All categories

3 years ago

Help me please past due! Complete all 5 problems in simplest form and tell me what you did

Mathematics

1 answer:

Sonbull [250]3 years ago

6 0

3/6= 1/2 because 3 is the half of 6 and 1 is the half of 2

5/20= 1/4 because 1 x 5= 5 and 4 x 5= 20

9/27= 1/3 because 1 x 9 =9 and 3 x 9=27

1/2 is the simplest form already

12/36= 1/3 because 1 x 12= 12 and 3 x 12= 36

You might be interested in

Sofia takes 10mg of a medicine whose concentration in blood decreases by a factor of one half every day. Let t be the number of

adelina 88 [10]

Answer:

$L = 10(\frac{1}{2} )^{t-4}$

Step-by-step explanation:

If t is the number of days since Sofia took the medication of 10 mg, then the amount of medication in her system after t days will be $S = 10(\frac{1}{2} )^{t}$ .

So, the concentration of medicine in blood decreases by a factor of one half every day.

If we count the number of days from today and Lexi takes 10 mg of the same medicine four days later then the amount of medicine left on her body after t days will be $L = 10(\frac{1}{2} )^{t-4}$ . (Answer)

3 0

3 years ago

Read 2 more answers

CosA.cosecA=cotA can you please help me with this

chubhunter [2.5K]

CscA = 1/sinA

CosACscA = cosA × 1/sinA = cosA/SinA = cotA

3 0

3 years ago

10.2 = 0.6y what is y

AVprozaik [17]

Answer:

y = 17

Step-by-step explanation:

isolate the variable by dividing each side by factors that don't contain the variable.

6 0

2 years ago

Use l'Hôpital's Rule to find the following limit. ModifyingBelow lim With x right arrow 0StartFraction 3 sine (x )minus 3 x Ove

steposvetlana [31]

Answer:

$\lim_{x \to 0} \frac{3sinx-3x}{7x^3}=-\frac{1}{14}$

Step-by-step explanation:

The limit is:

$\lim_{x \to 0} \frac{3sinx-3x}{7x^3}=\frac{0}{0}$

so, you have an indeterminate result. By using the l'Hôpital's rule you have:

$\lim_{x \to 0} \frac{a(x)}{b(x)}= \lim_{x \to 0} \frac{a'(x)}{b'(x)}$

by replacing, and applying repeatedly you obtain:

$\lim_{x \to 0} \frac{3sinx-3x}{7x^3}= \lim_{x \to 0}\frac{3cosx-3}{21x^2}= \lim_{x \to 0}\frac{-3sinx}{42x}= \lim_{x \to 0}\frac{-3cosx}{42}\\\\ \lim_{x \to 0} \frac{3sinx-3x}{7x^3}=\frac{-3cos0}{42}=-\frac{1}{14}$

hence, the limit of the function is -1/14

8 0

3 years ago

Q1: Consider the slope-intercept form of the equation of the line representing the sports car's distance from Officer Franklin's

lozanna [386]

Since you didn't give the actual equation, I can only help with the mean of each part.

Q1: The coefficient of x is called the slope. That is the number in front of the x.
Q2: The constant term is the y-intercept. That is the number at the end of the equation.

4 0

2 years ago