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3 years ago

Plz help I’m stuck on this question

Mathematics

1 answer:

Leona [35]3 years ago

4 0

Answer:

the answer is C

Step-by-step explanation:

i hope it will help you

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Simplified version of -12x + 3y = -3?

damaskus [11]

20 I think!

Look it up just to be sure

4 0

3 years ago

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If an intravenous solution containing 123 mg of a drug substance in each 250-mL bottle is to be administered at the rate of 200

12345 [234]

Answer:

24.39mL of the solution would be given per hour.

Step-by-step explanation:

This problem can be solved by direct rule of three, in which there are a direct relationship between the measures, which means that the rule of three is a cross multiplication.

The first step to solve this problem is to see how many mg of the solution is administered per hour.

Each minute, 200 ug are administered. 1mg has 1000ug, so

1mg - 1000 ug

xmg - 200 ug

$1000x = 200$

$x = \frac{200}{1000}$

$x = 0.2mg$

In each minute, 0.2 mg are administered. Each hour has 60 minutes. How many mg are administered in 60 minutes?

1 minute - 0.2 mg

60 minutes - x mg

$x = 60*0.2$

$x = 12mg$

In an hour, 12 mg of the drug is administered. In 250 mL, there is 123 mg of the drug. How many ml are there in 12 mg of the drug.

123mg - 250mL

12 mg - xmL

$123x = 250*12$

$x = \frac{250*12}{123}$

$x = 24.39$ mL

24.39mL of the solution would be given per hour.

7 0

3 years ago

The image shows a geometric representation of the function f(x)=x^2-2x-6 written in standard form. What is the function written

Degger [83]

Y=(x-1)^2-7 This is this function in vertex form

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3 years ago

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5. The box plot summarizes the test scores for 100 students: u 55 60 65 70 75 80 85 90 95 100 Which term best describes the shap

ASHA 777 [7]

C would be the correct answer

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2 years ago

LORAN is a long range hyperbolic navigation system. Suppose two LORAN transmitters are located at the coordinates (−60,0) and (6

USPshnik [31]

LORAN follows an hyperbolic path.

The equation of the hyperbola is: $\mathbf{\frac{x^2}{2500} + \frac{y^2}{1100} = 1}$

The coordinates are given as:

$\mathbf{(x,y) = (-60,0)\ (60,0)}$

The center of the hyperbola is

$\mathbf{(h,k) = (0,0)}$

The distance from the center to the focal points is given as:

$\mathbf{c = 60}$

Square both sides

$\mathbf{c^2 = 3600}$

The distance from the receiver to the transmitters is given as:

$\mathbf{2a = 100}$

Divide both sides by 2

$\mathbf{a = 50}$

Square both sides

$\mathbf{a^2 = 2500}$

We have:

$\mathbf{b^2 = c^2 - a^2}$

This gives

$\mathbf{b^2 = 3600 - 2500}$

$\mathbf{b^2 = 1100}$

The equation of an hyperbola is:

$\mathbf{\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1}$

So, we have:

$\mathbf{\frac{(x - 0)^2}{2500} + \frac{(y - 0)^2}{1100} = 1}$

$\mathbf{\frac{x^2}{2500} + \frac{y^2}{1100} = 1}$

Hence, the equation of the hyperbola is: $\mathbf{\frac{x^2}{2500} + \frac{y^2}{1100} = 1}$