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3 years ago

If cot xº = 3/4 what is the value of b?

Mathematics

1 answer:

balandron [24]3 years ago

3 0

Answer:

$\cot(x) = \frac{3}{4} = \frac{10.5}{2b}$

$4 \times 10.5 = 2 b \times 3$

$42 = 6b$

$b = 7$

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The curves r1(t) = 2t, t2, t4 and r2(t) = sin t, sin 5t, 2t intersect at the origin. Find their angle of intersection, θ, correc

masya89 [10]

Answer:

Therefore the angle of intersection is $\theta =79.48^\circ$

Step-by-step explanation:

Angle at the intersection point of two carve is the angle of the tangents at that point.

Given,

$r_1(t)=(2t,t^2,t^4)$

and $r_2(t)=(sin t , sin5t, 2t)$

To find the tangent of a carve , we have to differentiate the carve.

$r'_1(t)=(2,2t,4t^3)$

The tangent at (0,0,0) is [ since the intersection point is (0,0,0)]

$r'_1(0)=(2,0,0)$ [ putting t= 0]

$|r'_1(0)|=\sqrt{2^2+0^2+0^2} =2$

Again,

$r'_2(t)=(cos t ,5 cos5t, 2)$

The tangent at (0,0,0) is

$r'_2(0)=(1 ,5, 2)$ [ putting t= 0]

$|r'_1(0)|=\sqrt{1^2+5^2+2^2} =\sqrt{30}$

If θ is angle between tangent, then

$cos \theta =\frac{r'_1(0).r'_2(0)}{|r'_1(0)|.|r'_2(0)|}$

$\Rightarrow cos \theta =\frac{(2,0,0).(1,5,2)}{2.\sqrt{30} }$

$\Rightarrow cos \theta =\frac{2}{2\sqrt{30} }$

$\Rightarrow cos \theta =\frac{1}{\sqrt{30} }$

$\Rightarrow \theta =cos^{-1}\frac{1}{\sqrt{30} }$

$\Rightarrow \theta =79.48^\circ$

Therefore the angle of intersection is $\theta =79.48^\circ$ .

8 0

3 years ago

What is the sum of 1 8 + 5 16 + 3 8 ?

Alexus [3.1K]

Answer:

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Could i have help??? There is 2 questions.

liberstina [14]

First one:
3x+5<23
3x<18 (subtract 5 from both sides)
x<6
So A: less than 6 cars

Second one:
3(x-4)+5(2x+1)
3x-12+10x+5 (distribute)
So the answer is C

Hope this helps

6 0

3 years ago

The length of a rectangle is (2x+5) units and the breadth is (3x-1) units .Find the perimeter and the area of the rectangle.

jolli1 [7]

Answer:

Perimeter = 10x + 8 units.

Area = 6x^2 + 13x - 5 units^2.

Step-by-step explanation:

The perimeter is 2*length + 2*breadth

= 2(2x + 5) + 2(3x - 1)

= 4x + 10 + 6x - 2

= 10x + 8 units.

The area = length * breadth

= (2x + 5)(3x - 1)

= 6x^2 - 2x + 15x - 5

= 6x^2 + 13x - 5 units^2.

7 0

3 years ago

HELP with these questions

zlopas [31]

Step-by-step explanation:

transform the parent graph of f(x) = ln x into f(x) = - ln (x - 4) by shifting the parent graph 4 units to the right and reflecting over the x-axis

(???, 0): 0 = - ln (x - 4)

$\frac{0}{-1} = \frac{-ln (x - 4)}{-1}$

0 = ln (x - 4)

$e^{0} = e^{ln (x - 4)}$

1 = x - 4

+4 +4

5 = x

(5, 0)

(???, 1): 1 = - ln (x - 4)

$\frac{0}{-1} = \frac{-ln (x - 4)}{-1}$

1 = ln (x - 4)

$e^{1} = e^{ln (x - 4)}$

e = x - 4

+4 +4

e + 4 = x

6.72 = x

(6.72, 1)

Domain: x - 4 > 0

+4 +4

x > 4

(4, ∞)

Vertical asymptotes: there are no vertical asymptotes for the parent function and the transformation did not alter that

No vertical asymptotes

*************************************************************************

transform the parent graph of f(x) = 3ˣ into f(x) = - 3ˣ⁺⁵ by shifting the parent graph 5 units to the left and reflecting over the x-axis

Domain: there is no restriction on x so domain is all real number

(-∞, ∞)

Range: there is a horizontal asymptote for the parent graph of y = 0 with range of y > 0. the transformation is a reflection over the x-axis so the horizontal asymptote is the same (y = 0) but the range changed to y < 0.

(-∞, 0)

Y-intercept is when x = 0:

f(x) = - 3ˣ⁺⁵

= - 3⁰⁺⁵

= - 3⁵

= -243

Horizontal Asymptote: y = 0 (explanation above)

5 0

3 years ago