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frosja888 [35]
3 years ago
11

the parking garage holds 300 cars on each level. there are 4 levels in the garage. how many cars can the parking garage hold in

all
Mathematics
2 answers:
Leviafan [203]3 years ago
8 0
We must simply multiply 300 by 4. This will show us how many cars are in the entire garage.

300 * 4 = 1200

There are 1200 cars in the parking garage.
Arturiano [62]3 years ago
3 0
Is each part of the garage is full there would be 1200 card in the parking garage
:) because 300 times 4 = 1200
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Diano4ka-milaya [45]
By using a coordinate system I believe you can find the position of any objects on a flat surface.

If you have an eraser on your table and would like to know its position, you could make your own x and y axis and see in which quadrant your object is in.
your eraser could be 2 units in the x direction (horizontal) and 5 units in the y direction (vertical). 

Now you can use this 'x and y' axis that you have drawn to locate any object. 
If you want to be accurate, you should draw your axes with a meter ruler and choose your point of origin.

Hope I answered your question.

8 0
2 years ago
A survey was conducted at Giant Supermarket where 50 students were randomly chosen and asked
Dima020 [189]

Answer:

The answer for both apples and bananas is 21

The answer for only apples or only bananas is 45

Step-by-step explanation:

36 plus 30 minus 45

50-5 is 45

only apples or bananas is 15 + 9

8 0
3 years ago
The ratio of the number of volleyball to the number of basketballs in the P.E room is 4:7 .there are 21 basketballs . How many v
densk [106]
Let volleyball be v and let basketball be b.

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v : 21 = 4 : 7

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Hope this helps.
8 0
3 years ago
I NEED HELP WITH ME MATH
ruslelena [56]

ok so lets start of with the fact that the whole thing is 21 units.

So we do 21-9=12 so then we now know that the rectangle is 12 so then we do 12*8 units which is= 96 . Now the triangles. so the first one we know that its 8*3 and times it by 1/2 because two triangles is equal to a rectangle so the first triangle is 12 and now the second. its 9-3= 6 then its 6*8*1/2 which is equal to 24 so now the final answer is

96+12+24 which is equal to 132 so i'm guessing it 132 square units

5 0
2 years ago
Not sure what to do here can someone please help
kumpel [21]

Answer:

  x = 2

Step-by-step explanation:

These equations are solved easily using a graphing calculator. The attachment shows the one solution is x=2.

__

<h3>Squaring</h3>

The usual way to solve these algebraically is to isolate radicals and square the equation until the radicals go away. Then solve the resulting polynomial. Here, that results in a quadratic with two solutions. One of those is extraneous, as is often the case when this solution method is used.

  \sqrt{x+2}+1=\sqrt{3x+3}\qquad\text{given}\\\\(x+2)+2\sqrt{x+2}+1=3x+3\qquad\text{square both sides}\\\\2\sqrt{x+2}=(3x+3)-(x+3)=2x\qquad\text{isolate the root term}\\\\x+2=x^2\qquad\text{divide by 2, square both sides}\\\\x^2-x-2=0\qquad\text{write in standard form}\\\\(x-2)(x+1)=0\qquad\text{factor}

The solutions to this equation are the values of x that make the factors zero: x=2 and x=-1. When we check these in the original equation, we find that x=-1 does not work. It is an extraneous solution.

  x = -1: √(-1+2) +1 = √(3(-1)+3)   ⇒   1+1 = 0 . . . . not true

  x = 2: √(2+2) +1 = √(3(2) +3)   ⇒   2 +1 = 3 . . . . true . . . x = 2 is the solution

__

<h3>Substitution</h3>

Another way to solve this is using substitution for one of the radicals. We choose ...

  u=\sqrt{x+2}\qquad\text{requires $u\ge0$}\\\\u^2-2=x\qquad\text{solve for x}\\\\u+1=\sqrt{3(u^2-2)+3}\qquad\text{substitute for x in the original equation}\\\\(u+1)^2=3u^2-3\qquad\text{square both sides, simplify a little}\\\\2u^2-2u-4=0\qquad\text{subtract $(u+1)^2$}\\\\2(u-2)(u+1)=0\qquad\text{factor}

Solutions to this equation are ...

  u = 2, u = -1 . . . . . . the above restriction on u mean u=-1 is not a solution

The value of x is ...

  x = u² -2 = 2² -2

  x = 2 . . . . the solution to the equation

_____

<em>Additional comment</em>

Using substitution may be a little more work, as you have to solve for x in terms of the substituted variable. It still requires two squarings: one to find the value of x in terms of u, and another to eliminate the remaining radical. The advantage seems to be that the extraneous solution is made more obvious by the restriction on the value of u.

6 0
1 year ago
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