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DerKrebs [107]
3 years ago
8

Find the width (w)of a rectangular room with a length (l) of 60 meters and an area (a) of 2,400 square meters

Mathematics
1 answer:
sdas [7]3 years ago
6 0

Answer:

The width of the room would be 40 meters

Step-by-step explanation:

To find this, start with the area formula.

A = LW

Now input the known length and area.

2400 = 60W

Now divide both sides by 60

40 = W

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Solve 5m − 3h = 45 for h. Show your work.
mezya [45]

5m − 3h = 45

Subtract 5m from both sides

-3h = 45 - 5m

Divide both sides by -3

h = (45 - 5m)/ -3

h = -15 + 5m/3

5 0
3 years ago
Read 2 more answers
Steve and Elizabeth decide to become partners in a children’s cooking school. They need $80,000 for the franchise. They invest i
PilotLPTM [1.2K]

Answer:

The share of Steve  is $34,285 .71

The share of Elizabeth  is  $45,714.28

Step-by-step explanation:

The total investment of Steve and Elizabeth in cooking school = $80, 000

The ratio of investment of  Steve and Elizabeth  = 3: 4

Let us assume the common factor in the ratio is m.

So, the ratio of  Steve = 3 m

and the ratio of  Steve = 4 m

So, according to the question:

⇒ 3 m + 4 m   =  80,000

or, 7 m = 80,000

⇒ m = 80,000 / 7 = 11,428.57

or, m = 11,428.57

⇒ The share of Steve = 3 m = 3 x (11,428.57)  = 34,285 .71

⇒ The share of Elizabeth = 4 m = 4 x (11,428.57)  = 45,714.28

4 0
3 years ago
AB is tangent to circle O at B. What is the length of the radius r
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check the picture below

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4 0
3 years ago
60000ft² divided by 3 floors and 2 buildings is? How much each floor for each building?
9966 [12]
10000ft<span>² for each floor.

60000 divided by 3 = 20000
20000 divided by 2 = 10000

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3 0
3 years ago
Find the equation of the lines parallel and perpendicular to the line 5x+2y=12 through the point (-2,3)
muminat

Answer:

The equation of line parallel to given line and passing through points        ( - 2 , 3 ) is 5 x + 2 y + 4 = 0

The equation of line perpendicular to given line and passing through points ( - 2 , 3 ) is 2 x - 5 y + 19 = 0

Step-by-step explanation:

Given equation of line as :

5 x + 2 y = 12

or, 2 y = - 5 x + 12

or , y = \frac{-5}{2} x + \frac{12}{2}

Or, y = \frac{-5}{2} x + 6

∵ Standard equation of line is give as

y = m x + c

Where m is the slope of line and c is the y-intercept

Now, comparing given line equation with standard eq

So, The slope of the given line = m = \frac{-5}{2}

Again,

The other line if passing through the points (- 2 , 3 ) And  is parallel to given line

So, for parallel lines condition , the slope of both lines are equal

Let The slope of other line = M

So,  M = m = \frac{-5}{2}

∴ The equation of line with slope M and passing through points ( -2 , 3) is

y = M x + c

Now , satisfying the points

So, 3 = \frac{-5}{2} × ( - 2 ) + c

or, 3 =  \frac{10}{2} + c

Or, 3 = 5 + c

∴  c = 3 - 5 = - 2

c = - 2

So, The equation of line with slope  \frac{-5}{2}  and passing through points ( -2 , 3)

y =  \frac{-5}{2} x - 2

or, 2 y = - 5 x - 4

I.e 5 x + 2 y + 4 = 0

<u>Similarly</u>

The other line if passing through the points (- 2 , 3 ) And  is perpendicular  to given line

So, for perpendicular lines condition,the products of slope of both lines = - 1

Let The slope of other line = M'

So,  M' × m = - 1

Or, M' ×  \frac{-5}{2} = - 1

Or, M' = \frac{-1}{\frac{-5}{2}}

Or, M' =  \frac{2}{5}

∴ The equation of line with slope M and passing through points ( -2 , 3) is

y = M' x + c'

Now , satisfying the points

So, 3 = \frac{2}{5} × ( - 2 ) + c'

or, 3 =  \frac{- 4}{5} + c'

Or, 3 × 5 = - 4 + 5× c'

∴  5 c' = 15 + 4

or, 5 c' = 19

Or, c' =  \frac{19}{5}

So, The equation of line with slope  \frac{2}{5}  and passing through points ( -2 , 3)

y =  \frac{2}{5} x +  \frac{19}{5}

y =  \frac{2 x + 19}{5}

Or, 5 y = 2 x + 19

Or, 2 x - 5 y + 19 = 0

Hence The equation of line parallel to given line and passing through points ( - 2 , 3 ) is 5 x + 2 y + 4 = 0

And  The equation of line perpendicular to given line and passing through points ( - 2 , 3 ) is 2 x - 5 y + 19 = 0

Answer

4 0
3 years ago
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