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Kamila [148]
1 year ago
5

A local high school held a two-day fundraiser selling lunches. The equations shown represent the amount of money the school coll

ected during the two days.
{
12x + 4y = 152
32x + 12y = 420
12x + 4y = 15232x + 12y = 420


In the equations,
x
x
represents the cost of a chicken lunch and
y
y
represents the cost of a vegetarian lunch. Which is the cost of the vegetarian lunch?
Mathematics
1 answer:
Likurg_2 [28]1 year ago
5 0
It will be a kid getting hit in the leg by the kids in the lunch table so it will be 4
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E varies directly with the square root of C. If E=40 when C=25, find: C when E = 10.4
sukhopar [10]

Answer: C = 1.69

Step-by-step explanation:

E is proportional to √C

To remove proportionality, introduce a constant (k).

E = k × √C

From question,

E = 40 and C = 25

So,

40 = k ×√25

40 = k × 5

k = 8

Now,

C = ?

E = 10.4

k = 8

E = k × √C

10.4 = 8 × √C

10.4 / 8 = √C

( 10.4 / 8 ) ^ 2 = C

C = 1.69

6 0
2 years ago
Read 2 more answers
Legolas shoots 3 arrows at once from his bow. He has 177 arrows. How many times can Legolas shoot his bow before he needs more a
Nataliya [291]

Answer:

59

Step-by-step explanation:

3x=177

3x represents the # of arrow he can shoot at once every time

177 represents the total # of arrow he has

3 0
1 year ago
The graphs of the quadratic functions f(x) = 6 – 10x2 and g(x) = 8 – (x – 2)2 are provided below. Observe there are TWO lines si
natta225 [31]

Answer:

a) y = 7.74*x + 7.5

b)  y = 1.148*x + 6.036

Step-by-step explanation:

Given:

                                  f(x) = 6 - 10*x^2

                                  g(x) = 8 - (x-2)^2

Find:

(a) The line simultaneously tangent to both graphs having the LARGEST slope has equation

(b) The other line simultaneously tangent to both graphs has equation,

Solution:

- Find the derivatives of the two functions given:

                                f'(x) = -20*x

                                g'(x) = -2*(x-2)

- Since, the derivative of both function depends on the x coordinate. We will choose a point x_o which is common for both the functions f(x) and g(x). Point: ( x_o , g(x_o)) Hence,

                                g'(x_o) = -2*(x_o -2)

- Now compute the gradient of a line tangent to both graphs at point (x_o , g(x_o) ) on g(x) graph and point ( x , f(x) ) on function f(x):

                                m = (g(x_o) - f(x)) / (x_o - x)

                                m = (8 - (x_o-2)^2 - 6 + 10*x^2) / (x_o - x)

                                m = (8 - (x_o^2 - 4*x_o + 4) - 6 + 10*x^2)/(x_o - x)

                                m = ( 8 - x_o^2 + 4*x_o -4 -6 +10*x^2) /(x_o - x)

                                m = ( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x)

- Now the gradient of the line computed from a point on each graph m must be equal to the derivatives computed earlier for each function:

                                m = f'(x) = g'(x_o)

- We will develop the first expression:

                                m = f'(x)

                                ( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x) = -20*x

Eq 1.                          (-2 - x_o^2 + 4*x_o + 10*x^2) = -20*x*x_o + 20*x^2

And,

                              m = g'(x_o)

                              ( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x) = -20*x

                              -2 - x_o^2 + 4*x_o + 10*x^2 = -2(x_o - 2)(x_o - x)

Eq 2                       -2 - x_o^2 + 4*x_o+ 10*x^2 = -2(x_o^2 - x_o*(x + 2) + 2*x)

- Now subtract the two equations (Eq 1 - Eq 2):

                              -20*x*x_o + 20*x^2 + 2*x_o^2 - 2*x_o*(x + 2) + 4*x = 0

                              -22*x*x_o + 20*x^2 + 2*x_o^2 - 4*x_o + 4*x = 0

- Form factors:       20*x^2 - 20*x*x_o - 2*x*x_o + 2*x_o^2 - 4*x_o + 4*x = 0

                              20*x*(x - x_o) - 2*x_o*(x - x_o) + 4*(x - x_o) = 0

                               (x - x_o)(20*x - 2*x_o + 4) = 0  

                               x = x_o   ,     x_o = 10x + 2    

- For x_o = 10x + 2  ,

                               (g(10*x + 2) - f(x))/(10*x + 2 - x) = -20*x

                                (8 - 100*x^2 - 6 + 10*x^2)/(9*x + 2) = -20*x

                                (-90*x^2 + 2) = -180*x^2 - 40*x

                                90*x^2 + 40*x + 2 = 0  

- Solve the quadratic equation above:

                                 x = -0.0574, -0.387      

- Largest slope is at x = -0.387 where equation of line is:

                                  y - 4.502 = -20*(-0.387)*(x + 0.387)

                                  y = 7.74*x + 7.5          

- Other tangent line:

                                  y - 5.97 = 1.148*(x + 0.0574)

                                  y = 1.148*x + 6.036

6 0
1 year ago
At 9am you have run 2 miles. At 9:24 you have run 5 miles. What is your running rate in minutes per mile?
Drupady [299]

Answer:

8minutes per mile

Step-by-step explanation:

You ran 3 miles since 9am divide 24 by 3 to get 8 minutes

3 0
2 years ago
Brian invests £1850 into his bank account. He receives 2.7% per year simple interest. How much will Brian have after 3 years? Gi
Rama09 [41]

Answer:

£1999.85

Step-by-step explanation:

A = P(1 + rt)

Where:

A = Amount after t years

P = Initial amount invested = £1850

r = Interest rate = 2.7%

t = Time in years = 3 years

Calculation:

First, converting R percent to r a decimal

r = R/100 = 2.7%/100 = 0.027 per year.

Solving our equation:

A = £1850(1 + (0.027 × 3))

A = £1999.85

Therefore, Brian will have £1999.85 after 3 years.

3 0
2 years ago
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