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3 years ago

Consider this expression x^2-(3x+y)+16

Mathematics

1 answer:

Brilliant_brown [7]3 years ago

6 0

Answer:

x^2-3x-y+16?

Step-by-step explanation:

simplest form already, but can be converted into a proper polynomial.

x^2-3x-y+16

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Suppose you have two urns with poker chips in them. Urn I contains two red chips and four white chips. Urn II contains three red

Neporo4naja [7]

Answer:

Multiple answers

Step-by-step explanation:

The original urns have:

Urn 1 = 2 red + 4 white = 6 chips
Urn 2 = 3 red + 1 white = 4 chips

We take one chip from the first urn, so we have:

The probability of take a red one is : $\frac{1}{3}$ (2 red from 6 chips(2/6=1/2))

For a white one is: $\frac{2}{3}$ (4 white from 6 chips(4/6=(2/3))

Then we put this chip into the second urn:

We have two possible cases:

First if the chip we got from the first urn was white. The urn 2 now has 3 red + 2 whites = 5 chips
Second if the chip we got from the first urn was red. The urn two now has 4 red + 1 white = 5 chips

If we select a chip from the urn two:

In the first case the probability of taking a white one is of: $\frac{2}{5}$ = 40% ( 2 whites of 5 chips)
In the second case the probability of taking a white one is of: $\frac{1}{5}$ = 20% ( 1 whites of 5 chips)

This problem is a dependent event because the final result depends of the first chip we got from the urn 1.

For the fist case we multiply :

$\frac{4}{6}$ x $\frac{2}{5}$ = $\frac{4}{15}$ = 26.66% ( $\frac{4}{6}$ the probability of taking a white chip from the urn 1, $\frac{2}{5}$ the probability of taking a white chip from urn two)

For the second case we multiply:

$\frac{1}{3}$ x $\frac{1}{5}$ = $\frac{1}{30}$ = .06% ( $\frac{1}{3}$ the probability of taking a red chip from the urn 1, $\frac{1}{5}$ the probability of taking a white chip from the urn two)

8 0

4 years ago

Somebody help me with this <br><br> 30 points ☺️

brilliants [131]

Answer:

4r- 21

Step-by-step explanation:

4 - r= r and r+ 21/= R21 and Simplified Is -4 and R21

8 0

3 years ago

Geometry Math

Lostsunrise [7]

Answer:

Step 3: The measure of angle DAB is 60°. Reason: Since DAE is a right angle, and EAB is 30 degrees, 90 - 30 is 60. (Just find the reason that is related to this statement).

Step 4: The measure of angle ABC is 130°. Reason: Same-side Interior Angles Theorem

Step-by-step explanation:

6 0

3 years ago

Ken's fitness member ship cost $100 to join plus a $25 monthly fee. Ken also used a coupon to get 10% off the monthly fee. What

GREYUIT [131]

The equation that models the total cost is y = 100 + 22.5m

Step-by-step explanation:

Fitness member ship cost = $100

Monthly fee = $25

Discount = 10% on monthly fee

To calculate new monthly fee after discount,

25 - (25*10%) --> 25 - 2.5 --> $22.5

Number of months = m

Fee in m months = 22.5*m = 22.5m

<u>Equation:</u>

(x) is the monthly fee $22.5

(m) is the number of months

(y) is the total cost

So,

y = mx + c

y = m(22.5) + 100

y = 100 + 22.5m

Therefore, the equation that models the total cost is y = 100 + 22.5m

Keyword: Equation

Learn more about equations at

#LearnwithBrainly

5 0

3 years ago

In a G.P the difference between the 1st and 5th term is 150, and the difference between the

liubo4ka [24]

Answer:

Either $\displaystyle \frac{-1522}{\sqrt{41}}$ (approximately $-238$ ) or $\displaystyle \frac{1522}{\sqrt{41}}$ (approximately $238$ .)

Step-by-step explanation:

Let $a$ denote the first term of this geometric series, and let $r$ denote the common ratio of this geometric series.

The first five terms of this series would be:

$a$ ,
$a\cdot r$ ,
$a \cdot r^2$ ,
$a \cdot r^3$ ,
$a \cdot r^4$ .

First equation:

$a\, r^4 - a = 150$ .

Second equation:

$a\, r^3 - a\, r = 48$ .

Rewrite and simplify the first equation.

$\begin{aligned}& a\, r^4 - a \\ &= a\, \left(r^4 - 1\right)\\ &= a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) \end{aligned}$ .

Therefore, the first equation becomes:

$a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) = 150$ ..

Similarly, rewrite and simplify the second equation:

$\begin{aligned}&a\, r^3 - a\, r\\ &= a\, \left( r^3 - r\right) \\ &= a\, r\, \left(r^2 - 1\right) \end{aligned}$ .

Therefore, the second equation becomes:

$a\, r\, \left(r^2 - 1\right) = 48$ .

Take the quotient between these two equations:

$\begin{aligned}\frac{a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right)}{a\cdot r\, \left(r^2 - 1\right)} = \frac{150}{48}\end{aligned}$ .

Simplify and solve for $r$ :

$\displaystyle \frac{r^2+ 1}{r} = \frac{25}{8}$ .

$8\, r^2 - 25\, r + 8 = 0$ .

Either $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ or $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ .

Assume that $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = -\frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= -\frac{1522\sqrt{41}}{41} \approx -238\end{aligned}$ .

Similarly, assume that $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = \frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= \frac{1522\sqrt{41}}{41} \approx 238\end{aligned}$ .

4 0

3 years ago