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UNO [17]
3 years ago
7

Mary is 2 times older than Bob and Bob is 5 years older than Sally. Sally is 10 years old how old is Bob

Mathematics
1 answer:
s344n2d4d5 [400]3 years ago
8 0

Answer:

Bob is 15 years old.

Step-by-step explanation:

If Sally is 10 years old and Bob is five years older than Sally then we just need to add 10+5=15.

Hope this helps!!

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30 POINTS please help me!
Stels [109]
To prove a similarity of a triangle, we use angles or sides.

In this case we use angles to prove

∠ACB = ∠AED (Corresponding ∠s)
∠AED = ∠FDE (Alternate ∠s)

∠ABC = ∠ADE (Corresponding ∠s)
∠ADE = ∠FED (Alternate ∠s)

∠BAC = ∠EFD (sum of ∠s in a triangle)

Now we know the similarity in the triangles.

But it is necessary to write the similar triangle according to how the question ask.

The question asks " ∆ABC is similar to ∆____. " So we find ∠ABC in the prove.

∠ABC corressponds to ∠FED as stated above.
∴ ∆ABC is similar to ∆FED

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7 0
3 years ago
PLEASE HELP!!<br> Convert each linear equation to standard form.<br> y-4= -2(x+3)
hjlf
So first, this is point slope form. The first step is to distribute.)
y-4=-2x-6
Then add 4 to both sides.
y=-2x-2
Then add 2x and it becomes
2x+y=-2
4 0
3 years ago
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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}=

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* 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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This should be the answer
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Step-by-step explanation:

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