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Masja [62]
3 years ago
11

Nia is three years older than half Jordans age , if Jordan is eight years old , how old is Nia?

Mathematics
1 answer:
Kobotan [32]3 years ago
8 0
Hello! 

If Nia is three years older than half of Jordan's age, we must divide Jordan's age by 2 (to find half of his age).

Jordan is eight years old:
8 divided by 2 equals 4

4 years old is half of Jordan's age. Now, to find Nia's age, we must add 3 to half of Jordan's age (4).

4+3= 7
Nia is 7 years old. 

I hope this helps answer your question! Have a great day!
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Evaluate the function y=1/2(x)-4 for each of the given domain values? PLZ HELP ME
erik [133]

Answer:

c. -13/4.

d. -13/3.

Step-by-step explanation:

c. f(3/2) = (1/2)(3/2) - 4

= 3 / 4 - 4

= 0.75 - 4

= -3.25

= -3 and 1/4

= -13/4.

d. f(-2/3) = (1/2)(-2/3) - 4

= -2/6 - 4

= -1/3 - 4

= -1/3 - 12/3

= -13/3.

Hope this helps!

6 0
3 years ago
What is the distance between (-4,-8) and (-4,0) round to the nearest tenth if needed​
mixer [17]

Answer:

8

Step-by-step explanation:

Distance between x coordinates: 0 units

Distance between y coordinates: 8 units

sqrt(0^2+8^2)

8 units

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3 years ago
If the product of the extremes is 8,then the geometric mean is 2 4 2√2
luda_lava [24]

Answer:

the correct answer choice would be choice C. or 2√2

Step-by-step explanation:

I have the same assignment in odyssey and other brainly user helped me solve this answer :)

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3 years ago
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Write 55% as a decimal and as a fraction.
Sedaia [141]
The answer for this question is 0.55%

4 0
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Someone help me out please
uysha [10]

1 step (B): raise both sides of the equation to the power of 2.

(\sqrt{x+3}-\sqrt{2x-1})^2=(-2)^2,\\   (x+3)-2\sqrt{x+3}\cdot \sqrt{2x-1}+(2x-1)=4,\\ 3x+2-2\sqrt{x+3}\cdot \sqrt{2x-1}=4.

2 step (A): simplify to obtain the final radical term on one side of the equation.

-2\sqrt{x+3}\cdot \sqrt{2x-1}=4-3x-2,\\ -2\sqrt{x+3}\cdot \sqrt{2x-1}=2-3x,\\ 2\sqrt{x+3}\cdot \sqrt{2x-1}=3x-2.

3 step (F): raise both sides of the equation to the power of 2 again.

(2\sqrt{x+3}\cdot \sqrt{2x-1})^2=(3x-2)^2,\\ 4(x+3)(2x-1)=(3x-2)^2.

4 step (E): simplify to get a quadratic equation.

4(2x^2-x+6x-3)=(3x)^2-2\cdot 3x\cdot 2+2^2,\\ 8x^2+20x-12=9x^2-12x+4,\\ x^2-32x+16=0.

5 step (D): use the quadratic formula to find the values of x.

D=(-32)^2-4\cdot 16=1024-64=960, \\ \sqrt{D} =8\sqrt{5} ,\\ x_{1,2}=\dfrac{32\pm 8\sqrt{5}}{2} =16\pm 4\sqrt{5}.

6 step (C): apply the zero product rule.

x^2-32x+16=(x-16-4\sqrt{5}) (x-16+4\sqrt{5}) ,\\ (x-16-4\sqrt{5}) (x-16+4\sqrt{5}) =0,\\ x_1=16+4\sqrt{5} ,x_2=16-4\sqrt{5}.

Additional 7 step: check these solutions, substituting into the initial equation.

3 0
3 years ago
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