All categories

3 years ago

The sequence below shows the total number of days Francisco had used his

Mathematics

1 answer:

drek231 [11]3 years ago

3 0

1 week because ye. 47

You might be interested in

A colleague has been tutoring six students in 11th grade to prepare for the ACT. This colleague has asked you to evaluate the pe

marissa [1.9K]

We have been given that a colleague has been tutoring six students in 11th grade to prepare for the ACT. Student scores were as follows: 20, 18, 16, 15, 23, 20. We are asked to find the mean of the ACT scores.

We will use mean formula to solve our given problem.

$\text{Mean}=\frac{\text{Sum of terms}}{\text{Number of terms}}$

$\text{Mean}=\frac{20+18+16+15+23+20}{6}$

$\text{Mean}=\frac{112}{6}$

$\text{Mean}=18.66666$

Upon rounding to nearest whole number, we will get:

$\text{Mean}\approx 19$

Therefore, the mean of the ACT scores is 19 and option 'c' is the correct choice.

5 0

3 years ago

Does anybody know this??????

aniked [119]

The correct answer is 196 or Option A.
The explanation is just as complicated as the question.

7 0

3 years ago

Read 2 more answers

What is 40,023,032 in expanded form

Tanya [424]

40,023,032 = (4 x 1000000000) + (0 x 100000000) + (0 x 10000000) + (0 x 1000000) + (2 x 100000) + (3 x 10000) + (0 x 1000) + (0 x 100) + (3 x 10) + (2 x 1)

4 0

3 years ago

2 tan 30°<br>II<br>1 + tan- 300

shusha [124]

Question:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$

Answer:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= sin(60^{\circ})$

Step-by-step explanation:

Given

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$

Required

Simplify

In trigonometry:

$tan(30^{\circ}) = \frac{1}{\sqrt{3}}$

So, the expression becomes:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2 * \frac{1}{\sqrt{3}}}{1 + (\frac{1}{\sqrt{3}})^2}$

Simplify the denominator

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2 * \frac{1}{\sqrt{3}}}{1 + \frac{1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{1 + \frac{1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{ \frac{3+1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{ \frac{4}{3}}$

Express the fraction as:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2}{\sqrt 3} / \frac{4}{3}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2}{\sqrt 3} * \frac{3}{4}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{1}{\sqrt 3} * \frac{3}{2}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3}{2\sqrt 3}$

Rationalize

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3}{2\sqrt 3} * \frac{\sqrt{3}}{\sqrt{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3\sqrt{3}}{2* 3}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\sqrt{3}}{2}$

In trigonometry:

$sin(60^{\circ}) = \frac{\sqrt{3}}{2}$

Hence:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= sin(60^{\circ})$

3 0

3 years ago

Jules kicks a soccer ball off the ground and into the air with an initial velocity of 25 feet

hichkok12 [17]

Answer:

Set up your height equation, then factor or use quadratic formula to find when h(t)=0

Step-by-step explanation:

Your equation will be -16t^2+vt +h where v is the initial velocity and h is the starting height. Now either factor or quadratic equation, whichever is easier for you.

Remember that the ball is on the ground when h(t)=0 since that is the height. There will be two zeros, one is a negative number so would be before you kicked the ball, the other one will be when the ball comes back down.

5 0

3 years ago