Ask question

Ask question

All categories

3 years ago

12

A senior paid $3.47, $9.50 and $2.50 for lunch during a basketball tournament. What was the average amount he paid over three da

ys?

1 answer:

cestrela7 [59]3 years ago

4 0

Answer:

The average amount he paid over 3 days is $15.47.

Step-by-step explanation:

You might be interested in

HELP PLEASE HELP WILL GIVE POINTS PLEASE!!!!!!!!!

sdas [7]

Answer: positive association and linear association

Step-by-step explanation:

4 0

3 years ago

(x^2-2x-4)(x^2-3x-5)

Montano1993 [528]

Answer:

−x4+5x3−3x2−14x+40

Step-by-step explanation:

7 0

3 years ago

ra1l [238]

Answer:

True

Step-by-step explanation:

$x^{2} -6x=9$

subtract $9$ from both sides

$x^{2} -6x-9$

<em>plz mark me brainliest.</em> ;)

5 0

3 years ago

Which rectangular prism has a surface area of 62 square units? [Note: Art is not drawn to scale.]

Murljashka [212]

A rectangular prism with a length of 6, width of 2, and height of 4.

3 0

3 years ago

Read 2 more answers

points C,D, and E are collinear on CE, and CD:DE = 3/5. C is located at (1,8), D is located at (4,5), and E is located at (x,y).

kap26 [50]

Answer:

The point E is located at (9,0)

x=9, y=0

Step-by-step explanation:

we have that

Points C,D, and E are collinear on CE

Point D is between point C and point E

we know that

$CE=CD+DE$ -----> equation A (by addition segment postulate)

$\frac{CD}{DE}=\frac{3}{5}$

$CD=\frac{3}{5}DE$ ------> equation B

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

<em>Find the distance CD </em>

we have

C(1,8), D(4,5)

substitute in the formula

$CD=\sqrt{(5-8)^{2}+(4-1)^{2}}$

$CD=\sqrt{(-3)^{2}+(3)^{2}}$

$CD=\sqrt{18}\ units$

<em>Find the distance DE</em>

substitute the value of CD in the equation B and solve for DE

$\sqrt{18}=\frac{3}{5}DE$

$DE=\frac{5\sqrt{18}}{3}\ units$

<em>Find the distance CE</em>

$CE=CD+DE$

we have

$DE=\frac{5\sqrt{18}}{3}\ units$

$CD=\sqrt{18}\ units$

substitute the values in the equation A

$CE=\sqrt{18}+\frac{5\sqrt{18}}{3}$

$CE=\frac{8\sqrt{18}}{3}$

<em>Applying the formula of distance CE</em>

we have

$CE=\frac{8\sqrt{18}}{3}$

C(1,8), E(x,y)

substitute in the formula of distance

$\frac{8\sqrt{18}}{3}=\sqrt{(y-8)^{2}+(x-1)^{2}}$

squared both sides

$128=(y-8)^{2}+(x-1)^{2}$ -----> equation C

<em>Applying the formula of distance DE</em>

we have

$DE=\frac{5\sqrt{18}}{3}\ units$

D(4,5), E(x,y)

substitute in the formula of distance

$\frac{5\sqrt{18}}{3}=\sqrt{(y-5)^{2}+(x-4)^{2}}$

squared both sides

$50=(y-5)^{2}+(x-4)^{2}$ -----> equation D

we have the system

$128=(y-8)^{2}+(x-1)^{2}$ -----> equation C

$50=(y-5)^{2}+(x-4)^{2}$ -----> equation D

Solve the system by graphing

The intersection point both graphs is the solution of the system

The solution is the point (9,0)

therefore

The point E is located at (9,0)

see the attached figure to better understand the problem

6 0

3 years ago