What is the correct expression to show the cost after d days?

A football stadium sells regular and box seating. There are twelve times as many

Vera_Pavlovna [14]

Answer:

There are 801 box seats and 9612 regular seats.

explanation:

Let the number of box seats be x,

Then the regular seats is 12x

The sum of seats equals to 10,413

Solve:

12x + x ➙ 10,413

13x ➙ 10413

x ➙ 801

There are 801 box seats.

Find for regular seats:

➙ 12(x)

➙ 12(801)

➙ 9612

There are 9612 regular seats.

6 0

2 years ago

Read 2 more answers

A group of students made tacos for their class lunch. The number of tacos each student made is shown in the table.

nevsk [136]

Answer:

-the ratio of the number of tacos Jackie made to the number of tacos Marta made

-the ratio of the number of tacos Marta made to the number of tacos Manny made

-the ratio of the number of tacos Manny made to the number of tacos Marta made

-the ratio of the number of tacos Vinny and Marta made to the number of tacos Vinny and Evelyn made

Step-by-step explanation:

There are 4 that apply! Jackie and many both made 6 and Marta made 4 so the ratio is 6:4! Vinny and Marty's combined is 6 and Vinny and evelyn's combined is 4 so the ratio is 6:4!

8 0

3 years ago

Find the slope of a line given two or more points (2,4)(5,3)

masha68 [24]

Answer:

-1/3 (can u plz give me the brainliest answer???)

Step-by-step explanation:

change in y over the change in x. (or rise over run).

-1/3

Have a nice day!!!!!!!!!!!

KA

7 0

3 years ago

Read 2 more answers

Prove that the segments joining the midpoint of consecutive sides of an isosceles trapezoid form a rhombus.

sergiy2304 [10]

Answer:

See explanation

Step-by-step explanation:

a) To prove that DEFG is a rhombus, it is sufficient to prove that:

All the sides of the rhombus are congruent: $|DG|\cong |GF| \cong |EF| \cong |DE|$
The diagonals are perpendicular

Using the distance formula; $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$|DG|=\sqrt{(0-(-a-b))^2+(0-c)^2}$

$\implies |DG|=\sqrt{a^2+b^2+c^2+2ab}$

$|GF|=\sqrt{((a+b)-0)^2+(c-0)^2}$

$\implies |GF|=\sqrt{a^2+b^2+c^2+2ab}$

$|EF|=\sqrt{((a+b)-0)^2+(c-2c)^2}$

$\implies |EF|=\sqrt{a^2+b^2+c^2+2ab}$

$|DE|=\sqrt{(0-(-a-b))^2+(2c-c)^2}$

$\implies |DE|=\sqrt{a^2+b^2+c^2+2ab}$

Using the slope formula; $m=\frac{y_2-y_1}{x_2-x_1}$

The slope of EG is $m_{EG}=\frac{2c-0}{0-0}$

$\implies m_{EG}=\frac{2c}{0}$

The slope of EG is undefined hence it is a vertical line.

The slope of DF is $m_{DF}=\frac{c-c}{a+b-(-a-b)}$

$\implies m_{DF}=\frac{0}{2a+2b)}=0$

The slope of DF is zero, hence it is a horizontal line.

A horizontal line meets a vertical line at 90 degrees.

Conclusion:

Since $|DG|\cong |GF| \cong |EF| \cong |DE|$ and $DF \perp FG$ , DEFG is a rhombus

b) Using the slope formula:

The slope of DE is $m_{DE}=\frac{2c-c}{0-(-a-b)}$

$m_{DE}=\frac{c}{a+b)}$

The slope of FG is $m_{FG}=\frac{c-0}{a+b-0}$

$\implies m_{FG}=\frac{c}{a+b}$

5 0

3 years ago

Ben bowled 124 and 188 in his first two games. What must he bowl in his third game to have an average of at least 160?

Svetlanka [38]

Answer:

x ≥ 168

Step-by-step explanation:

(124 + 188 + x) / 3 ≥ 160 => 312 + x ≥ 480 => x ≥ 480 - 312 ≥ 168

x ≥ 168

God with you!!!

3 0

3 years ago