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3 years ago

Write the equation of the line that passes through the points (4,7) and (8,6). Put

Mathematics

1 answer:

____ [38]3 years ago

8 0

Y=-1/4x+8

First find the slope

Slope= y-y1/x-x1

So (6-7)/(8-4) = -1/4

Now take that slope and use one of the points and plug them into the point slope formula to solve for your y intercept.

Point slope formula: y=mx+b

So 7=-1/4(4)+b OR 6=-1/4(8)+b

Either way you’ll get b=8!

Put em together and you get: y=-1/4x+8

Ps you can double check your answer by plugging in one or both of the points into your equation to see if it works :)

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Adult humans have approximately 2.6 red blood cells and 3.92 white blood cells. About how many times greater is the number of re

kifflom [539]

Answer:

1.5 rounded to the nearest tenth

Step-by-step explanation:

To find how much greater WBC are from RBC we do,

3.92 ÷ 2.6

= 1.50769230769

Which is 1.5 rounded to the nearest tenth.

Thus,

there are approximately 1.5 times WBC than RBC.

Hope this helps :)

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3 years ago

How to do this and what's the answer

Darya [45]

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3 years ago

Find the value of x.

enyata [817]

The value of x is 13

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3 years ago

15. Oscar wants to rent a boat. The rental shop charges and initial fee of $12 plus $8 per hour to rent the boat. Write a equati

lbvjy [14]

C=8h+12
the 8h represents the $8 per hour for renting the boat, and h represents however many hours you rent the boat for. The +12 is the initial $12 fee. This is in slope intercept form, just using different variables than y=mx+b

5 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