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Suppose parametric equations for the line segment between (6,9) and (0,0) have the form: x = a+bt y = c+dt If the parametric cur

olasank [31]

What is 0.05 Equal to 0.5.

6 0

2 years ago

Write an equation of the circle with center (-6, 2) and diameter 10.

marshall27 [118]

Solution:

We know that the formula to write the equation of a circle is (x – h)² + (y – k)² = r², where h is the x coordinate, k is the y-coordinate, r is the radius of circle, and the center of circle is (-6,2).

Finding the radius of the circle.

Diameter: 10 units
Radius: 10/2 units = 5 units

Creating the equation:

(x – h)² + (y – k)² = r²
=> (x + 6)² + (y – 2)² = 5²

The equation is (x + 6)² + (y – 2)² = 5².

Learn more: brainly.com/question/26573763

6 0

1 year ago

2 tan 30° II 1 + tan- 300

shusha [124]

Question:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$

Answer:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= sin(60^{\circ})$

Step-by-step explanation:

Given

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$

Required

Simplify

In trigonometry:

$tan(30^{\circ}) = \frac{1}{\sqrt{3}}$

So, the expression becomes:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2 * \frac{1}{\sqrt{3}}}{1 + (\frac{1}{\sqrt{3}})^2}$

Simplify the denominator

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2 * \frac{1}{\sqrt{3}}}{1 + \frac{1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{1 + \frac{1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{ \frac{3+1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{ \frac{4}{3}}$

Express the fraction as:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2}{\sqrt 3} / \frac{4}{3}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2}{\sqrt 3} * \frac{3}{4}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{1}{\sqrt 3} * \frac{3}{2}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3}{2\sqrt 3}$

Rationalize

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3}{2\sqrt 3} * \frac{\sqrt{3}}{\sqrt{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3\sqrt{3}}{2* 3}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\sqrt{3}}{2}$

In trigonometry:

$sin(60^{\circ}) = \frac{\sqrt{3}}{2}$

Hence:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= sin(60^{\circ})$

3 0

3 years ago

Will choose brainliest! Please help! (This is Khan Academy)

NeTakaya

Answer:

Option B. A = (5/6)^-⅛

Step-by-step explanation:

From the question given above, we obtained:

(5/6)ˣ = A¯⁸ˣ

We can obtain the value of A as follow:

(5/6)ˣ = A¯⁸ˣ

Cancel x from both side

5/6 = A¯⁸

Recall:

M¯ⁿ = 1/Mⁿ

A¯⁸ = 1/A⁸

Thus,

5/6 = 1/A⁸

Cross multiply

5 × A⁸ = 6

Divide both side by 5

A⁸ = 6/5

Take the 8th root of both sides

A = ⁸√(6/5)

Recall

ⁿ√M = M^1/n

Thus,

⁸√(6/5) = (6/5)^⅛

Therefore,

A = (6/5)^⅛

Recall:

(A/B)ⁿ = (B/A)¯ⁿ

(6/5)^⅛ = (5/6)^-⅛

Therefore,

A = (5/6)^-⅛

8 0

3 years ago

Please help me! (this is from khan academy 6th grade math!)

Svetach [21]

Answer:

1st → 1st

2nd → 3rd

3rd → 2nd

Step-by-step explanation:

just divide the first set of student wishes to the second set of students in the same question. the 1st for example

x-ray to all students is 6/ 36= 1/6 = 1:6

for all students, you add all the students in the table

8 0

1 year ago