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Nutka1998 [239]

2 years ago

11

One side of a rectangle is parallel to the x-axis. One vertex of the rectangle is located at (5,2) and another vertex at (1,4).

What are the coordinates of the other two vertices?

1 answer:

harina [27]2 years ago

8 0

Answer:

(1,2) and (5,4)

Step-by-step explanation:

Hello There!

I had made a graph to make it easier to understand.

The points given to us are represented by the green points

The points that would make a rectangle with lines that are parallel with the x axis would be (1,2) and (5,4) (represented by the red points)

As you can see the points added had created a rectangle with sides that are parallel with the x axis

This meets all of the requirements therefore the other two vertices would have the points (1,2) and (5,4)

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50 is one tenth of what number?

Stella [2.4K]

Answer:

500

Step-by-step explanation:

Well you find one tenth of a number by dividing it by 10

So to do the opposite, you can do 50*10 which is 500

Or if you really wanted to you could go through each option dividing them all by 10 till you got 50

5 0

3 years ago

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For each order. Determine weather is a solution to x = -5

Yakvenalex [24]

First number is x and second number is y
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4 0

2 years ago

The volume of a cylinder is 176 π cm and its height is 11 π cm.

Liula [17]

Answer:

r = sqrt(16/pi)

Step-by-step explanation:

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2 years ago

I am trying to figure this question: <br><br> 1 + tan^2A = Sec^2A

chubhunter [2.5K]

Answer:

see explanation

Step-by-step explanation:

Using the Pythagorean identity

cos²A + sin²A = 1 ( divide terms by cos²A )

$\frac{cos^2A}{cos^2A}$ + $\frac{sin^2A}{cos^2A}$ = $\frac{1}{cos^2A}$ , that is

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3 0

3 years ago

Give the properties for the equation -2x 2 - y + 10x - 7 = 0.

Sladkaya [172]

Given:

The quadratic equation is

$-2x^2-y+10x-7=0$

To find:

The vertex of the given quadratic equation.

Solution:

If a quadratic function is $f(x)=ax^2+bx+c$ , then

$Vertex=\left(\dfrac{-b}{2a},f(\dfrac{-b}{2a})\right)$

We have,

$-2x^2-y+10x-7=0$

It can be written as

$-2x^2+10x-7=y$

$y=-2x^2+10x-7$ ...(i)

Here, $a=-2,b=10,c=-7$ .

$\dfrac{-b}{2a}=\dfrac{-10}{2(-2)}$

$\dfrac{-b}{2a}=\dfrac{-10}{-4}$

$\dfrac{-b}{2a}=\dfrac{5}{2}$

Putting $x=\dfrac{5}{2}$ in (i), we get

$y=-2(\dfrac{5}{2})^2+10(\dfrac{5}{2})-7$

$y=-2(\dfrac{25}{4})+\dfrac{50}{2}-7$

$y=\dfrac{-50}{4}+25-7$

$y=\dfrac{-25}{2}+18$

On further simplification, we get

$y=\dfrac{-25+36}{2}$

$y=\dfrac{11}{2}$

So, the vertex of the given quadratic equation is $\left(\dfrac{5}{2},\dfrac{11}{2}\right)$ .

Therefore, the correct option is A.

3 0

2 years ago