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3 years ago

N^2+100=20n Please show work

1 answer:

7 0

Subtract 20n from both sides of the equation.
n^2+100-20n=0
factor using the perfect square rule.
(n-10)^2=0
set the n-10 equal to 0.
n-10=0
add 10 to both sides of the equation.
n=0

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Determine whether the equation 3(4g+6)=2(6g+9) has one solution, no solution, or infinitely many solutions. The equation has . Q

taurus [48]

So firstly, foil 3(4g + 6) and 2 (6g + 9): $12g+18=12g+18$

Now since the quantity is the same on both sides of this equation, this means that <u>this equation has infinite solutions.</u>

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3 years ago

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Wright an Expression to represent: three minus the product of two and a number z

Nookie1986 [14]

Answer:

2z-3

Step-by-step explanation:

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3 years ago

How many solutions exist for the system of equations graphed below?

Nat2105 [25]

Answer:

no solutions

Step-by-step explanation:

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3 years ago

Describe and correct the error a student made in finding the midpoint of CD with C(4,5) and D(1,4).

Andru [333]

Using the given information, the midpoint of CD is (2.5, 4.5)

<h3>Midpoint of a line </h3>

From the question, we are to determine the midpoint of line CD

The midpoint of a line is given by the formula,

[(x₁ + x₂)/2 , (y₁ + y₂)/2]

From the given information, we are to find the midpoint of line CD with coordinates C(4,5) and D(1,4)

∴ x₁ = 4

x₂ = 1

y₁ = 5

y₂ = 4

Putting the parameters into the formula for midpoint, we get

[(x₁ + x₂)/2 , (y₁ + y₂)/2]

[(4 + 1)/2 , (5 + 4)/2]

(5/2 , 9/2)

(2.5, 4.5)

Hence, the midpoint of CD is (2.5, 4.5)

Learn more on Midpoint of a line here: brainly.com/question/896396

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2 years ago

Please math , help

Furkat [3]

Answer:

As both the mean and standard deviation are in the desired ranges, the tool passes the technical control.

Step-by-step explanation:

Mean of the batch:

The mean of the batch is the sum of all values divided by the number of items. So

$M = \frac{199+204+201+197+195+204+200}{7} = 200$

Mean in the desired interval.

Standard deviation:

Square root of the sum of the difference squared between each term and the mean, divided by the number of items. So

$S = \sqrt{\frac{(199-200)^2+(204-200)^2+(201-200)^2+(197-200)^2+(195-200)^2+(204-200)^2+(200-200)^2}{7}} = 1.18$

As both the mean and standard deviation are in the desired ranges, the tool passes the technical control.

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3 years ago