Suppose n(U) = 150, n(A) = 35, and n(B) = 89. (a) If n(A U B) = 90, find n(A n B).

Three people invest in a treasure dive, each investing the amount listed below. The dive results in 40 gold coins. Apportion tho

garri49 [273]

Answer:

Alice: 20 coins, Ben: 16 coins, and Carlos: 4 coins.

Step-by-step explanation:

We know that the dive results in 40 gold coins.

We know that:

Alice: $7,600 Ben: $5,900 Carlos: $1,400.

So they have $ 14900 in total.

Now we will divide that money by 40.

$x=\frac{14900}{40}\\\\x=372.5$

So 1 gold coin costs $ 372.5. Now we will calculate how much they can get for the money they have.

$A=\frac{7600}{372.5}=20.4\\\\B=\frac{5900}{372.5}=15.8\\\\C=\frac{1400}{372.5}=3.75\\$

We conclude that:

Alice: 20 coins, Ben: 16 coins, and Carlos: 4 coins.

6 0

3 years ago

Divide the following polynomial expression. (n^4 +5n^3+9n^2) / 9n^3

Serhud [2]

Dhjdjdjdjdjdjndjdjdjdjdjdjjsjdjdjdjd

6 0

3 years ago

Pls help...also ill mark brainlest

pentagon [3]

Answer:

D

Step-by-step explanation:

I think it is D

4 0

2 years ago

Parallelogram PQRS has diagonals that are perpendicular. Based on that, which statement is most correct?

viva [34]

Answer:

B- Parallelogram PQRS is also a rhombus.

Step-by-step explanation:

Given

Parallelogram PQRS with perpendicular diagonals

Required

Which of the option is true

(a) PQRS can be a rectangle

A rectangle do not have perpendicular diagonals.

Hence, (a) is false

If (a) is false, then (d) is also false

(b) PQRS can a rhombus

The diagonals of a rhombus are not perpendicular.

So (b) is true

No need to check for (c), since only option is true

4 0

3 years ago

Read 2 more answers

PLZ HELP ME ☻ <img src="https://tex.z-dn.net/?f=%5C%5B%5Cfrac%7Bxy%7D%7Bx%20%2B%20y%7D%20%3D%201%2C%20%5Cquad%20%5Cfrac%7Bxz%7D%

Yanka [14]

Answer:

$x=\frac{12}{7} \\y=\frac{12}{5} \\z=-12$

Step-by-step explanation:

Let's re-write the equations in order to get the variables as separated in independent terms as possible \:

First equation:

$\frac{xy}{x+y} =1\\xy=x+y\\1=\frac{x+y}{xy} \\1=\frac{1}{y} +\frac{1}{x}$

Second equation:

$\frac{xz}{x+z} =2\\xz=2\,(x+z)\\\frac{1}{2} =\frac{x+z}{xz} \\\frac{1}{2} =\frac{1}{z} +\frac{1}{x}$

Third equation:

$\frac{yz}{y+z} =3\\yz=3\,(y+z)\\\frac{1}{3} =\frac{y+z}{yz} \\\frac{1}{3}=\frac{1}{z} +\frac{1}{y}$

Now let's subtract term by term the reduced equation 3 from the reduced equation 1 in order to eliminate the term that contains "y":

$1=\frac{1}{y} +\frac{1}{x} \\-\\\frac{1}{3} =\frac{1}{z} +\frac{1}{y}\\\frac{2}{3} =\frac{1}{x} -\frac{1}{z}$

Combine this last expression term by term with the reduced equation 2, and solve for "x" :

$\frac{2}{3} =\frac{1}{x} -\frac{1}{z} \\+\\\frac{1}{2} =\frac{1}{z} +\frac{1}{x} \\ \\\frac{7}{6} =\frac{2}{x}\\ \\x=\frac{12}{7}$

Now we use this value for "x" back in equation 1 to solve for "y":

$1=\frac{1}{y} +\frac{1}{x} \\1=\frac{1}{y} +\frac{7}{12}\\1-\frac{7}{12}=\frac{1}{y} \\ \\\frac{1}{y} =\frac{5}{12} \\y=\frac{12}{5}$

And finally we solve for the third unknown "z":

$\frac{1}{2} =\frac{1}{z} +\frac{1}{x} \\\\\frac{1}{2} =\frac{1}{z} +\frac{7}{12} \\\\\frac{1}{z} =\frac{1}{2}-\frac{7}{12} \\\\\frac{1}{z} =-\frac{1}{12}\\z=-12$

8 0

3 years ago