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2 years ago

Of the pets in the pet show, 5/6 are cats. 4/5 of the cats are calico cats. What fraction of the pets are calico cats?

Mathematics

2 answers:

kondaur [170]2 years ago

4 0

Answer:

8=135.2

Step-by-step explanation:

if it help u can follow me

aliya0001 [1]2 years ago

3 0

Answer:8=135.2

Step-by-step explanation:

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PROBLEM. 11.1: Rewrite These! Rewrite each quotient as a sum or a difference.

Jet001 [13]

Answer:

1. (4x-10)/2

4x/2 - 10/2

2x - 5

2. (1-50x)/-2

1/-2 -50x/-2

25x - 1/2

3. 5(x + 10)/25

(5x + 50)/25

5x/25 + 50/25

1/5 x + 2

4.(-0.2x + 5)/2

-0.2x/2 + 5/2

-0.1x + 2.5

3 0

3 years ago

Which situation could be solved using the equation -4+4=0

navik [9.2K]

we are supposed to explain

Which situation could be solved using the equation $-4+4=0$

As we can see in the given equation , one number is -4 and the other number is 4

When we add two number of same value but opposite in sign we get zero.

This equivalent to the real life situation of paying a post paid bills.

Suppose you are using a post paid service , and the bills gets generated at the end of the month.

Suppose the bills amount is $x.

Once you pay that bills , then total outstanding again becomes $0.

Because $-x+x=0$

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