Luke buys a certain brand of cereal that cost $5

Which of the following fractions can be represented by a terminating decimal?

Karo-lina-s [1.5K]

Answer:

B) 13/8

Step-by-step explanation:

First we need to understand what a terminating decimal is: It is a decimal that has a finite number of decimal values that are not zero, meaning that its decimal values end at some point

Let's go through our possible answers with trial and error:

A) 8/9

8/9 = 0.888888888888...9 (Incorrect)

B) 13/8

13/8 = 1.625 (Correct)

C) 4/3

4/3 = 1.3333333333...4 (Incorrect)

D) 6/11

6/11 = 0.545454545454...54 (Incorrect

13/8 is our answer because in decimal form it is a terminating decimal.

8 0

2 years ago

i have been trying to solve this compound and double angle question please help me find the answer to these question guys

natali 33 [55]

Answer:

These type of questions are super tricky b/c you have to remember all the different versions of the identities, and then they put the question in some odd form, I feel like this should land math professors in jail , for dishonesty , b/c it's really a form of "how tricky can I make a question and still have a way to solve it" anyway,

Step-by-step explanation:

a)

next the question asks 1-cos 2A and this is total abuse of notation. the way this should be written is 1- cos( 2A) so we know that the A is part of the cosine functions input... btw.. in any computer program, it would never ever let you get away with that top form of the expression. :/ anyway... I keep ranting.. huh... sorry :P

1-cos(2A) is an odd form of the identity 1/2(1-cos(2A) = $sin^{2}$ (A) the 1/2 is missing but we can add that pretty easy, we just have to remember to take it out too. I usually forget to do that. and my professor marks me off completely, totally wrong, but I just miss one small thing :/ anyway....

our 1-cos(2A) needs the 1/2 added to it. or if we move that 1/2 to the other side it looks like 2* $sin^{2}$ (A) = 1-cos(2A) and this is that "odd" from of the identity that I was talking about.

next let's deal with sin(2A) it has an identity of 2 sin(A)cos(A) which is really nice for us b/c it will cancel out the 2 in then numerator for us, nice !

now our fraction looks like [2* $sin^{2}$ (A)] / 2 sin(A)cos(A)

so cancel out one of the sines

2*sin(A) / 2 cos(A)

cancel the 2s

Sin(A) / Cos(A) = Tan(A)

nice it worked out :P

b)

by the above that we just worked out, then

Tan(15) = Sin(15) / Cos(15)

I had to look up what sin of 15 is b/c it's not one of those special angles but it does have an exact form of

Sin(15) = (√3 - 1) / 2√2

Cos(A) = (√3 + 1) / 2√2

you can use rule of Cos(A-B) = Cos(A)Cos(B)+Sin(A)Sin(B) to get the above and a similar rule for Sin(A-B)

back to our problem, the 2√2 will cancel out

then we have

Tan(15) = (√3 - 1) /(√3 + 1)

in the form that is above that's exact, the roots could be approximated but i'll just leave that in the form that is exact. Most math professors like that form.

4 0

3 years ago

Given the graph below, which of the following statements is true?

Nesterboy [21]

To be a function, it has to pass the vertical line test, which means a vertical line drawn anywhere in the graph should cut the graph only once, not more.

To be a one-to-one function, it has to pass the vertical line test as well as the horizontal line test, which means that a horizontal line drawn anywhere in the graph should cut the graph only once, not more.

Of course, it passes the vertical line test, so its a function.

Does it pass the horizontal line test? NO! As shown in the attached picture, the blue line (horizontal line) cuts the graph at 3 places. We therefore eliminate choices A and B. We can eliminate choice C because nowhere is it required a function to pass through the origin for it to be one-to-one. Hence, the answer is D (blue line shows this).

ANSWER: D