*PLEASE ANSWER, DIFFICULT QUESTION*

Please help 4 math pages with questions. 1 has all the circle graphs, and all the other 3 have each circle graphs question. Plea

oksano4ka [1.4K]

Answer:

Favorite chip Answers v

Step-by-step explanation:

1. How many were surveyed? 82+8+18+62=100

2. How many said Jalepeno was their favorite? 8/100 0r 2/25 or 8%

3. How many said BBQ was their favorite?62/100 or 31/50 or 62%

4. How would it effect your final answer if the percentages did not add up to 100%? This means that your percentages are wrong. In a circle graph the percentages always equal 100%!

5. Is it possible to determine a missing percentage in a circle graph? Yes, we can always assume the total percentage is 100 so you add your given percentages together and then subtract that total from 100 to find the missing percentage.

8 0

2 years ago

1. Find 3 rational numbers between −1//3 and 1/−5

Alborosie

Step-by-step explanation:

-1×5 1×3

------ -----

3×5 -5×3

=

-5/15 -3/15

-5/15×4/4 ;

-3/15×4/4

=

-20/60 ;

-12/60

you can write any 8 rational no. between them

please mark as brainliest answer as it will also give you 3 pts

8 0

3 years ago

Pls help A bag contains 40 colored tiles. There are 12 red, 17 blue, 8 yellow and 3 green. One tile is drawn at

aleksley [76]

D blue I think, it has the most tiles in the bag 17/40 while all the other fractions are smaller decreasing their chances

4 0

3 years ago

Please help with any of this Im stuck and having trouble with pre calc is it basic triogmetric identities using quotient and rec

german

How I was taught all of these problems is in terms of r, x, and y. Where sin = y/r, cos = x/r, tan = y/x, csc = r/y, sec = r/x, cot = x/y. That is how I will designate all of the specific pieces in each problem.

#3

Let's start with sin here. $\frac{2\sqrt{5}}{5} = \frac{2}{\sqrt{5}}$ Therefore, because sin is y/r, r = $\sqrt{5}$ and y = +2. Moving over to cot, which is x/y, x = -1, and y = 2. We know y has to be positive because it is positive in our given value of sin. Now, to find cos, we have to do x/r.

cos = $\frac{-1}{\sqrt{5}} = \frac{-\sqrt{5}}{5}$

#4

Let's start with secant here. Secant is r/x, where r (the length value/hypotenuse) cannot be negative. So, r = 9 and x = -7. Moving over to tan, x must still equal -7, and y = $4\sqrt{2}$ . Now, to find csc, we have to do r/y.

csc = $\frac{9}{4\sqrt{2}} = \frac{9\sqrt{2}}{8}$

The pythagorean identities are

sin^2 + cos^2 = 1,

1 + cot^2 = csc^2,

tan^2 + 1 = sec^2.

#5

Let's take a look at the information given here. We know that cos = -3/4, and sin (the y value), must be greater than 0. To find sin, we can use the first pythagorean identity.

sin^2 + (-3/4)^2 = 1

sin^2 + 9/16 = 1

sin^2 = 7/16

sin = $\sqrt{7/16}$ = $\frac{\sqrt{7}}{4}$

Now to find tan using a pythagorean identity, we'll first need to find sec. sec is the inverse/reciprocal of cos, so therefore sec = -4/3. Now, we can use the third trigonometric identity to find tan, just as we did for sin. And, since we know that our y value is positive, and our x value is negative, tan will be negative.

tan^2 + 1 = (-4/3)^2

tan^2 + 1 = 16/9

tan^2 = 7/9

tan = - $\sqrt{7/9} = \frac{-\sqrt{7}}{3}$

#6

Let's take a look at the information given here. If we know that csc is negative, then our y value must also be negative (r will never be negative). So, if cot must be positive, then our x value must also be negative (a negative divided by a negative makes a positive). Let's use the second pythagorean identity to solve for cot.

1 + cot^2 = $(\frac{-\sqrt{6}}{2})^{2}$

1 + cot^2 = 6/4

cot^2 = 2/4

cot = $\frac{\sqrt{2}}{2}$

tan = $\sqrt{2}$

Next, we can use the third trigonometric identity to solve for sec. Remember that we can get tan from cot, and cos from sec. And, from what we determined in the beginning, sec/cos will be negative.

$(\frac{2}{\sqrt{2}})^2$ + 1 = sec^2