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lesya692 [45]
3 years ago
7

Is -12.5 greater than -12? will brainlist

Mathematics
2 answers:
notka56 [123]3 years ago
7 0

Answer:

No

Step-by-step explanation:

negative numbers work the opposite of how positive numbers work. Think about it like this. -12 is 12 units less than 0 and -12.5 is 12.5 units less than 0 which means that -12.5 is less than -12

Ivenika [448]3 years ago
5 0

Answer:

no because the bigger the negative the smaller the parts are... so therefore is would be smaller than -12 because its -0.5 smaller the negative sign makes all the bigger numbers smaller than 1

Step-by-step explanation:

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Step-by-step explanation:

1. 7y=14

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so the equation has one solution

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Six times a number is at least -24
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Write and equation of the translated or rotated graph in general form (picture below)
WINSTONCH [101]

Answer:

The answer is hyperbola; (x')² - (y')² - 16 = 0 ⇒ answer (a)

Step-by-step explanation:

* At first lets talk about the general form of the conic equation

- Ax² + Bxy + Cy²  + Dx + Ey + F = 0

∵ B² - 4AC < 0 , if a conic exists, it will be either a circle or an ellipse.

∵ B² - 4AC = 0 , if a conic exists, it will be a parabola.

∵ B² - 4AC > 0 , if a conic exists, it will be a hyperbola.

* Now we will study our equation:

 xy = -8

∵ A = 0 , B = 1 , C = 0

∴ B² - 4 AC = (1)² - 4(0)(0) = 1 > 0

∴ B² - 4AC > 0

∴ The graph is hyperbola

* The equation xy = -8

∵ We have term xy that means we rotated the graph about

  the origin by angle Ф

∵ Ф = π/4

∴ We rotated the x-axis and the y-axis by angle π/4

* That means the point (x' , y') it was point (x , y)

- Where x' = xcosФ - ysinФ and y' = xsinФ + ycosФ

∴ x' = xcos(π/4) - ysin(π/4) , y' = xsin(π/4) + ycos(π/4)

∴ x' = x/√2 - y/√2 = (x - y)/√2

∴ y' = x/√2 + y/√2 = (x + y)/√2

* Lets substitute x' and y' in the 1st answer

∵ (x')² - (y')² - 16 = 0

∴ (\frac{x-y}{\sqrt{2}})^{2}-(\frac{x+y}{\sqrt{2}})^{2}=

 ( \frac{x^{2}-2xy+y^{2}}{2})-(\frac{x^{2}+2xy+y^{2}}{2})-16=0

* Lets open the bracket

∴ \frac{x^{2}-2xy+y^{2}-x^{2}-2xy-y^{2}}{2}-16=0

* Lets add the like terms

∴ \frac{-4xy}{2}-16=0

* Simplify the fraction

∴ -2xy - 16 = 0

* Divide the equation by -2

∴ xy + 8 = 0

∴ xy = -8 ⇒ our equation

∴ Answer (a) is our answer

∴ The answer is hyperbola; (x')² - (y')² - 16 = 0

* Look at the graph:

- The black is the equation (x')² - (y')² - 16 = 0

- The purple is the equation xy = -8

- The red line is x'

- The blue line is y'

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3 years ago
Read 2 more answers
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

This y-value that we just found is our k-value.

Next, we are going to set up our equation in vertex form. As a reminder, vertex form is: y=a(x-h)^{2} +k

a: 1

h: 3

k: -25

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

Hope this helps!

3 0
3 years ago
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