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aksik [14]
3 years ago
14

Which description is the basis of proof that sin^2 x + cos^2 x= 1 ?

Mathematics
1 answer:
Vika [28.1K]3 years ago
8 0
Answer: Choice B
Equation of a circle with (0,0) as its center and radius = 1

---------------------------------------------------------------

Explanation:

Recall that the unit circle is the equation 
x^2 + y^2 = 1
This is a circle with center (h,k) = (0,0) and radius r = 1
It is in the form 
(x-h)^2 + (y-k)^2 = r^2

We can write the first equation mentioned in the form sin^2(t)+cos^2(t) = 1
Using 
x = cos(t)
y = sin(t)
t = angle

The point P(x,y) on the unit circle has an x coordinate of cos(t) and a y coordinate of sin(t). We can see this through drawing a right triangle. 
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Work out<br> 2/5<br> of 60<br> Pls work this out. It’s urgent!
xxTIMURxx [149]
Answer is 24.

firsty remember OF= MULTIPLY

NOW we can calculate this easily

here the equation is given

2/5×60. u can solve this in given ways below….…..

1……….

u can simply divided 60 by 5. which is equal to 12. and now the equation remain 2×12 . & it's equal to 24. which is our answer.

2……….

u can write 2/5 in decimals by dividing this . so 2/5 is equal to 0.4

and now u r left with 0.4×60

so after multiplying it comes 24.0

and that is equal to 24. (because after decimal zeros has no value)

3……….

and u also can multiply 2 by 60 . it occurs 120 then devide it by 5 . the answer comes 24.
5 0
2 years ago
A sample survey interviews SRSs of 500 female college students and 550 male college students. Researchers want to determine whet
stira [4]

Answer:

Option (d) and (e)

Step-by-step explanation:

We are given that a sample survey interviews SRS of 500 female college students and 550 male college students. And Researchers want to determine whether there is a difference in the proportion of male and female college students who worked for pay last summer.

In all, 410 of the females and 484 of the males say they worked for pay last summer.

From this, Null Hypothesis, H_0 : p_1 = p_2 {means proportion of male and female college students who worked for pay last summer are same}

Alternate Hypothesis, H_1 : p_1\neq p_2 {means there is a difference in the proportion of male and female college students who worked for pay last summer}

Now, since we know that results were statistically significant at the 1% level.

So, if p-value is less than the significance level ⇒ we will reject H_0

     if p-value is more than the significance level ⇒ we will not reject H_0

From the options given it is sure that P-value is less than 1%,i.e.;

<em>P-value is less than the significance level, so we will reject null hypothesis and conclude that there is convincing evidence that the proportion of all male college students in the study who worked for pay last summer is different from the proportion of all female college students in the study who worked for pay last summer. </em>

Also, option (d) and (e) are same in my opinion.

4 0
3 years ago
Original:45 new:30 percent of change
Sveta_85 [38]

<u>Answer:</u>

<h2>-33.33%</h2>

<u>Explanation:</u>

30-45 = -15

percent of change = 100(-15/45) = -33.33%

6 0
3 years ago
5x + 2y = -10<br> 3x + y = 66<br><br> please show work
liq [111]

Answer:

x=142, y=-360. (142, -360).

Step-by-step explanation:

5x+2y=-10

3x+y=66

----------------

5x+2y=-10

-2(3x+y)=-2*66

-----------------------

5x+2y=-10

-6x-2y=-132

-------------------

-x=-142

x=142

3(142)+y=66

426+y=66

y=66-426

y=-360

8 0
2 years ago
Read 2 more answers
The graph shows the volume of oil (L litres) in a tank at time t seconds. a)find the gradient of the graph b)explain what this g
Alisiya [41]

Answer/Step-by-step explanation:

a. Using two points on the line (0, 2) and (4, 12), find the gradient:

Gradient = ∆y/∆x = (12 - 2) / (4 - 0) = 10/4 = 2.5

Gradient = 2.5

b. The gradient, 2.5, represents the unit rate. That is, the volume of the oil in the tank per second.

i.e. 2.5 litres per second

c. The value, L = 2 represents the y-intercept, which is the starting value or initial volume of oil in the tank at 0 secs.

4 0
2 years ago
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