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Stels [109]
3 years ago
6

Two batches of soup will serve 8 people. Lonnie needs to make enough soup for 120 people. If the recipe calls for 2 onions for o

ne batch, how many onions will Lonnie need?
Mathematics
2 answers:
Inessa05 [86]3 years ago
5 0
<span><em>"2 batches of soup will serve                            8 people."</em>
<u>Then x </u></span><span><u>batches of soup will serve                         120 people         
</u>120.2 = 8.x
x = 30 batches of soup will serve 120 people.

<em>"2 onions for                1 batch"</em>
<u>Then y onions for         30 batches                                                          
</u>30.2 = 1.y
y = 60 onions<u>




</u></span>
zlopas [31]3 years ago
5 0
2 batches - 8 people 
2 onions - 1 batch
x onions- 120 people
From proportion
y batches-120 people
2batches - 8 people
after cross multiplication
120/8=15*2=30 batches 
x onions - 30 batches
2 onions- 1 batches
30 batches *2=60 onions
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Answer:

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Step-by-step explanation:

Simplify the following:

(sqrt(14))/(sqrt(18))

Hint: | Simplify radicals.

sqrt(18) = sqrt(2×3^2) = 3 sqrt(2):

(sqrt(14))/(3 sqrt(2))

Hint: | Multiply numerator and denominator of (sqrt(14))/(3 sqrt(2)) by sqrt(2).

Rationalize the denominator. (sqrt(14))/(3 sqrt(2)) = (sqrt(14))/(3 sqrt(2))×(sqrt(2))/(sqrt(2)) = (sqrt(14) sqrt(2))/(3×2):

(sqrt(14) sqrt(2))/(3×2)

Hint: | Multiply 3 and 2 together.

3×2 = 6:

(sqrt(14) sqrt(2))/6

Hint: | For a>=0, sqrt(a) sqrt(b) = sqrt(a b). Apply this to sqrt(14) sqrt(2).

sqrt(14) sqrt(2) = sqrt(14×2):

(sqrt(14×2))/6

Hint: | Multiply 14 and 2 together.

14×2 = 28:

(sqrt(28))/6

Hint: | Simplify radicals.

sqrt(28) = sqrt(2^2×7) = 2 sqrt(7):

(2 sqrt(7))/6

Hint: | In (2 sqrt(7))/6, divide 6 in the denominator by 2 in the numerator.

2/6 = 2/(2×3) = 1/3:

Answer:  (sqrt(7))/3

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pishuonlain [190]

Complete question:

A police car is located 50 feet to the side of a straight road. A red car is driving along the road in the direction of the police car and is 130 feet up the road from the location of the police car. The police radar reads that the distance between the police car and the red car is decreasing at a rate of 75 feet per second. How fast is the red car actually traveling along the road.

Answer:

The red car is traveling along the road at 80.356 ft/s

Step-by-step explanation:

Given

Police car is 50 feet side off the road

Red car is 130 feet up the road

Distance between them is decreasing at the rate of 75 feet per sec

Let x be how far the police is off the road.

Let y be how far the red car is up the road.  

Let h be the distance between the police and the red car.

This forms a right triangle so we can use the Pythagorean theorem, to solve for h

h² = x² + y²

h² = 50² + 130²

h² = 19400

h = √19400

h = 139.284 ft

Again;

Let dx/dt be how fast the police is traveling toward the road.

Let dy/dt  be how fast the red car is traveling along the road.

Let dh/dt be how fast the distance between the police and the car is decreasing.

Recall that, h² = x² + y² (now differentiate with respect to time, t)

2h(dh/dt) = 2x(dx/dt) + 2y(dy/dt)

divide through by 2

h(dh/dt) = x(dx/dt) + y(dy/dt)

since the police car is not, dx/dt = 0

and dy/dt is the how fast is the red car actually traveling along the road

139.284(75) = 50(0) + 130(dy/dt)

10446.3 = 0 + 130(dy/dt)

dy/dt = 10446.3 / 130

dy/dt = 80.356 ft/s

Therefore, the red car is traveling along the road at 80.356 ft/s

6 0
4 years ago
Help its geometry...............
Digiron [165]

m GH = 57''

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tan 39° = opposite / adjacent

tan 39° = m GH / m HJ

tan 39° *  m HJ = m GH

m HJ = m GH / tan 39°

m HJ ≈ 57'' / 0.810

m HJ ≈ 70.4''    <——<span>—    measure of the segment HJ</span>

________


sin 39° = opposite / hypotenuse

sin 39° = m GH / m GJ

sin 39° * m GJ = m GH

m GJ = m GH / sin 39°

m GJ = 57'' / 0.629

m GJ ≈ 90.6''    <———    measure of the segment GJ

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So the perimeter is

p  =  m GH  +  m HJ  +  m GJ

p  =  57''  +  70.4''  +  90.6''

p  =  218''    <——<span>—    this is the answer.</span>


I hope this helps. =)


Tags:  <em>perimeter right triangle sine cosine tangent sin cos tan trig trigonometry geometry</em>

7 0
3 years ago
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