What is the solution to this system?

Let
A (3, 1)
B (0, 4)
C(3, 7)
D (6, 4)

step 1
find the distance AB
d=√[(y2-y1)²+(x2-x1)²]------> dAB=√[(4-1)²+(0-3)²]-----> dAB=√18 cm

step 2
find the distance CD
d=√[(y2-y1)²+(x2-x1)²]------> dCD=√[(4-7)²+(6-3)²]-----> dCD=√18 cm

step 3
find the distance AD
d=√[(y2-y1)²+(x2-x1)²]------> dAD=√[(4-1)²+(6-3)²]-----> dAD=√18 cm

step 4
find the distance BC
d=√[(y2-y1)²+(x2-x1)²]------> dBC=√[(7-4)²+(3-0)²]-----> dBC=√18 cm

step 5
find slope AB and CD
m=(y2-y1)/(x2-x1)
mAB=-1
mCD=-1
AB and CD are parallel and AB=CD

step 6
find slope AD and BC
m=(y2-y1)/(x2-x1)
mAD=1
mBC=1
AD and BC are parallel and AD=BC
and
AB and AD are perpendicular
BC and CD are perpendicular

therefore
the shape is a square wit length side √18 cm

area of a square=b²
b is the length side of a square
area of a square=(√18)²------> 18 cm²

the answer is
18 cm²

see the attached figure

6 0

3 years ago

assume that Congress has approved a new tax to fund scientific research to find alternative fuel sources. everyone will pay $200

Tresset [83]

The answer would be regressive.

4 0

3 years ago

The equation y = \large 1\frac{1}{2}x represents the number of cups of dried fruit, y, needed to make x pounds of granola. Deter

Natasha2012 [34]

Answer:

$(1\frac{1}{2},1)$ - False

$(4,6)$ - True

$(18,12)$ -- False

$(0,0)$ -- True

$(2\frac{1}{2},3\frac{3}{4})$ -- True

Step-by-step explanation:

The points are

$(1\frac{1}{2},1)$ , $(4,6)$ , $(18,12)$ , $(0,0)$ and $(2\frac{1}{2},3\frac{3}{4})$ ---- missing from the question

Given

$y = 1\frac{1}{2}x$

Required

Determine if each of the points would be on $y = 1\frac{1}{2}x$

To do this, we simply substitute the value of x and of each point in $y = 1\frac{1}{2}x$ .

(a) $(1\frac{1}{2},1)$

In this case;

$x = 1\frac{1}{2}$ and $y = 1$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 1\frac{1}{2}$

$y = \frac{3}{2} * \frac{3}{2}$

$y = \frac{9}{4}$

$y = 2\frac{1}{4}$

The point $(1\frac{1}{2},1)$ won't be on the graph because the corresponding value of y for $x = 1\frac{1}{2}$ is $y = 2\frac{1}{4}$

(b) $(4,6)$

In this case;

$x = 4$

$y = 6$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 4$

$y = \frac{3}{2} * 4$

$y = \frac{3* 4}{2}$

$y = \frac{12}{2}$

$y = 6$

The point $(4,6)$ would be on the graph because the corresponding value of y for $x = 4$ is $y = 6$

(c) $(18,12)$

In this case:

$x = 18;y = 12$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 18$

$y = \frac{3}{2} * 18$

$y = \frac{3* 18}{2}$

$y = \frac{54}{2}$

$y = 27$

The point $(18,12)$ wouldn't be on the graph because the corresponding value of y for $x = 18$ is $y = 12$

(d) $(0,0)$

In this case;

$x =0; y = 0$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 0$

$y = 0$

The point $(0,0)$ would be on the graph because the corresponding value of y for $x = 0$ is $y = 0$

(e) $(2\frac{1}{2},3\frac{3}{4})$

In this case:

$x = 2\frac{1}{2}$ ; $y = 3\frac{3}{4}$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 2\frac{1}{2}$

$y = \frac{3}{2} * \frac{5}{2}$

$y = \frac{15}{4}$

$y = 3\frac{3}{4}$

The point $(2\frac{1}{2},3\frac{3}{4})$ would be on the graph because the corresponding value of y for $x = 2\frac{1}{2}$ is $y = 3\frac{3}{4}$

3 0

2 years ago