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2 years ago

Simplify..................................

Mathematics

1 answer:

balu736 [363]2 years ago

7 0

Answer:

idk this stuff yet

Step-by-step explanation:

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If h(x)=x-7 and g(x)=x ^2, which expression is equivalent to (g•h)(5)

pashok25 [27]

x^3 ‐ 7x^2

Explanation:

x^2(x-7) = x^3 ‐ 7x^2

7 0

2 years ago

Let P and Q be polynomials with positive coefficients. Consider the limit below. lim x→[infinity] P(x) Q(x) (a) Find the limit i

jenyasd209 [6]

Answer:

If the limit that you want to find is $\lim_{x\to \infty}\dfrac{P(x)}{Q(x)}$ then you can use the following proof.

Step-by-step explanation:

Let $P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}$ and $Q(x)=b_{m}x^{m}+b_{m-1}x^{n-1}+\cdots+b_{1}x+b_{0}$ be the given polinomials. Then

$\dfrac{P(x)}{Q(x)}=\dfrac{x^{n}(a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n})}{x^{m}(b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m})}=x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}$

Observe that

$\lim_{x\to \infty}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\dfrac{a_{n}}{b_{m}}$

and

$\lim_{x\to \infty} x^{n-m}=\begin{cases}0& \text{if}\,\, nm\end{cases}$

Then

$\lim_{x\to \infty}=\lim_{x\to \infty}x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\begin{cases}0 & \text{if}\,\, nm \end{cases}$

3 0

2 years ago

Divide 81 ft in the ratio 5 :4

Mumz [18]

Answer:

Rr = 51 : 68

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Sahil got 28 questions right on the math test. Angelina got 7 more wrong answers than Sahil. There where 40 questions on the tes

Mariana [72]

[A+7= 28] is the correct equation I believe, though I may be wrong.

7 0

2 years ago

Read 2 more answers

How are exponential functions related to logarithmic functions?<br> Model that with an example.

scoundrel [369]

Exponential functions are related to logarithmic functions in that they are inverse functions. Exponential functions move quickly up towards a [y] infinity, bounded by a vertical asymptote (aka limit), whereas logarithmic functions start quick but then taper out towards an [x] infinity, bounded by a horizontal asymptote (aka limit).
If we use the natural logarithm (ln) as an example, the constant "e" is the base of ln, such that:
ln(x) = y, which is really stating that the base (assumed "e" even though not shown), that:
${e}^{y} = x$
if we try to solve for y in this form it's nearly impossible, that's why we stick with ln(x) = y
but to find the inverse of the form:
${e}^{y} = x$
switch the x and y, then solve for y:
${e}^{x} = y$
So the exponential function is the inverse of the logarithmic one, f(x) = ln x

3 0

3 years ago

Simplify.................................. ​

Simplify..................................