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Len [333]
2 years ago
3

Liberty Middle School has 600 students. In Anna’s class, 3 out of 8 students walk to school. How many students at the school can

be expected to walk to school?
Mathematics
1 answer:
Alexeev081 [22]2 years ago
3 0

Answer:

um

Step-by-step explanation:

3/8×600

=225 student walk home

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(a) Find a vector-parametric equation r⃗ 1(t)=⟨x(t),y(t),z(t)⟩r→1(t)=⟨x(t),y(t),z(t)⟩ for the shadow of the circular cylinder x2
motikmotik

Answer: (a) r1(t) = <2cost , 0 , 2sint>

(b) <2cost , (1 - 12sint - 10cost)/8 , 2sint>

Step-by-step explanation:

x2+z2=4

a)

Now, in the xz plane, we know that y = 0...

So, x^2 + z^2 = 4 will simply be a circle centered at (0,0)..

This can be easily parameterized as

x = 2cos(t)

z = 2sin(t)

So, the required parameterization is :

r1(t) = <2cost , 0 , 2sint>

b)

Cylinder : x^2 + z^2 = 4

Plane : 5x+8y+6z=1

Easily enough, the x^2 + z^2 = 4 can again be parameterized as

x = 2cost , z = 2sint

With this, we can find y using plane equation...

5x+8y+6z=1

5(2cost) + 8y + 6(2sint) = 1

8y = 1 - 12sint - 10cost

y = (1 - 12sint - 10cost)/8

So, the parameterization is :

<2cost , (1 - 12sint - 10cost)/8 , 2sint>

6 0
2 years ago
Select the correct answer. Gary throws a parachute toy from the roof of a building which has a height of 96 feet. The toy reache
Artyom0805 [142]

Answer:

Z

Step-by-step explanation:

The height needs to start at 96

8 0
2 years ago
A bright blue parallelogram is painted on a playground. It has a height of 16 feet and a length of 23 feet. If one gallon of pai
vlada-n [284]

Answer:3 gallons of paint

Step-by-step explanation:multiply 23 times 16 and then divide by 175.

6 0
2 years ago
ΔABC underwent a sequence of rigid transformations to give ΔA′B′C′. Which transformations might have taken place?
Papessa [141]

Answer: The correct option is second, a rotation 90^{\circ} clockwise about the origin followed by a reflection across the x-axis.

Explanation:

From the given figure it is noticed that the vertices of ΔABC are A(-6,4), B(-4,6), C(-2,2) and vertices of ΔA'B'C' are A'(4,-6), B(6,-4), C(2,-2).

It means if the point is P(x,y) then after transformation it will be P'(y,x).

If a point P(x,y) reflection across the y-axis followed by a reflection across the x-axis, then the image of point after transformation will be P'(-x,-y), therefore it is not the correct option.

If a shape is rotated 90^{\circ} clockwise about the origin then the  point P(x,y) will be P'(y,-x) and after that reflect across the x-axis, so the point after transformation will be P'(y,x), therefore it is the correct option.

If a shape is rotated 270^{\circ} clockwise about the origin then the  point P(x,y) will be P'(-y,x) and after that reflect across the x-axis, so the point after transformation will be P'(-y,-x), therefore it is not the correct option.

If a point P(x,y) reflection across the x-axis followed by a reflection across the y-axis, then the image of point after transformation will be P'(-x,-y), therefore it is not the correct option.

Hence, the correct option is second, a rotation 90^{\circ} clockwise about the origin followed by a reflection across the x-axis.

4 0
2 years ago
Read 2 more answers
Which of the following best explains how this relationship and the value of sin theta can be used to find the other trigonometri
Lena [83]

Answer:

OPtion I is right

Step-by-step explanation:

Once we know sin of an angle, and it lies in II quadrant, we know that

cos, sec, tan and cot would be negative but csc will be positive.

So use the fact that

sin^{2}t+cos^{2} t=1\\cost =-\sqrt{1-sin^2 t}

Thus cos is obtained using negative square root.

Now tan = sin/cos, and sec = 1/cos:

cot =1/tan and csc =1/sin

Thus all value can be obtained easily

So option I

8 0
2 years ago
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