Identify the sequence as arithmetic, geometric, both, or neither. <br><br> 7 9 11 13

The map shows the location of the airport and a warehouse in a city. Though not displayed on the map, there is also a factory 10

kipiarov [429]

step 1

find the coordinates of the factory

the factory is 104 miles due north of the warehouse

the coordinates of the warehouse are (39,-52)

coordinates of the factory------> (39,-52+104)------> (39,52)

step 2

find the distance warehouse to airport

let

A coordinates of warehouse-------> A( 39,-52)

B coordinates of airport---------> B(-39,52)

C coordinates of factory-------> C(39,52)

distance AB=√[(52+52)²+(-39-39)²]------> distance AB=130 miles

step 3

find the distance airport to factory

B(-39,52)

C(39,52)

distance BC=√[(52-52)²+(39+39)²]-----> distance BC=39 miles

the total number of miles=distance AB+distance BC---> 130+39

the total number of miles=169 miles

<em>Thus</em> the answer is 169

4 0

2 years ago

Read 2 more answers

I NEED HELP WITH THIS PLEASEEE!!!!!

Galina-37 [17]

Answers:
The area for the square on the left is 81m^2. This is found because all sides of this square are the same, so the length and width are the same. Just multiply 9 x 9.

The area for the triangle is 31.5m^2. We find the left side is 9 meters because the triangle shares the same side as the square on the left side. We also find the bottom side is 7 because that is the length of each side of that square because all sides on a square is the same. We then multiply 9 times 7, getting 63, we divide this by 2 because it’s a triangle.

The square on the right has the area of 225 because both length and width is 15.

4 0

2 years ago

Help me simplify the following expression, show steps please

malfutka [58]

$\frac{2x^3y^3}{4y^2} = \frac{x^3y^3}{2y^2} = \frac{x^3y}{2}$

4 0

2 years ago

Y=x^2-6x-16 in vertex form

satela [25.4K]

Answer:

$y=(x-3)^{2} -25$

Step-by-step explanation:

The standard form of a quadratic equation is $y=ax^{2} +bx+c$

The vertex form of a quadratic equation is $y=a(x-h)^{2} +k$

The vertex of a quadratic is (h,k) which is the maximum or minimum of a quadratic equation. To find the vertex of a quadratic, you can either graph the function and find the vertex, or you can find it algebraically.

To find the h-value of the vertex, you use the following equation:

$h=\frac{-b}{2a}$

In this case, our quadratic equation is $y=x^{2} -6x-16$ . Our a-value is 1, our b-value is -6, and our c-value is -16. We will only be using the a and b values. To find the h-value, we will plug in these values into the equation shown below.

$h=\frac{-b}{2a}$ ⇒ $h=\frac{-(-6)}{2(1)}=\frac{6}{2} =3$

Now, that we found our h-value, we need to find our k-value. To find the k-value, you plug in the h-value we found into the given quadratic equation which in this case is $y=x^{2} -6x-16$

$y=x^{2} -6x-16$ ⇒ $y=(3)^{2} -6(3)-16$ ⇒ $y=9-18-16$ ⇒ $y=-25$