All categories

3 years ago

Which of these changes in today's world is a product of advances in biology?

1 answer:

5 0

"Increased crop yields" is the one change in today's world among the choices given in the question that <span>is a product of advances in biology. The correct option among all the options that are given in the question is the first option. I hope that this is the answer that has come to your desired help.</span>

You might be interested in

Which one is distributive property

faust18 [17]

Answer:

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

Help please..................

Ivahew [28]

(f+g)(x) = f(x) + g(x)
(f+g)(x) = [ f(x) ] + [ g(x) ]
(f+g)(x) = [ 3x-2 ] + [ 2x+1 ]
(f+g)(x) = (3x+2x) + (-2+1)
(f+g)(x) = 5x - 1

Answer is choice B

8 0

3 years ago

How many nonzero terms of the Maclaurin series for ln(1 x) do you need to use to estimate ln(1.4) to within 0.001?

Vilka [71]

Answer:

The estimate of In(1.4) is the first five non-zero terms.

Step-by-step explanation:

From the given information:

We are to find the estimate of In(1 . 4) within 0.001 by applying the function of the Maclaurin series for f(x) = In (1 + x)

So, by the application of Maclurin Series which can be expressed as:

$f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2 f"(0)}{2!}+ \dfrac{x^3f'(0)}{3!}+... \ \ \ \ \ --- (1)$

Let examine f(x) = In(1+x), then find its derivatives;

f(x) = In(1+x)

$f'(x) = \dfrac{1}{1+x}$

f'(0) $= \dfrac{1}{1+0}=1$

f ' ' (x) $= \dfrac{1}{(1+x)^2}$

f ' ' (x) $= \dfrac{1}{(1+0)^2}=-1$

f ' ' '(x) $= \dfrac{2}{(1+x)^3}$

f ' ' '(x) $= \dfrac{2}{(1+0)^3} = 2$

f ' ' ' '(x) $= \dfrac{6}{(1+x)^4}$

f ' ' ' '(x) $= \dfrac{6}{(1+0)^4}=-6$

f ' ' ' ' ' (x) $= \dfrac{24}{(1+x)^5} = 24$

f ' ' ' ' ' (x) $= \dfrac{24}{(1+0)^5} = 24$

Now, the next process is to substitute the above values back into equation (1)

$f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2f' \ '(0)}{2!}+\dfrac{x^3f \ '\ '\ '(0)}{3!}+\dfrac{x^4f '\ '\ ' \ ' \(0)}{4!}+\dfrac{x^5f' \ ' \ ' \ ' \ '0)}{5!}+ ...$

$In(1+x) = o + \dfrac{x(1)}{1!}+ \dfrac{x^2(-1)}{2!}+ \dfrac{x^3(2)}{3!}+ \dfrac{x^4(-6)}{4!}+ \dfrac{x^5(24)}{5!}+ ...$

$In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...$

To estimate the value of In(1.4), let's replace x with 0.4

$In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...$

$In (1+0.4) = 0.4 - \dfrac{0.4^2}{2}+\dfrac{0.4^3}{3}-\dfrac{0.4^4}{4}+\dfrac{0.4^5}{5}- \dfrac{0.4^6}{6}+...$

Therefore, from the above calculations, we will realize that the value of $\dfrac{0.4^5}{5}= 0.002048$ as well as $\dfrac{0.4^6}{6}= 0.00068267$ which are less than 0.001

Hence, the estimate of In(1.4) to the term is $\dfrac{0.4^5}{5}$ is said to be enough to justify our claim.

∴

The estimate of In(1.4) is the first five non-zero terms.

8 0

2 years ago

Water flows from a faucet at a constant rate. Assume that 6 gallons of water are already in a tub by the time we notice the fauc

Thepotemich [5.8K]

The equation would be y=3.2x+6

6 0

3 years ago

At the bottom is a completed addition problem, with all the digits replaced by letters. Every letter represents a single digit a

Rashid [163]

Answer:

T = 5 (H = 7, M = 1); T = 7 (H = 4, M = 2); T = 9 (H = 1, M = 3)

Step-by-step explanation:

We know MH+MH+MH=TM

MH is a two digit number, and TM is also a two digit number.

MH+MH+MH=3*(MH)<100

3M<=9, meaning that M could be 1, 2, or 3.

If M=1, MH+MH+MH=3(10+H)=30+3H=TM=10*T+1

30+3H=10T+1, so 3H must include 1 in one's digit.

The only possibility is that 3H = 21, then H = 7

MH+MH+MH=17+17+17=TM=51, T = 5

If M=2, MH+MH+MH=3(20+H)=60+3H=TM=10T+2

60+3H=10T+2, so 3H must include 2 in one's digit.

The only possibility is 3H = 12, then H = 4

MH+MH+MH=24+24+24=72, T = 7

If M=3, MH+MH+MH=3(30+H)=90+3H=TM=10T+3

90+3H=10T+3, so 3H must include 3 in one's digit.

The only possibility is 3H = 3, then H =1

MH+MH+MH=31+31+31=93, T=9

4 0

3 years ago