All categories

2 years ago

In rectangle PQRS, PR = 18x – 28 and QS = x + 380. Find the value of x and the length of each diagonal.

Mathematics

2 answers:

Flauer [41]2 years ago

5 0

Diagonals of a rectangle are congruent.

18x - 28 = x + 380

Solving for x, I get 17.

So, x = 17.

PR = 18x - 28

PR = 18(17) - 28

PR = 278

QS = x + 380

QS = 17 + 380

QS = 397

Len [333]2 years ago

5 0

Question-

In rectangle PQRS, PR = 18x – 28 and QS = x + 380. Find the value of x and the length of each diagonal.

18x - 28 = x + 380

Answer-

Solving for x, I get 17.

So, x = 17.

PR = 18x - 28

PR = 18(17) - 28

PR = 278

QS = x + 380

QS = 17 + 380

QS = 397

You might be interested in

2.26-30 3x+y=6 (0,_) (_,0) (3,_) (6,_) (5,_)

bekas [8.4K]

Answer:

Step-by-step explanation:

In ordered pairs (a,b) a is the x value and b is a y value.

if we have 3x+y=6

if x=0, y=6 --> (0,6)

if y=0, x=2 -->(2,0)

if x=3, y=-3--> (3, -3)

if x=6, y= -12 --->(6, -12)

if x=6, y= -9 ---> (5, -9)

7 0

2 years ago

A circle has the equation 2x²+12x+2y²−16y−150=0.

KonstantinChe [14]

Answer: B. The coordinates of the center are (-3,4), and the length of the radius is 10 units.

Step-by-step explanation:

The equation of a circle in the center-radius form is:

$(x-h)^{2} +(y-k)^{2}=r^{2}$ (1)

Where $(h,k)$ are the coordinates of the center and $r$ is the radius.

Now, we are given the equation of this circle as follows:

$2x^{2}+12x+2y^{2}-16y-150=0$ (2)

And we have to write it in the format of equation (1). So, let's begin by applying common factor 2 in the left side of the equation:

$2(x^{2}+6x+y^{2}-8y-75)=0$ (3)

Rearranging the equation:

$x^{2}+6x+y^{2}-8y=75$ (4)

$(x^{2}+6x)+(y^{2}-8y)=75$ (5)

Now we have to complete the square in both parenthesis, in order to have a perfect square trinomial in the form of $(a\pm b)^{2}=a^{2}\pm+2ab+b^{2}$ :

For the first parenthesis:

$x^{2}+6x+b^{2}$

We can rewrite this as:

$x^{2}+2(3)x+b^{2}$

Hence in this case $b=3$ and $b^{2}=9$ :

$x^{2}+2(3)x+3^{2}=x^{2}+6x+9=(x+3)^{2}$

For the second parenthesis:

$y^{2}-8y+b^{2}$

We can rewrite this as:

$y^{2}-2(4)y+b^{2}$

Hence in this case $b=-3$ and $b^{2}=9$ :

$y^{2}-2(4)y+4^{2}=y^{2}-8y+16=(y-4)^{2}$

Then, equation (5) is rewritten as follows:

$(x^{2}+6x+9)+(y^{2}-8y+16)=75+9+16$ (6)

Note we are adding 9 and 16 in both sides of the equation in order to keep the equality.

Rearranging:

$(x-3)^{2}+(y-4)^{2}=100$ (7)

At this point we have the circle equation in the center radius form $(x-h)^{2} +(y-k)^{2}=r^{2}$

Hence:

$h=-3$

$k=4$

$r=\sqrt{100}=10$

8 0

3 years ago

Is this correct ? Please help me

NNADVOKAT [17]

Answer:

Median: 8

Minimum: 3

Maximum: 15

First quartile: 6

Third quartile: 12.5

Interquartile Range: 6.5

Outliers: none

Step-by-step explanation:

6 0

2 years ago

Which kind of map would be used to show population in the countries of the world?

yaroslaw [1]

It would be a special purposes map :)

3 0

2 years ago

A rectangle is 5cm longer that it is broad. It perimeter is 26cm. Find the length and breadth of the rectangle

Maslowich

I hope this helps you

length w+5

width w

26=2 (w+5+w)

13=2w+5

2w=8

w=4

length =4+5=9

3 0

2 years ago