\dfrac1{x(2x+3)} = \dfrac ax + \dfrac b{2x+3}$

Combining the fractions on the right gives

$\dfrac1{x(2x+3)} = \dfrac{a(2x+3) + bx}{x(2x+3)}$

$\implies 1 = (2a+b)x + 3a$

$\implies \begin{cases}3a=1 \\ 2a+b=0\end{cases} \implies a=\dfrac13, b = -\dfrac23$

so that

$\dfrac1{x(2x+3)} = \boxed{\dfrac13 \left(\dfrac1x - \dfrac2{2x+3}\right)}$

2. a. The given ODE is separable as

$x(2x+3) \dfrac{dy}dx} = y \implies \dfrac{dy}y = \dfrac{dx}{x(2x+3)}$

Using the result of part (1), integrating both sides gives

$\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + C$

Given that y = 1 when x = 1, we find

$\ln|1| = \dfrac13 \left(\ln|1| - \ln|5|\right) + C \implies C = \dfrac13\ln(5)$

so the particular solution to the ODE is

$\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + \dfrac13\ln(5)$

We can solve this explicitly for y :

$\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3| + \ln(5)\right)$

$\ln|y| = \dfrac13 \ln\left|\dfrac{5x}{2x+3}\right|$

$\ln|y| = \ln\left|\sqrt[3]{\dfrac{5x}{2x+3}}\right|$

$\boxed{y = \sqrt[3]{\dfrac{5x}{2x+3}}}$

2. b. When x = 9, we get

$y = \sqrt[3]{\dfrac{45}{21}} = \sqrt[3]{\dfrac{15}7} \approx \boxed{1.29}$

8 0

2 years ago

V x² + 9 = x-3<br> links will be reported.

brilliants [131]

Answer:

x−12

x^2

Step-by-step explanation:

4 0

2 years ago

I really need help with this pls help me

Agata [3.3K]

I think it is the first one -2

6 0

2 years ago

Read 2 more answers

What is the equivalent to (3x squared)cubed ?

Amiraneli [1.4K]

$(3x^2)^3=(3^1x^2)^3 =3^1^*^3x^2^*^3=3^3x^6=3*3*3*x^6=27x^6$

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3 years ago

What is the simplified form of the quantity y−squared minus y minus 12 over the quantity y−squared minus 2y minus 15?

makvit [3.9K]

(y^2-y-12)/(y^2-2y-15) factor both the numerator and denominator...

(y^2-4y+3y-12)/(y^2-5y+3y-15)

(y(y-4)+3(y-4))/(y(y-5)+3(y-5))

((y-4)(y+3))/((y-5)(y+3)) so the (y+3) terms cancel leaving

(y-4)/(y-5)

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2 years ago