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fgiga [73]
3 years ago
8

What is my GPA (Highschool) *The first class is a B*The last class is a C*

Mathematics
1 answer:
Rudiy273 years ago
3 0
B is your GPA great job kid
You might be interested in
Se uma pessoa comprar um aparelho eletrônico em 5 prestações mensais de 304,00 quanto pagará por esse aparelho ?
kobusy [5.1K]

Answer:

1520

Step-by-step explanation:

Pela pergunta acima, sabemos que:

1 parcela mensal = 304

5 parcelas mensais = x

Multiplicação cruzada

x = 5 prestações mensais × 304/1 prestações mensais

= 1520

Portanto, o valor que ele pagou pelo dispositivo é 1520

6 0
3 years ago
C=<br> 5/9<br> (F−32)<br> i need help with this question its half of my grade plssss thxx
n200080 [17]

Answer:

c=5/9f+-160/9

I just simplifed

Please give brainliest

havea nice day :D

3 0
3 years ago
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'

6 0
3 years ago
Read 2 more answers
Solve for x:<br>7x(x+1.8)=0<br><br>Need help ASAP.
andrew11 [14]

Answer:

i got you

Step-by-step explanation:

for x1= -1.8 and x2=0

3 0
3 years ago
Read 2 more answers
Which is true regarding the graphed function f(x)?
Sladkaya [172]
<h2>Explanation:</h2>

Hello! Remember you have to write complete questions in order to get good and exact answers. Here I'll assume the graphed function comes from:

f(x)=\left(x-4\right)^{2}+5

So this is the equation of a parabola that opens upward and whose vertex lies on the point:

V(4,5)

The graph of this function is shown below. Which is true regarding the graphed function f(x)?

  • It is true that the domain is the set of all real numbers because this is a polynomial function.

  • It is true that the range is the set of all real numbers such that y ≥ 5

  • It is true that this parabola opens upward

  • It is true that it doesn't cut the x-axis

6 0
3 years ago
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