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PtichkaEL [24]
3 years ago
10

two sporting goods stores are having discount sales on footballs. At one store, a basketball is on sale for 15% off the regular

price of $22.95. At the other store, the same kind of basketball is on sale for 20% off the regular price of $24.75. What is the difference between the sale prices of the two stores?
Mathematics
1 answer:
CaHeK987 [17]3 years ago
6 0
Sale price for store A:
(100-15)/100×22.95
85/100×22.95
=$19.5075
~$19.5

Sale Price for store B:
(100-20)/100×24.75
80/100×24.75
=%$19.8
Thus the difference in the sale prices will be:
19.8-19.5
=$0.3

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Answer:

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Step-by-step explanation:

\frac{ \{ \frac{3}{4} -  \frac{5}{2}  \} }{ \{ \frac{3}{5}  -  \frac{3}{2}  \}}   =  \frac{( -  \frac{7}{4} )}{ (-  \frac{9}{10} )}  \\  \\   =  \frac{7}{4}  \div  \frac{9}{10}

use the reciprocal of 9/10 :

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3 years ago
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Jason is pulling a box across the room. He is pulling with a force of 19 newtons and his arm is making a 30 angle with the horiz
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We are to solve for the vertical and horizontal component of the force he is pulling with.

1. The vertical component of Jason's force is 9.5Newtons.

2.The horizontal component of Jason's force is 2.67Newtons.

3. Therefore, the net force in the vertical direction is -12.5Newtons.

4.Therefore, Resultant, R is 72.11

Question 1:

If Jason is pulling with a force of 19 Newtons and his arm is at an angle of 30° with the horizontal.

By resolving Jason's force of pull in the vertical direction, Fy = 19 × Sin 30° = 9.5 Newtons.

The <em>vertical</em> component of Jason's force is 9.5 Newtons.

Question 2:

Also,If Jason is pulling with a force of 14 Newtons and his arm is at an angle of 79° with the <em>horizontal</em>.

By resolving Jason's force of pull in the <em>horizontal</em> direction, Fx = 14 × Cos79° = 2.67 Newtons.

The horizontal component of Jason's force is 2.67Newtons.

Question 3:

If Jason is pulling with a force of 23Newtons and his arm is at an angle of 30° with the <em>horizontal</em>.

By resolving Jason's force of pull in the <em>vertical</em> direction, Fy = 23 × Sin 30° = 11.5Newtons.

The vertical component of Jason's force of pull is, 11.5Newtons.

If the box weighs 24Newtons.

By treating up as the +ve vertical direction and down as the -ve vertical direction,

Therefore, the weight of the box acts in the -ve vertical direction, while Jason's vertical force component acts in the +ve vertical direction.

Therefore, the net force in the vertical direction is = 11.5Newtons + (-24Newtons)

Therefore, the net force in the vertical direction is -12.5Newtons.

Question 4:

If there are two forces at right angles to eachother, one of magnitude, 68 and the other of magnitude, 24.

The resultant force on the object can be obtained by Pythagoras theorem (Triangle law of forces).

Resultant, R = √(68²+24²) = √5200.

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