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2 years ago

6

Katie and jennifer are playing a game. katie and jennifer each started with 100 points. at the end of each turn katies points do

ubled. at the end of each turn jennifers points increased by 200. at the end of which turn will katie have more points than jennifer

1 answer:

ella [17]2 years ago

3 0

It is the fourth round and this is because in round one Katie has 200 and Jenny will have 300.In round two katie will have 400 and Jennifer will have 500.In round 3 Katie will have 600 and and jennifer will have 800.I

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The bricklayer bought 7500 bricks to brick a wall. He calculated that every horizontal layer of bricks will contain 321 bricks.

statuscvo [17]

There will be 23 layers of brick, because 7500/321 = 23 with 117 leftover.

6 0

3 years ago

I need help in these questions

Salsk061 [2.6K]

Answer:

see explanation

Step-by-step explanation:

All of these questions use the external angle theorem, that is

The external angle of a triangle is equal to the sum of the 2 opposite interior angles.

18

∠3 = 43° + 22° = 65°

19

∠2 + 71 = 92 ( subtract 71 from both sides )

∠2 = 21°

20

90 + ∠4 = 123 ( subtract 90 from both sides )

∠4 = 33°

21

2x - 15 + x - 5 = 148

3x - 20 = 148 ( add 20 to both sides )

3x = 168 ( divide both sides by 3 )

x = 56

Hence ∠ABC = x - 5 = 56 - 5 = 51°

22

2x + 27 + 2x - 11 = 100

4x + 16 = 100 ( subtract 16 from both sides )

4x = 84 ( divide both sides by 4 )

x = 21

Hence ∠JKL = 2x - 11 = (2 × 21) - 11 = 42 - 11 = 31°

3 0

3 years ago

A sheet of paper is 8.5 in X 11 inches if you cut 2 in squared from each corner what is the area if the paper

IceJOKER [234]

A=length x height. First we want to know the area of the original sheet of paper. 8.5 x 11= 93.5 inches squared. Then we take two squares with side length 2 inches. The area of each of these squares is 2 x 2, or 4 inches. Since there are two, the total area we take away is 8 inches. We subtract 8 inches from 93.5 since we took it away, and the final area of the paper is 85.5 inches^{2}

Hope this helped!

3 0

2 years ago

Write an equation of a line that has a slope of –5 and a y-intercept of 2.

Bingel [31]

Answer:

y-2 = -5x

5x+y = 2 is the required equation of a line.

4 0

2 years ago

<img src="https://tex.z-dn.net/?f=%5Cleft%20%5C%7B%20%7B%7Bx%2By%3D1%7D%20%5Catop%20%7Bx-2y%3D4%7D%7D%20%5Cright.%20%5C%5C%5Clef

brilliants [131]

Answer:

<em>(a) x=2, y=-1</em>

<em>(b) x=2, y=2</em>

<em>(c)</em> $\displaystyle x=\frac{5}{2}, y=\frac{5}{4}$

<em>(d) x=-2, y=-7</em>

Step-by-step explanation:

<u>Cramer's Rule</u>

It's a predetermined sequence of steps to solve a system of equations. It's a preferred technique to be implemented in automatic digital solutions because it's easy to structure and generalize.

It uses the concept of determinants, as explained below. Suppose we have a 2x2 system of equations like:

$\displaystyle \left \{ {{ax+by=p} \atop {cx+dy=q}} \right.$

We call the determinant of the system

$\Delta=\begin{vmatrix}a &b \\c &d \end{vmatrix}$

We also define:

$\Delta_x=\begin{vmatrix}p &b \\q &d \end{vmatrix}$

And

$\Delta_y=\begin{vmatrix}a &p \\c &q \end{vmatrix}$

The solution for x and y is

$\displaystyle x=\frac{\Delta_x}{\Delta}$

$\displaystyle y=\frac{\Delta_y}{\Delta}$

(a) The system to solve is

$\displaystyle \left \{ {{x+y=1} \atop {x-2y=4}} \right.$

Calculating:

$\Delta=\begin{vmatrix}1 &1 \\1 &-2 \end{vmatrix}=-2-1=-3$

$\Delta_x=\begin{vmatrix}1 &1 \\4 &-2 \end{vmatrix}=-2-4=-6$

$\Delta_y=\begin{vmatrix}1 &1 \\1 &4 \end{vmatrix}=4-3=3$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-6}{-3}=2$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{3}{-3}=-1$

The solution is x=2, y=-1

(b) The system to solve is

$\displaystyle \left \{ {{4x-y=6} \atop {x-y=0}} \right.$

Calculating:

$\Delta=\begin{vmatrix}4 &-1 \\1 &-1 \end{vmatrix}=-4+1=-3$

$\Delta_x=\begin{vmatrix}6 &-1 \\0 &-1 \end{vmatrix}=-6-0=-6$

$\Delta_y=\begin{vmatrix}4 &6 \\1 &0 \end{vmatrix}=0-6=-6$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-6}{-3}=2$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{-6}{-3}=2$

The solution is x=2, y=2

(c) The system to solve is

$\displaystyle \left \{ {{-x+2y=0} \atop {x+2y=5}} \right.$

Calculating:

$\Delta=\begin{vmatrix}-1 &2 \\1 &2 \end{vmatrix}=-2-2=-4$

$\Delta_x=\begin{vmatrix}0 &2 \\5 &2 \end{vmatrix}=0-10=-10$

$\Delta_y=\begin{vmatrix}-1 &0 \\1 &5 \end{vmatrix}=-5-0=-5$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-10}{-4}=\frac{5}{2}$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{-5}{-4}=\frac{5}{4}$

The solution is

$\displaystyle x=\frac{5}{2}, y=\frac{5}{4}$

(d) The system to solve is

$\displaystyle \left \{ {{6x-y=-5} \atop {4x-2y=6}} \right.$

Calculating:

$\Delta=\begin{vmatrix}6 &-1 \\4 &-2 \end{vmatrix}=-12+4=-8$

$\Delta_x=\begin{vmatrix}-5 &-1 \\6 &-2 \end{vmatrix}=10+6=16$

$\Delta_y=\begin{vmatrix}6 &-5 \\4 &6 \end{vmatrix}=36+20=56$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{16}{-8}=-2$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{56}{-8}=-7$

The solution is x=-2, y=-7

4 0

3 years ago