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3 years ago

What is 14 + (8y + 2) = Y=5

Mathematics

2 answers:

Irina18 [472]3 years ago

8 0

Answer:

not sure but i think 64 ?

ch4aika [34]3 years ago

4 0

14+((8x5)+2)=14+42=56

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-4x+5+9x-2= How do you simplify this

Black_prince [1.1K]

-4x+9x=5x

5-2=3

answer is 5x+3

I hope I am right

6 0

3 years ago

Consider the partial set of values for the functions f(x) and g(x): what is the value of f/g(1)

Alborosie

9514 1404 393

Answer:

-4

Step-by-step explanation:

(f/g)(1) = f(1)/g(1) = 4/-1 = -4

Read f(1) and g(1) from the table. They are listed in the column with x=1.

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3 years ago

What are the coordinates of the image of the given point B (-10,-6) under a dilation with the center at the original and

Elodia [21]

Answer:

D. -5, -3

Step-by-step explanation:

-10 / 1/2 = -5

-6 / 1/2 = -3

also / means divided

if you wouldn't mind giving me brainliest, i'm just trying to rank up.

5 0

3 years ago

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If there are 44 countries in Europe, out of 195 countries in the world, what percent of countries are in Europe?

frutty [35]

Answer:

There are 44 countries in Europe today, according to the United Nations. The full list is shown in the table below, with current population and subregion (based on the United Nations official statistics).

6 0

3 years ago

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<img src="https://tex.z-dn.net/?f=%20%20%5Crm%5Csum%20%5Climits_%7Bn%20%3D%200%7D%5E%7B%20%5Cinfty%20%7D%20%20%5Carcsin%20%5Clar

sineoko [7]

Recall that over an appropriate domain,

$\arcsin(x) \pm \arcsin(y) = \arcsin\left(x \sqrt{1-y^2} \pm y \sqrt{1-y^2}\right)$

Let $x=\frac1{n+1}$ and $y=\frac1{n+2}$ (these belong to the "appropriate domain", so the identity holds). We have

$\dfrac{\sqrt{n+3}}{(n+2)\sqrt{n+1}} = \dfrac1{n+1} \sqrt{\dfrac{(n+3)(n+1)}{(n+2)^2}} = \dfrac1{n+1} \sqrt{1 - \dfrac1{(n+2)^2}} = x \sqrt{1-y^2}$

and

$\dfrac{\sqrt n}{(n+1)\sqrt{n+2}} = \dfrac1{n+2} \sqrt{\dfrac{n(n+2)}{(n+1)^2}} = \dfrac1{n+2} \sqrt{1 - \dfrac1{(n+1)^2}} = y \sqrt{1-x^2}$

Then the sum telescopes, as

$\displaystyle \sum_{n=0}^\infty \arcsin\left(x \sqrt{1-y^2} - y \sqrt{1-x^2}\right) = \sum_{n=0}^\infty \left( \arcsin(x) - \arcsin(y) \right) \\\\ = \left(\arcsin(1) - \arcsin\left(\frac12\right)\right) + \left(\arcsin\left(\frac12\right) - \arcsin\left(\frac13\right)\right) + \cdots \\\\ = \arcsin(1) = \boxed{\frac\pi2}$

3 0

2 years ago