What is the midpoint of OA ?A: (2m, n)


B: (m, n)


C: (m – n, 0)


D: (m, 2n)
What is the midpoint of OA ? A: (2m, n) - 1

Answers

Answer 1
Answer:

Answer:

(m,n)

Step-by-step explanation:

Since O is at the origin, we can take the coordinates of point A and divide them by 2 to find the midpoint

( 2m/2,2n/2)

The midpoint is at (m,n)


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Can someone help me in this trig question, please? thanks A person is on the outer edge of a carousel with a radius of 20 feet that is rotating counterclockwise around a point that is centered at the origin. What is the exact value of the position of the rider after the carousel rotates 5pi/12

Answers

The exact value of the position of the rider after the carousel rotates 5π/12 is 5 (-√2 + √6), 5(√2 + √6).

The position

Since the position of the carousel is (x, y) = (20cosθ, 20sinθ) and we need to find the position when θ = 5π/12 = 5π/12 × 180 = 75°

So, substituting the value of θ into the positions, we have

(20cos75°, 20sin75°)

The value of 20cos75°

20cos75° = 20cos(45 + 30)

Using the compound angle formula

cos(A + B) = cosAcosB - sinAsinB

With A = 45 and B = 30

cos(45 + 30) = cos45cos30 - sin45sin30

= 1/√2 × √3/2 - 1/√2 × 1/2

= 1/2√2(√3 - 1)

= 1/2√2(√3 - 1) × √2/√2

= √2(√3 - 1)/4

= (√6 - √2)/4

= (-√2 + √6)/4

So, 20cos75° = 20 × (-√2 + √6)/4

= 5 (-√2 + √6)

The value of 20sin75°

20sin75° = sin(45 + 30)

Using the compound angle formula

sin(A + B) = sinAcosB + cosAsinB

With A = 45 and B = 30

sin(45 + 30) = sin45cos30 + cos45sin30

= 1/√2 × √3/2 + 1/√2 × 1/2

= 1/2√2(√3 + 1)

= 1/2√2(√3 + 1) × √2/√2

= √2(√3 + 1)/4

= (√6 + √2)/4

= (√2 + √6)/4

So, 20sin75° = 20 × (√2 + √6)/4

= 5(√2 + √6)

Thus, (20cos75°, 20sin75°) = 5 (-√2 + √6), 5(√2 + √6).

So, the exact value of the position of the rider after the carousel rotates 5π/12 is 5 (-√2 + √6), 5(√2 + √6).

Learn more about position here:

brainly.com/question/11001232

\bf \textit{the position of the rider is clearly }20cos\left( (5\pi )/(12) \right)~~,~~20sin\left( (5\pi )/(12) \right)\n\n-------------------------------\n\n\cfrac{5}{12}\implies \cfrac{2+3}{12}\implies \cfrac{2}{12}+\cfrac{3}{12}\implies \cfrac{1}{6}+\cfrac{1}{4}\n\n\n\textit{therefore then }\qquad \cfrac{5\pi }{12}\implies \cfrac{1\pi }{6}+\cfrac{1\pi }{4}\implies \cfrac{\pi }{6}+\cfrac{\pi }{4}\n\n-------------------------------

\bf \textit{Sum and Difference Identities}\n\nsin(\alpha + \beta)=sin(\alpha)cos(\beta) + cos(\alpha)sin(\beta)\n\ncos(\alpha + \beta)= cos(\alpha)cos(\beta)- sin(\alpha)sin(\beta)\n\n-------------------------------\n\ncos\left( (\pi )/(6)+(\pi )/(4) \right)=cos\left( (\pi )/(6)\right)cos\left((\pi )/(4) \right)-sin\left( (\pi )/(6)\right)sin\left((\pi )/(4) \right)

\bf cos\left( (\pi )/(6)+(\pi )/(4) \right)=\cfrac{√(3)}{2}\cdot \cfrac{√(2)}{2}-\cfrac{1}{2}\cdot \cfrac{√(2)}{2}\implies \cfrac{√(6)}{4}-\cfrac{√(2)}{4}\implies \boxed{\cfrac{√(6)-√(2)}{4}}\n\n\nsin\left( (\pi )/(6)+(\pi )/(4) \right)=sin\left( (\pi )/(6)\right)cos\left( (\pi )/(4) \right)+cos\left( (\pi )/(6)\right)sin\left((\pi )/(4) \right)

\bf sin\left( (\pi )/(6)+(\pi )/(4) \right)=\cfrac{1}{2}\cdot \cfrac{√(2)}{2}+\cfrac{√(3)}{2}\cdot \cfrac{√(2)}{2}\implies \cfrac{√(2)}{4}+\cfrac{√(6)}{4}\implies \boxed{\cfrac{√(2)+√(6)}{4}}\n\n-------------------------------\n\n20\left( \cfrac{√(6)-√(2)}{4} \right)\implies 5(-√(2)+√(6))\n\n\n20\left( \cfrac{√(2)+√(6)}{4} \right)\implies 5(√(2)+√(6))

I need help w this soon

Answers

Answer:

18

Step-by-step explanation:

Answer:

18 because it's asking for the tw

Which of the following scenarios could be represented by the expression 6x+11y? A. The total weight of a box containing x 11-ounce packages of almonds and y 6-ounce packages of peanuts

B. The cost for x children and y adults to go to the movies if the adult tickets are $11 and the child tickets are $6

C. The number of kids in a school class if there are 6 girls and 11 boys

D. The total number of marbles in a jar if there are 11 green and 6 red

Answers

The scenarios that best represent the relationship between the variables would be B. It not only properly defines the variables but also, uses the correct pairs of numbers to go along with the variables of x and y.






Nick is solving the equation 3x^2=20-7x with the quadratic formula which values could He use for a, b and c

Answers

Answer:

C. a = 3

b = 7

c = -20

Step-by-step explanation:

x = -b+/- (sq) b^2 - 4ac.

Well the way to plug the numbers in is by using this equation: Ax^2 + Bx + C = 0

S if you use your equation : 3x^2 = 20 - 7x

First you have to move everything to one side.

So it'll look like this:

3x^2 = 20 - 7x

-20      -20

3x^2 - 20 = -7x

+7x  +7x

3x^2 - 20 + 7x = 0

don't forget that... Ax^2 +Bx+ C= 0

3 = A

7 = B

-20 = C

What is 15 - 1/6n = 1/6n - 1

Answers

15-1/6n=1/6n-1
15+1=1/6n+1/6n
16=1+1/6n
16=2/6n
16=1/3n
16*3/1=n
therefore, n=48.........

Answer:

n=48

Step-by-step explanation:

(5,?) is on the line below. Find the other half of the coordinate.
y = 2/3x + 7

Answers

We need to plug 5 in place of x into the equation of the line and find the y-value.

y = (2/3)x + 7

y = (2/3)5 + 7

y = 10/3 + 7

y = 10/3 + 21/3

y = 31/3

y = 10 1/3

The point is (5, 10 1/3) OR (5, 31/3).
(x,y) = (5, ?), so x = 5. Substitute 5 into the function for x to find the value of y when x = 5. y = 2/3 (5) + 7 = (2 x 5) / 3 + 7 = 10/3 + 7 = 10/3 + 21/3 = 31/3, which equals 10 1/3 or 10.3 repeating.