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Nadya [2.5K]
3 years ago
6

At what points will the line y=x intersect the unit circle x^2+y^2=1

Mathematics
2 answers:
maw [93]3 years ago
8 0
Graph it or solve 2x^2 = 1, x=+-1/Sqrt(2) = +-0.707
and the same for y.
Points are (-.707, -.707) and (.707,.707) or (1/Sqrt(2).....
irina [24]3 years ago
7 0
I dont know what grade are you in
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Iliana is painting a picture. She has green, red, yellow, purple, orange, and blue paint. She wants her painting to have four di
ivanzaharov [21]

The number of ways for which she could pick four colours if green must be one of them is; 10 ways.

<h3>How many ways can she picks four colours if green must be there?</h3>

It follows from the task that there are 6 colours in total that she could pick from.

Hence, since she needs four colours with green being one of them, it follows that she only has 3 colours to pick from 5.

Hence, the numbers of possible combinations is; 5C3 = 10 ways.

Read more on combinations;

brainly.com/question/2280043

#SPJ1

5 0
2 years ago
Can Anyone please solve this? Will mark brainiest!!!!
sergey [27]

Answer:

  (a) see attached

  (b) CD = 29

Step-by-step explanation:

<h3>(a) </h3>

See the attachment for a diagram.

In the diagram, we have shown point E on the plane so that BE is parallel to AC.

__

<h3>(b)</h3>

Triangle BDE is a right triangle with hypotenuse BD = 25 and leg BE = AC = 15. The length DE is given by the Pythagorean theorem as ...

  DE = √(BD^2 -BE^2) = √(625 -225) = 20

Triangle CDE is a right triangle with hypotenuse CD and legs CE = 21 and DE = 20. The length of the hypotenuse is also given by the Pythagorean theorem as ...

  CD = √(CE^2 +DE^2) = √(21^2 +20^2) = √841

  CD = 29

8 0
2 years ago
What is the simplified form of the fifth root of x to the fourth power times the fifth root of x to the fourth power
elena55 [62]
I believe the following is your problem (if not do rectify me). If so, then:

⁵√x⁴ .⁵√x⁴
1st method: 
⁵√x⁴ .⁵√x⁴ = x⁴/⁵ . x⁴/⁵ = x⁽⁴/⁵ +x⁴/⁵⁾ = x⁸/⁵ = ⁵√x⁸ = ⁵√(x⁵.x³) = x. ⁵√x³
2nd method:
⁵√x⁴ . ⁵√x⁴ = ⁵√(x⁴. x⁴) = ⁵√(x⁴⁺⁴) = ⁵√x⁸ = x .⁵√x³ 
3 0
3 years ago
Read 2 more answers
HELP ASAP!!!!!!!!!!!!!!
yawa3891 [41]

Answer:

10

Step-by-step explanation:

just try to make the lines even and get the answer 10

7 0
3 years ago
Read 2 more answers
You have given an equal sided triangle with side length a. A straight line connects the center
GarryVolchara [31]

Answer:

Where α is an acute angle (first figure)

The area of the shaded triangle = ((√3)·a²/4)·sin(α)·csc(120 - α))

Where α is an obtuse angle (second figure)

The required area of the shaded region = (√3)·a²/4 + (√3)·a²/4)·sin(α)·sec(α + π/6)

Step-by-step explanation:

Where α is an acute angle (first figure)

The given parameters are;

The given triangle = Equilateral Triangle

Let the sides of the equilateral triangle = 2·a

Therefore;

The measure of each interior angles of the given triangle = 60°

Let c represent the side of the shaded triangle opposite ∠α and b represent the side of the shaded triangle opposite ∠60° and c, represent the third side of the shaded triangle, we have;

The sides of the equilateral triangle = 2·a

By sine rule, we have;

c/sin(α) = b/sin(60°) = a/sin(180 - (60 + α)) = a/sin(120 - α))

b = sin(60°) × a/sin(120 - α)) = (√3)/2 × a/sin(120 - α))

The area of the shaded triangle = 1/2 × a × b × sin(α) = 1/2 × a × (√3)/2 × a/sin(120 - α)) × sin(α) = ((√3)·a²/4)·sin(α)·csc(120 - α))

The area of the shaded triangle = ((√3)·a²/4)·sin(α)·csc(120 - α))

Where α is an obtuse angle (second figure)

The required area of the shaded region = The area of the equilateral triangle - The area of the small unshaded triangle, with base side a and interior angles, (180° - α), 60° and ((180 - (180° - α) - 60°) = ) α - 60°

The area of the unshaded triangle is found as follows;

By sine rule, we have;

c/sin(180° - α) = b/sin(60°) = a/sin(α - 60°)

b = sin(60°) × a/sin(α - 60°) = (√3)/2 × a/sin(α - 60°)

The area of the unshaded triangle = 1/2 × a × b × sin(α) = 1/2 × a × (√3)/2 × a/sin(α - 60°) × sin(α) = -((√3)·a²/4)·sin(α)·sec(α + π/6)

The area of the shaded triangle =  -((√3)·a²/4)·sin(α)·sec(α + π/6)

The required area of the shaded region = 1/2×a²·sin(60°)  - (-((√3)·a²/4)·sin(α)·sec(α + π/6))

The required area of the shaded region = (√3)·a²/4 + (√3)·a²/4)·sin(α)·sec(α + π/6)

4 0
3 years ago
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