Please help me with both of these.

The moon has a mass of 7.35x1022, while earth has a mass of 5.97x1024 Kg. What is the combined mass of the moon and earth.

Veseljchak [2.6K]

Answer:

6.0435*10²⁴

Step-by-step explanation:

Moon: 7.35*10²²

Earth: 5.97*10²⁴

7.35*10²² = 0.0735*10²⁴

then:

0.0735*10²⁴ + 5.97*10²⁴ = (0.0735 + 5.97)*10²⁴

= 6.0435 *10²⁴

5 0

3 years ago

Factor the algebraic expression below in terms of a single trigonometric function. sin 2x + sin x - 2

Pie

Answer:

The factored form is (sin x +2)(sin x-1)

Step-by-step explanation:

We have been given the trigonometric function $\sin^2 x +\sinx-2$

We can factor this by AC method. In AC method we multiply the term a and c and then write the middle term b in such a way that the sum/difference is equal to the product 'ac'

Using the method, we can write sinx as 2sinx -sinx

$\sin^2 x +2\sinx-\sin x-2$

Now, we group the first two terms and the last two terms

$(\sin^2 x +2\sinx)+(-\sin x-2)$

Now, we take GCF from each group

$\sin x(\sin x +2)-1(\sin x+2)$

Factor out (sinx+2)

$(\sin x +2)(\sin x-1)$

Therefore, the factored form is (sin x +2)(sin x-1)

7 0

3 years ago

Read 2 more answers

What number is equalivent the to 14/5?

Airida [17]

2.8 because 14/5 equals 2 80/100.

5 0

2 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