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max2010maxim [7]
3 years ago
9

Use the distributive property to rewrite each expression. -6(r-8)

Mathematics
2 answers:
Rainbow [258]3 years ago
8 0
<span>-6(r - 8)                               Original Mathematical Expression.

-6 * r + - 6 * -8                     Distribute.

-6r + 48                               Multiply.

48 - 6r                                Switch around expression.

Answer: -6r + 48 or 48 - 6r</span>
aksik [14]3 years ago
6 0
-6(r-8)
= -6r + 48
thankyou
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A carnival charges two different prices for admissions. Adults cost $4 while children cost $1.50. If a total of $5050 was collec
GenaCL600 [577]

Answer:

Step-by-step explanation:

We have to have 2 different equations to solve this.  One equation will represent the number of tickets sold while the other represents the money collected when the tickets were sold.

We know that adult tickets + children tickets = 2200 tickets.

That's the "number of tickets" equation.  Let's call adult tickets "a" and children's tickets "c".  So a + c = 2200

Now if each adult costs $4, then the expression that represents that as a cost is 4a.  If there is 1 adult, the cost is $4(1) = $4; if there are 2 adults, the cost is $4(2) = $8; if there are 3 adults, the cost is $4(3) = $12, etc.

The same goes for the children's tickets.  If each child's ticket is $1.50, then the expression that represents the cost of a child's ticket is 1.5c (we don't need the 0 at the end; it doesn't change anything to drop it off).  The total money brought in from the cost of these tickets was $5050, so

4a + 1.5c = 5050

Let's solve the first equation for a.  If a + c = 2200, then a = 2200 - c.  Sub that into the second equation and solve it for c:

4(2200 - c) + 1.5c = 5050 and

8800 - 4c + 1.5c = 5050 and

-2.5c = -3750 so

c = 1500

That means that there were 1500 children's tickets sold.  If a + c = 2200, then a + 1500 = 2200 so

a = 2200 - 1500 so

a = 700

There were 1500 children's tickets sold and 700 adult tickets sold.

7 0
3 years ago
A car is on a driveway that is inclined 10 degrees to the horizontal. A force of 490 lb is required to keep the car from rolling
lara31 [8.8K]

Answer:

a) weight  of the car = 2816,1 lbs

b) 2773 lbs

Step-by-step explanation:

The equilibrium force is 490 lbs. That force keep the car at rest, then

∑ Fy  =  0     and  ∑Fx  =  0

Forces acting on the car:

The external force   490 lbs

weight of the car   uknown

Normal force

sin∠10°  =  0,174

cos∠10° = 0.985

∑Fx  =  0         mg*sin10°- 490 = 0      ∑Fy =  0      mg*cos10° - N  =  0

mg*0,174= 490

mg  =  490 / 0,174

mg = 2816,1 lbs

weight  of the car = 2816,1 lbs

The Normal force

mg*cos10° - N  =  0        2816,1 * 0,985 = N

N = 2773 lbs

Then equal force in magnitude and in opposite direction will car exets on the driveway

3 0
3 years ago
Can you find the sum of this triangle?<br>please ​
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Answer:

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julsineya [31]

Answer:

(7,5)

Step-by-step explanation:

{2x + 4y = 34

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4y = 27 - 7

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Kobotan [32]

Answer:

The answer is between first option and last option.

Step-by-step explanation:

That my Contributions to this questions

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