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2 years ago

Can you answer this plzz and the pic that goes with it???????? Thx soo much!!!!!

Mathematics

2 answers:

kkurt [141]2 years ago

5 0

Answer: c. t > 82

Step-by-step explanation:

Let t represent the number of tickets that the committee would sell.

The dance committee of Pine Bluff Middle School earns $72 from a bake sale and will earn $4 for each ticket they sell to the Spring Fling dance. This means that the revenue that they would generate from selling x tickets is

72 + 4x

The dance will cost $400.

To make profit, revenue must be greater than cost. Therefore, the inequality to determine the number of tickets the committee could sell to have money left over after they pay for this year's dance is

72 + 4x > 400

4x > 400 - 72

4x > 428

x > 328/4

x > 82

nikitadnepr [17]2 years ago

5 0

Answer:

Inequality

72 + 4t > 400

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Jenna’s method: Mia’s method:

boyakko [2]

Answer:

Mia's method have one advantage that the steps became less than Jenna's method.

Step-by-step explanation:

Jenna’s method :

$5(30 + 4)$

$=(5)(30) + (5)(4)$

$=150 + 20$

$=170$

Mia’s method :

$5(30 + 4)$

$=(5)(34)$

$=170$

To find : Explain why both Jenna and Mia arrived at the same answer and the advantage of one method over the other ?

Solution :

The method of both were different but correct.

One apply distributive property and other apply direct calculation.

So, both Jenna and Mia arrived at the same answer

But Mia's method have one advantage that the steps became less than Jenna's method.

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2 years ago

Read 2 more answers

If 36a=45/b, then ab=

Ronch [10]

Answer:

$1.25$

Step-by-step explanation:

$let \: a = x \: and \: b = y$

$36x = \frac{45}{y}$

$36xy = 45$

$xy = \frac{45}{36}$

$xy = 1.25$

$therefore \: ab \: is \: 1.25$

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2 years ago

Find the values for a, b, and c that complete the simplification.

lisov135 [29]

Answer: A: 6

B: 4

C: 2

Step-by-step explanation:

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2 years ago

Greg is trying to solve a puzzle where he has to figure out two numbers, x and y. Three less than two-third of x is greater than

Fittoniya [83]

Inequation 1:

$\frac{2}{3}x-3 \geq y$

to plot the pairs (x, y) for which the inequation holds, draw the line $y=\frac{2}{3}x-3$

then pick a point in either side of the line. If that point is a solution of the inequation, than color that region of the line, if that point is not a solution, then color the other part of the line.

we do the same for the second inequation. Then the solution, is the region of the x-y axes colored in both cases.

inequation 2:

$y+ \frac{2}{3}x\ \textless \ 4$

$y\ \textless \ - \frac{2}{3} x+ 4
$

draw the lines

i) $y=\frac{2}{3}x-3$ use points (0, -3), (3, -1)

ii) $y=- \frac{2}{3} x+ 4$ use points ( 0, 4), (3, 2)

let's use the point P(3, 3) to see what region of the lines need to be coloured:

$\frac{2}{3}x-3 \geq y$ ;
$\frac{2}{3}(3)-3 \geq 3$
$2-3 \geq 3$ , not true so we color the region not containing this point

$y+ \frac{2}{3}x\ \textless \ 4$
$(3)+ \frac{2}{3}(3)\ \textless \ 4$
$3+ 5\ \textless \ 4$ not true, so we color the region not containing the point (3, 3)

The graph representing the system of inequalities is the region colored both red and blue, with the blue line not dashed, and the red line dashed.

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2 years 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

2 years ago