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3 years ago

11

You’re given side AB with a length of 6 centimeters and side BC with a length of 5 centimeters. The measure of angle A is 30°. H

ow many triangles can you construct using these measurements?
a. 0
b. 1
c. 2
d. infinitely many

1 answer:

Helen [10]3 years ago

3 0

The full thing would be 90 but when you look the answer would be 2
because you divide it by the even number

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Mrs. Piper is driving Anna, Jo, and Mai home from school. All of them want to ride in the front seat. How can she make a fair de

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Answer:

I would say the answers would be A and B. This is because they each get a fair chance.

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3 years ago

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Which of the following proportions can be used to find the value of n?

MakcuM [25]

Answer:

B. $\frac{n}{11} =\frac{15}{5}$

Step-by-step explanation:

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which has the greater volume? 1.a piece of paper rolled into a cylinder lengthwise, or the same piece of paper rolled into a cyl

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Neither?
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2 years ago

Find an equation for the plane y=4x in cylindrical coordinates. (Type theta for θ in your answer.)

Paha777 [63]

Answer:

<h2><em>sinθ - 4cosθ = 0</em></h2>

Step-by-step explanation:

Given the equation of a plane in rectangular coordinates to be y = 4x.

The cylindrical coordinates of the axis is as given below;

x = rcosθ

y = rsinθ

z = z

Since there is no z coordinate in the equation of the plane given, we will only substitute x = rcosθ and y = rsinθ into the equation y = 4x and simply the result as shown;

y = 4x

rsinθ = 4( rcosθ)

rsinθ = 4rcosθ

sinθ = 4cosθ

sinθ - 4cosθ = 0

<em>Hence the equation for the plane in cylindrical coordinate is expressed as sinθ - 4cosθ = 0</em>

8 0

2 years ago

evaluate the line integral ∫cf⋅dr, where f(x,y,z)=5xi−yj+zk and c is given by the vector function r(t)=⟨sint,cost,t⟩, 0≤t≤3π/2.

meriva

We have

$\displaystyle \int_C \vec f \cdot d\vec r = \int_0^{\frac{3\pi}2} \vec f(\vec r(t)) \cdot \dfrac{d\vec r}{dt} \, dt$

and

$\vec f(\vec r(t)) = 5\sin(t) \, \vec\imath - \cos(t) \, \vec\jmath + t \, \vec k$

$\vec r(t) = \sin(t)\,\vec\imath + \cos(t)\,\vec\jmath + t\,\vec k \implies \dfrac{d\vec r}{dt} = \cos(t) \, \vec\imath - \sin(t) \, \vec\jmath + \vec k$

so the line integral is equilvalent to

$\displaystyle \int_C \vec f \cdot d\vec r = \int_0^{\frac{3\pi}2} (5\sin(t) \cos(t) + \sin(t)\cos(t) + t) \, dt$

$\displaystyle \int_C \vec f \cdot d\vec r = \int_0^{\frac{3\pi}2} (6\sin(t) \cos(t) + t) \, dt$

$\displaystyle \int_C \vec f \cdot d\vec r = \int_0^{\frac{3\pi}2} (3\sin(2t) + t) \, dt$

$\displaystyle \int_C \vec f \cdot d\vec r = \left(-\frac32 \cos(2t) + \frac12 t^2\right) \bigg_0^{\frac{3\pi}2}$

$\displaystyle \int_C \vec f \cdot d\vec r = \left(\frac32 + \frac{9\pi^2}8\right) - \left(-\frac32\right) = \boxed{3 + \frac{9\pi^2}8}$

7 0

2 years ago