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Paul [167]
3 years ago
13

Miss milliman reserving a room at the hotel India for a trip to see a show at the Fox Theater Hotel India charge of $717 for a t

hree-night stay what would be the unit rate or price for a one night stay
Mathematics
1 answer:
sammy [17]3 years ago
4 0
717/3= 239
 


Divide 717 by 3 to get the price of one night.
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Help! complete the table with the missing measurements. the perimeter is 40ft is that helps
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mel-nik [20]

Answer:

40 students

Step-by-step explanation:

In this problem, one is given the ratio ( 8 : 3 ), since a ratio is another way of writing a fraction one can rewrite it as; \frac{8}{3} (8) represent the number of students who want to learn about magnets, and (3) represents the number of students who want to learn about magic tricks.

In the problem, it is also given that 25 more students wanted to learn about magnets compared to magic tricks. If one uses the parameter (x) to represent the factor by which the fraction \frac{8}{3} is reduced, then one can get the equation;

8x = 3x + 25

Use inverse operations to solve,

8x = 3x + 25

-3x

5x = 25

/5

x = 5

Substitute back in to find out how many students want to learn about magnets,

8x

= 8(5)

= 40

3 0
2 years ago
What is 18/25 simplified
Alekssandra [29.7K]
18/25 is already simplified. There is no number that can go in both 18 and 25.
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3 years ago
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3 years ago
Find the components of the vertical force Bold Upper FFequals=left angle 0 comma negative 10 right angle0,−10 in the directions
quester [9]

Solution :

Let $v_0$ be the unit vector in the direction parallel to the plane and let $F_1$ be the component of F in the direction of v_0 and F_2 be the component normal to v_0.

Since, |v_0| = 1,

$(v_0)_x=\cos 60^\circ= \frac{1}{2}$

$(v_0)_y=\sin 60^\circ= \frac{\sqrt 3}{2}$

Therefore, v_0 = \left

From figure,

|F_1|= |F| \cos 30^\circ = 10 \times \frac{\sqrt 3}{2} = 5 \sqrt3

We know that the direction of F_1 is opposite of the direction of v_0, so we have

$F_1 = -5\sqrt3 v_0$

    $=-5\sqrt3 \left$

    $= \left$

The unit vector in the direction normal to the plane, v_1 has components :

$(v_1)_x= \cos 30^\circ = \frac{\sqrt3}{2}$

$(v_1)_y= -\sin 30^\circ =- \frac{1}{2}$

Therefore, $v_1=\left< \frac{\sqrt3}{2}, -\frac{1}{2} \right>$

From figure,

|F_2 | = |F| \sin 30^\circ = 10 \times \frac{1}{2} = 5

∴  F_2 = 5v_1 = 5 \left< \frac{\sqrt3}{2}, - \frac{1}{2} \right>

                   $=\left$

Therefore,

$F_1+F_2 = \left< -\frac{5\sqrt3}{2}, -\frac{15}{2} \right> + \left< \frac{5 \sqrt3}{2}, -\frac{5}{2} \right>$

           $= = F$

3 0
2 years ago
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