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victus00 [196]
4 years ago
5

HELP NEEDED!!!!

Mathematics
1 answer:
strojnjashka [21]4 years ago
8 0

Answer:

D

Step-by-step explanation:

Wkx2di2jd1nd1do jf2kd2ifj2dkdu2dj2d

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Need help don’t understand
Xelga [282]

Answer:

2 real solutions

Step-by-step explanation:

To determine the nature of the solutions use the discriminant

Δ = b² - 4ac

• If b² - 4ac > 0 then 2 real solutions

• If b² - 4ac = 0 the 2 real and equal solutions

• If b² - 4ac < 0 then no real solutions

Given

x² + 9x + 20 = 0 ← in standard form

with a = 1, b = 9, c = 20, then

b² - 4ac

= 9² - (4 × 1 × 20 ) = 81 - 80 = 1

Since b² - 4ac > 0 then there are 2 real solutions

7 0
3 years ago
F(x)=6(-x+3), what is the value of f(-1)
IgorLugansk [536]

Replace x in the equation with -1 and solve:

6(-(-1) + 3)

Simplify the parentheses: 6(1 + 3)

Add inside the parentheses:

6(4)

Multiply

6 x 4 = 24

Answer : 24

3 0
2 years ago
Round 468986 to the nearest thousand
kotykmax [81]

Answer:

https://www.snapsolve.com/solutions/Round-468986-to-the-nearest-thousand-1693462594399233

well you can see from this link to find your answer

Step-by-step explanation:

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6 0
3 years ago
Read 2 more answers
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
a shopkeeper sold goods for rs 2400 and made a profit of 25% in the process. find his profit per cent if he had sold his goods f
miv72 [106K]

case 1,

Let the CP be ₹x,

SP = ₹2400

Profit = SP – CP

= 2400 – x

Profit % = {(2400–x)/ x} × 100%

According to the question,

{(2400–x)/ x} × 100 = 25

=> (2400–x)/ x= 25 /100

=> 100(2400–x) = 25x [ cross multiplication]

=> 240000 – 100x = 25x

=> 240000 = 25x + 100x

=> 240000 = 125x

=> 240000/125 = x

=> x = 1920

So, CP = ₹1920

case 2,

SP = ₹2040

Profit = SP – CP

= 2040 – 1920

= ₹120

profit % = 120/1920 × 100%

= 16%

<h3>Thus, his profit would be 16% if he had sold his goods for ₹2040.</h3>
6 0
3 years ago
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