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3 years ago

What is the perimeter of this rectangle?

Mathematics

1 answer:

BlackZzzverrR [31]3 years ago

7 0

The length of the rectangle would be X2 - X1
The height of the rectangle would be Y2-Y1

Perimeter = 2 X length + 2 times height

The perimeter would be 2(X2-X1) + 2(Y2-Y1)

The first choice would be the correct answer.

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What is f and g answers

Arte-miy333 [17]

F.
Paul: 5x+20 Monica: 10x+5

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3 years ago

Write 5,090,000 in expanded form

Kitty [74]

Answer: 5,090,000= 5,000,000 + 90,000

Step-by-step explanation:

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3 years ago

WILL GIVE BRAINLIEST IF CORRECT

SCORPION-xisa [38]

9514 1404 393

Answer:

5544 square units

Step-by-step explanation:

Translation does not change the dimensions of the figure. The two figures together tell you the right triangle has a base of 154 units and a height of 72 units.

The area of a triangle is given by ...

A = 1/2bh

A = 1/2(154 units)(72 units) = 5544 square units

The areas of each of the triangles are 5544 square units.

8 0

2 years ago

wrote an equation for the line. passing through (-5,7) and parallel to the line whise equation is 5x-3y-4=0

xeze [42]

Step-by-step explanation:

5x-3y-4=0---------(1)

Slope = - coefficient of X/ coefficient of y

= -5/(-3)

= 5/3

(x1,y1)= (-5,7)

Equation of the line,

Y-Y1= slope(X-X1)

Y-7= (5/3)×(X-(-5))

Y-7= (5x+25)/3

3y-7=5x+25

5x-3y+32=0 is the required eqn

3 0

3 years ago

What value of b will cause the system to have an infinite number of solutions?

irga5000 [103]

b must be equal to -6 for infinitely many solutions for system of equations $y = 6x + b$ and $-3 x+\frac{1}{2} y=-3$

<u>Solution: </u>

Need to calculate value of b so that given system of equations have an infinite number of solutions

$\begin{array}{l}{y=6 x+b} \\\\ {-3 x+\frac{1}{2} y=-3}\end{array}$

Let us bring the equations in same form for sake of simplicity in comparison

$\begin{array}{l}{y=6 x+b} \\\\ {\Rightarrow-6 x+y-b=0 \Rightarrow (1)} \\\\ {\Rightarrow-3 x+\frac{1}{2} y=-3} \\\\ {\Rightarrow -6 x+y=-6} \\\\ {\Rightarrow -6 x+y+6=0 \Rightarrow(2)}\end{array}$

Now we have two equations

$\begin{array}{l}{-6 x+y-b=0\Rightarrow(1)} \\\\ {-6 x+y+6=0\Rightarrow(2)}\end{array}$

Let us first see what is requirement for system of equations have an infinite number of solutions

If $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y+c_{2}=0$ are two equation

$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ then the given system of equation has no infinitely many solutions.

In our case,

$\begin{array}{l}{a_{1}=-6, \mathrm{b}_{1}=1 \text { and } c_{1}=-\mathrm{b}} \\\\ {a_{2}=-6, \mathrm{b}_{2}=1 \text { and } c_{2}=6} \\\\ {\frac{a_{1}}{a_{2}}=\frac{-6}{-6}=1} \\\\ {\frac{b_{1}}{b_{2}}=\frac{1}{1}=1} \\\\ {\frac{c_{1}}{c_{2}}=\frac{-b}{6}}\end{array}$

As for infinitely many solutions $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

$\begin{array}{l}{\Rightarrow 1=1=\frac{-b}{6}} \\\\ {\Rightarrow6=-b} \\\\ {\Rightarrow b=-6}\end{array}$

Hence b must be equal to -6 for infinitely many solutions for system of equations $y = 6x + b$ and $-3 x+\frac{1}{2} y=-3$

8 0

3 years ago