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3 years ago

Proofs how do you solve them

1 answer:

3 0

Proofs are... well... just that. They are your proof for why a certain property works or what you did to show your work. So essentially, you just do math like normal, a proof is showing what you did and why it works.

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Let f(x)=cos(arctanx). What is the range of f?

Alenkasestr [34]

-π/2 < arctan(x) < π/2

So cos(π/2) < cos(arctan(x)) < cos(0)

0 < cos(arctan(x)) < 1

4 0

3 years ago

Instructions in picture

Kipish [7]

Answer:

12)

(9y +7)=(2y +98)( Because vertically opposite angle is always equal)

9y - 2y = 98-7

9y - 2y = 98-7

7y= 91

y =13

therfore y = 13

(9y + 7)

9*13+7

117+7

124

(2y +98)

2*13+98

26+98

124

5 0

3 years ago

A family has an annual income of $25,200. of this, 1/4 is spent for food, 1/5 for housing, 1/10 for clothing, 1/9 for savi

Firlakuza [10]

To answer the question, you have to multiply $25,200 by 1/4 (portion allocated/ alloted for the tax -
$25,200 * 1/4 = 25,200/4 (then divide 25,200 by 4)
= $6,300
Therefore, the family spent $6,300 for taxes from their annual income of $25,200.

8 0

3 years ago

The children from a football club are put into rows in the sports hall. When put into rows of 9 children, there are 2 children l

g100num [7]

We are told that the children from a football club are put into rows in the sports hall. When put into rows of 9 children there are 2 children left over. When put into rows of 12 children there are 2 children left over.

We will find least number of children in football club by finding LCM of 9 and 12.

Multiples of 9 are: 9, 18, 27, 36, 45, 54, 63,...

Multiples of 12 are: 12, 24, 36, 48, 60, 72, 84,...

We can see that least common multiple of 9 and 12 is 36.

We are told that 2 children left over from putting them into 9 and 12 children per row. To find least number we will add 2 to 36.

$9\cdot4+2=36+2=38$

$12\cdot3+2=36+2=38$

$\text{Least number of children in football club}=36+2=38$

Therefore, the least number of children in the football club is 38.

5 0

3 years ago

Solve (4 - x ≤ -1) ∩ (2 + 3x ≥ 17).

belka [17]

$4-x \leq -1\ \ \ | subtract\ 4\ \ \ \ \ and\ \ \ \ 2+3x \geq 17\ \ \ | subtract\ 2\\\\
-x \leq -5\ |*(-1)\ \ \ \ and\ \ \ \ 3x \geq 15\ \ | divide\ by\ 3\\\\
x \geq 5\ \ \ \ \ and\ \ \ \ x \geq 5\\\\
Solution\ is\ x\in[5,+\infty)$

3 0

3 years ago

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