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LiRa [457]
3 years ago
6

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 <br> 1<br> 8<br> +<br> 5<br> 16<br> +<br> 3<br> 8<br> ?
Alexus [3.1K]

Answer:

41

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Could i have help??? <br><br> 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]

<u>Step-by-step explanation:</u>

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

          <u> +4 </u>  <u>    +4 </u>

             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

          <u> +4 </u>   <u>    +4 </u>

         e + 4 = x

          6.72 = x

(6.72, 1)

Domain: x - 4 > 0

                <u>  +4 </u>  <u>+4  </u>

               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  <em>(explanation above)</em>

5 0
3 years ago
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