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3 years ago

Rewrite each pair of unlike fractions as like fractions 1/2,1/4

1 answer:

5 0

Example:
1. 2/3,3/4
2. 3/5,1/2 
3. 2/5,1/3 
4. 5.6,4.5 
5. 1/2,2/3 
6. 3/7,2/3 
7. 1/3,3/10 
8. 2/5,3/7

Answer:

1. 8/12, 9/12
2. 6/10, 5/10 
3. 6/15, 5/15 
4. 25/30, 24/30 
5. 3/6, 4/6 
6. 9/21, 14/21 
7. 10/30, 9/30 
8. 14/35, 15/35

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50.625 miles per hour

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You are standing 30 meters from a hot air balloon that is preparing to take

Lerok [7]

Answer:

The distance to the top of the balloon is approximately 33.977 meters

Step-by-step explanation:

The (horizontal) distance from the hot air balloon from the observer, l = 30 m

The angle of elevation to the top of the balloon, θ = 28°

With the balloon standing upright, the distance to the top of the balloon, d, the height of the balloon, h and the horizontal distance to the balloon, l, form a right triangle, where;

d = The hypotenuse side

l = The leg adjacent to the reference (given) angle,

h = The leg opposite to the given angle, θ

By trigonometric ratios, we have;

$cos\angle X = \dfrac{Adjacent\ leg \ length}{Hypotenuse \ length}$

From which we get;

$cos\angle \theta = \dfrac{l}{d}$

$cos(28^{\circ}) = \dfrac{30 \ m}{d}$

$d = \dfrac{30 \ m}{cos(28^{\circ}) } \approx 33.977 \, m$

The distance to the top of the balloon, d ≈ 33.977 m

8 0

2 years ago

suppose the line segment whose endpoints are H(5,0) and I(-6,-3) is reflected over the y axis? what are the coordinates of H’I’?

padilas [110]

Answer:

see explanation

Step-by-step explanation:

Under a reflection in the y- axis

a point (x, y ) → (- x, y ), thus

H(5, 0 ) → H'(- 5, 0 )

I(- 6, - 3 ) → I'(6, - 3 )

7 0

2 years ago

The directex of Y=1/8x^2 is?

nexus9112 [7]

$\bf \textit{parabola vertex form with focus point distance}\\\\
\begin{array}{llll}
(y-{{ k}})^2=4{{ p}}(x-{{ h}}) \\\\
\boxed{(x-{{ h}})^2=4{{ p}}(y-{{ k}})} \\
\end{array}
\qquad 
\begin{array}{llll}
vertex\ ({{ h}},{{ k}})\\\\
{{ p}}=\textit{distance from vertex to }\\
\qquad \textit{ focus or directrix}
\end{array}\\\\
-------------------------------\\\\
y=\cfrac{1}{8}x^2\implies (y-0)=\cfrac{1}{8}(x-0)^2\implies 8(y-0)=(x-0)^2
\\\\\\
4(2)(y-0)=(x-0)^2\impliedby \textit{that means, p = 2}$

so... if you notice, the vertex is at h,k and that'd be the origin, 0,0

so...since the directrix is "p" units from the vertex, so it'd be 2 units from 0,0

now, the parabola has an equation with a positive leading term's coefficient, namely the 1/8 is positive, thus, the parabola is opening upwards, and the directrix is "outside" the parabola, so is below the vertex

that puts the directrix 2 units below 0,0

y = -2

6 0

3 years ago