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3 years ago

The graphs of the linear functions f and g are shown above. If h(x)=f(x)+g(x), then h′(x)=

Mathematics

1 answer:

SpyIntel [72]3 years ago

7 0

Based on the graph,

f(x) = 1/2x+3

g(x) = -x+4

so, h(x) = -1/2 x +7

Since h'(x) is the derivative or slope of h(x) and h is a line,

h'(x) = -1/2, Answer C.

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Luis went to the bank he deposited $20 into his savings account and withdrew $40 from his checking account. Write integers to de

aliina [53]

Answer:

Savings Account: x+20

Checking Account:x-40

7 0

3 years ago

Modeling Radioactive Decay In Exercise, complete the table for each radioactive isotope.

Travka [436]

Answer :

The amount after 1000 years will be, 5.19 grams.

The amount after 10000 years will be, 0.105 grams.

Step-by-step explanation :

Half-life = 1599 years

First we have to calculate the rate constant, we use the formula :

$k=\frac{0.693}{t_{1/2}}$

$k=\frac{0.693}{1599\text{ years}}$

$k=4.33\times 10^{-4}\text{ years}^{-1}$

Now we have to calculate the amount after 1000 years.

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = $4.33\times 10^{-4}\text{ years}^{-1}$

t = time passed by the sample = 1000 years

a = initial amount of the reactant = 8 g

a - x = amount left after decay process = ?

Now put all the given values in above equation, we get

$1000=\frac{2.303}{4.33\times 10^{-4}}\log\frac{8}{a-x}$

$a-x=5.19g$

Thus, the amount after 1000 years will be, 5.19 grams.

Now we have to calculate the amount after 10000 years.

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = $4.33\times 10^{-4}\text{ years}^{-1}$

t = time passed by the sample = 10000 years

a = initial amount of the reactant = 8 g

a - x = amount left after decay process = ?

Now put all the given values in above equation, we get

$10000=\frac{2.303}{4.33\times 10^{-4}}\log\frac{8}{a-x}$

$a-x=0.105g$

Thus, the amount after 10000 years will be, 0.105 grams.

4 0

3 years ago

Solve using Fourier series.

Olin [163]

With $2L=\pi$ , the Fourier series expansion of $f(x)$ is

$\displaystyle f(x)\sim\frac{a_0}2+\sum_{n\ge1}a_n\cos\dfrac{n\pi x}L+\sum_{n\ge1}b_n\sin\dfrac{n\pi x}L$
$\displaystyle f(x)\sim\frac{a_0}2+\sum_{n\ge1}a_n\cos2nx+\sum_{n\ge1}b_n\sin2nx$

where the coefficients are obtained by computing

$\displaystyle a_0=\frac1L\int_0^{2L}f(x)\,\mathrm dx$
$\displaystyle a_0=\frac2\pi\int_0^\pi f(x)\,\mathrm dx$

$\displaystyle a_n=\frac1L\int_0^{2L}f(x)\cos\dfrac{n\pi x}L\,\mathrm dx$
$\displaystyle a_n=\frac2\pi\int_0^\pi f(x)\cos2nx\,\mathrm dx$

$\displaystyle b_n=\frac1L\int_0^{2L}f(x)\sin\dfrac{n\pi x}L\,\mathrm dx$
$\displaystyle b_n=\frac2\pi\int_0^\pi f(x)\sin2nx\,\mathrm dx$

You should end up with

$a_0=0$
$a_n=0$
(both due to the fact that $f(x)$ is odd)
$b_n=\dfrac1{3n}\left(2-\cos\dfrac{2n\pi}3-\cos\dfrac{4n\pi}3\right)$

Now the problem is that this expansion does not match the given one. As a matter of fact, since $f(x)$ is odd, there is no cosine series. So I'm starting to think this question is missing some initial details.

One possibility is that you're actually supposed to use the even extension of $f(x)$ , which is to say we're actually considering the function

$\varphi(x)=\begin{cases}\frac\pi3&\text{for }|x|\le\frac\pi3\\0&\text{for }\frac\pi3$

and enforcing a period of $2L=2\pi$ . Now, you should find that

$\varphi(x)\sim\dfrac2{\sqrt3}\left(\cos x-\dfrac{\cos5x}5+\dfrac{\cos7x}7-\dfrac{\cos11x}{11}+\cdots\right)$

The value of the sum can then be verified by choosing $x=0$ , which gives

$\varphi(0)=\dfrac\pi3=\dfrac2{\sqrt3}\left(1-\dfrac15+\dfrac17-\dfrac1{11}+\cdots\right)$
$\implies\dfrac\pi{2\sqrt3}=1-\dfrac15+\dfrac17-\dfrac1{11}+\cdots$

as required.

5 0

3 years ago

An investor invested a total of $2,800 in two mutual funds. One fund earned 5% profit while the other earned 3% profit. If the i

Natasha_Volkova [10]

Answer:

The amount invested in the mutual fund that earned 5% was $1,000

The amount invested in the mutual fund that earned 3% was $1,800

Step-by-step explanation:

Let

x ----> the amount invested in the mutual fund that earned 5%

y ----> the amount invested in the mutual fund that earned 3%

we know that