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LiRa [457]
3 years ago
10

What is the value of 3^ -2 as a fraction

Mathematics
1 answer:
Rainbow [258]3 years ago
6 0

Answer:

3^-2 = 0.1111111111

0.1111111111= 1/9

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6x – 3y = 5<br>y – 2x= 8​
noname [10]

Answer:

PA GEN SOLISYON --> "NO SOLUTION"

Step-by-step explanation:

Whether you multiply the top equation by ⅓ to make "-y" and "2x", or multiply the bottom equation by 3 to make "3y" and "-6x", you will see that 9⅓ ≠ 0, or 24 ≠ 0, therefore the result is "NO SOLUTION".

6 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

4/2 + 1 = sec^2

2 + 1 = sec^2

sec^2 = 3

sec = -\sqrt{3}

cos = \frac{-\sqrt{3}}{3}

Hope this helps!! :)

3 0
3 years ago
Read 2 more answers
Usama needs to read 6 novels each month.
Delicious77 [7]

Answer:

78 novels

Step-by-step explanation:

N= novels

M= month

6N = 1M

? = 13M

6N × 13M= 78NM

78NM ÷ 1M

cancel M, therefore;

the number of novels Usama needs to read in 13 months is 78.

5 0
3 years ago
Read 2 more answers
Annie has $18.00 to spend at the grocery store. She is going to buy bags of nuts that cost $5.70 each and one box of cookies tha
Tcecarenko [31]

Answer: She can buy up to 2 bags of nuts

Step-by-step explanation:

Hi, to answer this question we have to write an inequality:

The product of the number of bags of nuts bought (n) and the price per bag (5.70); plus the price of one box of cookies (4.25) must be less or equal to Annie's money (18).

5.70n +4.25 ≤18

Solving for x:

5.70n ≤18 -4.25

5.70n ≤13.75

n ≤13.75/5.70

n ≤2.41  

n ≤2 (rounded)

She can buy up to 2 bags of nuts

8 0
3 years ago
Read 2 more answers
How do you reduce fractions?
Irina18 [472]
4/12=1/3
8/14=4/7
6/8=3/4
3/6=1/2
5/15=1/3
4 0
3 years ago
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