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Elenna [48]
2 years ago
14

the sum of two odd numbers is 198 both of the numbers are multiples of 3 one of the numbers is 5 times the other which equation

could be used to find the two numbers
Mathematics
1 answer:
Sidana [21]2 years ago
6 0
198/6=33
33X5=165, 165 is 5 times of 33.
165+33=198 the sum of 2 odd number 165 & 33 is odd numbers.
33 & 165 is multiple of 3
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The equation 4x2 – 24x + 4y2 + 72y = 76 is equivalent to
DerKrebs [107]

Answer:

Option 4 is correct.

The equation 4x^2 -24x + 4y^2 + 72y = 76 is equivalent to 4(x-3)^2 + 4(y+9)^2 =436

Step-by-step explanation:'

Given equation: 4x^2 -24x + 4y^2 + 72y = 76

First group the terms with x and those with y;

(4x^2-24x)+(4y^2+72y) = 76

Next, we complete the squares.

We can do this by adding a third term such that the x terms and the y terms are perfect squares.

For this we must either add the same value on the other side of the equation or subtract the same value on the same side so that the equality is maintained.

⇒4(x^2-6x) +4(y+18y) = 76

or

4(x^2 -6x +3^2 -3^2) + 4(y^2 +18y +9^2 -9^2) = 76

4(x^2-6x + 3^2) - 36 + 4(y^2+18y +9^2) - 324 = 76

4(x-3)^2 + 4(y+9)^2 - 360 =76

Add 360 on both sides we get;

4(x-3)^2 + 4(y+9)^2 =360 +76

Simplify:

4(x-3)^2 + 4(y+9)^2 =436

Therefore, the given equation is equivalent to 4(x-3)^2 + 4(y+9)^2 =436

5 0
2 years ago
Which of the following functions is graphed below?
dalvyx [7]

Answer:

The function graphed is y = Ix - 5I - 4 ⇒ answer D

Step-by-step explanation:

* Lets explain how to solve this problem

- From the graph

# The graph intersects the x-axis at x = 1 and x = 9

∴ The x-intercepts are 1 and 9

- We can find x-intercept by equate y by 0

* Lets equate the answers by zero to find the x-intercept

# y = Ix + 5I - 4

∵ Ix + 5I - 4 = 0

- Add 4 to both sides

∴ Ix + 5I = 4

- Remember in Ia + bI = c, then we have two answers :

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∵ Ix + 5I = 4

∴ x + 5 = 4 ⇒ subtract 5 from both sides

∴ x = -1

- OR

∴ x + 5 = -4 ⇒ subtract 5 from both sides

∴ x = -9

∴ The x-intercepts are -1 and -9 not the same with figure

# y = Ix - 5I + 4

∵ Ix - 5I + 4 = 0

- Subtract 4 from both sides

∴ Ix - 5I = -4

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∴ We can't solve this equation

# y = Ix + 5I + 4

∵ Ix + 5I + 4 = 0

- Subtract 4 from both sides

∴ Ix + 5I = -4

- Remember in Ia + bI = c , c can't be negative because the absolute

 value is always positive

∴ We can't solve this equation

# y = Ix - 5I - 4

∵ Ix - 5I - 4 = 0

- Add 4 to both sides

∴ Ix - 5I = 4

- Remember in Ia + bI = c, then we have two answers :

 a + b = c  <em>OR </em> a + b = -c

∵ Ix - 5I = 4

∴ x - 5 = 4 ⇒ add 5 to both sides

∴ x = 9

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∴ x - 5 = -4 ⇒ add 5 to both sides

∴ x = 1

∴ The x-intercepts are 1 and 9 the same with figure

* The function graphed is y = Ix - 5I - 4

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