What is the value of "c" in the following quadratic? (Make sure the equation is in

The Venn diagram shows the relationship between several sets of numbers.

svet-max [94.6K]

Answer: C

Step-by-step explanation: i’m doing the test too

8 0

2 years ago

I need help ASAP!!Please explain how to solve the problem

rjkz [21]

Answer:

( x - 2 )^2 + ( y - 1 )^2 = 1

Step-by-step explanation:

As i previously explained,

The general form of equation for a circle is ( x - h )^2 + ( y - k )^2 = r^2.

h and k are the co-ordinates of the center of circle and r is the radius

Here, We have (2,1) as the co-ordinates of the center of circle

Now,

( x - h )^2 + ( y - k )^2 = r^2

or,( x - 2 )^2 + ( y - 1 )^2 = r^2

The radius of the circle is 1 units(from figure)

So, put that in, we get

( x - 2 )^2 + ( y - 1 )^2 = 1^2

or, ( x - 2 )^2 + ( y - 1 )^2 = 1

You can simplify this or leave it here.

8 0

2 years ago

What is: 3✖️(-4)➗(-2) -9 ——————— (+6)

Elis [28]

The answer is - 23/6 it should be

4 0

3 years ago

Kate currently has an account balance of $7,194.66. She opened the account 21 years ago with a deposit of $2,978.41. If the inte

jolli1 [7]

Use the attached formula.
exp = [log (total / principal) / n*years]
where "n" is compounding periods per year
exp = [log (7,194.66 / 2,978.41) / (365*21)
exp = log ( 2.4156042989 ) / 7,665
exp = log ( 2.4156042989 ) / 7,665
exp = 0.38302579382 / 7,665
exp = 0.00004997074935681670 

rate = (10^exp -1)* n
rate = (10^0.00004997074935681670 -1) * n
rate = (1.0001150685 -1) * 365
rate = (.0001150685) * 365
rate = 0.0420000025 
rate = 4.20000025 %
rate = 4.2 %

Yes, it's just that easy. LOL

5 0

2 years ago

Read 2 more answers

Suppose you have a light bulb that emits 50 watts of visible light. (Note: This is not the case for a standard 50-watt light bul

natali 33 [55]

Answer:

0.049168726 light-years

Step-by-step explanation:

The apparent brightness of a star is

$\bf B=\displaystyle\frac{L}{4\pi d^2}$

where

L = luminosity of the star (related to the Sun)

d = distance in ly (light-years)

The luminosity of Alpha Centauri A is 1.519 and its distance is 4.37 ly.

Hence the apparent brightness of Alpha Centauri A is

$\bf B=\displaystyle\frac{1.519}{4\pi (4.37)^2}=0.006329728$

According to the inverse square law for light intensity

$\bf \displaystyle\frac{I_1}{I_2}=\displaystyle\frac{d_2^2}{d_1^2}$

where

$\bf I_1=$ light intensity at distance $\bf d_1$

$\bf I_2=$ light intensity at distance $\bf d_2$

Let $\bf d_2$ be the distance we would have to place the 50-watt bulb, then replacing in the formula

$\bf \displaystyle\frac{0.006329728}{50}=\displaystyle\frac{d_2^2}{(4.37)^2}\Rightarrow\\\\\Rightarrow d_2^2=\displaystyle\frac{0.006329728*(4.37)^2}{50}\Rightarrow d_2^2=0.002417564\Rightarrow\\\\\Rightarrow d_2=\sqrt{0.002417564}=\boxed{0.049168726\;ly}$

Remark: It is worth noticing that Alpha Centauri A, though is the nearest star to the Sun, is not visible to the naked eye.

7 0

3 years ago