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zalisa [80]
3 years ago
10

There are 4 veggie pizzas for every 5 cheese pizzas. How many cheese pizzas are there if 108 pizzas total are ordered? please sh

ow work
Mathematics
2 answers:
Andre45 [30]3 years ago
7 0

Answer:

6.25

Step-by-step explanation:

Katarina [22]3 years ago
5 0

Answer:

60 because [total of pizzas]/(veggie + cheese)= 12 ((108/(4+5) = 12)) so you take 5 (the amount of cheese pizzas) and multiply it by 12 to get 60

sorry if i explained it bad im not too good at explaining

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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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