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Snezhnost [94]
3 years ago
11

lisa sold 81 magazine subscriptions which is was 27% of her class fundraising goal.how many magazine subscripitons does her clas

s hope to cell
Mathematics
1 answer:
hodyreva [135]3 years ago
7 0
Set up a proportion.

27% = 0.27

81/0.27 = x/1

Multiply 1 to both sides:

x = (81 * 1)/0.27

Multiply:

x = 81/0.27

Divide:

x = 300

So they hope to sell 300 magazine subscriptions.
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In isosceles △ABC the segment BD (with D∈ AC ) is the median to the base AC . Find BD, if the perimeter of △ABC is 50m, and the
Morgarella [4.7K]

Let the equal sides of the isosceles Δ ABC be x.

Given that the perimeter of Δ ABC = 50m.

Therefore, 2x + AC = 50 --- (1)

It is also given that the perimeter of Δ ABD = 40m.

Therefore, x + BD + AD = 40

BD is the median of the Δ ABC. Therefore, D is the midpoint of AC.

So AD = CD.

Or, AD = \frac{1}{2} AC

Therefore, x + BD + \frac{1}{2} AC = 40

Multiply both sides by 2.

2x + 2BD + AC = 80

From (1), 2x + AC = 50.

Therefore, 2BD + 50 = 80

2BD = 80 - 50

2BD = 30

BD = 15m.

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Pleasantburg has a population growth model of P(t)=at2+bt+P0 where P0 is the initial population. Suppose that the future populat
yulyashka [42]

Answer:

The population will reach 34,200 in February of 2146.

Step-by-step explanation:

Population in t years after 2012 is given by:

P(t) = 0.8t^{2} + 6t + 19000

In what month and year will the population reach 34,200?

We have to find t for which P(t) = 34200. So

P(t) = 0.8t^{2} + 6t + 19000

0.8t^{2} + 6t + 19000 = 34200

0.8t^{2} + 6t - 15200 = 0

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

ax^{2} + bx + c, a\neq0.

This polynomial has roots x_{1}, x_{2} such that ax^{2} + bx + c = a(x - x_{1})*(x - x_{2}), given by the following formulas:

x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}

x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}

\bigtriangleup = b^{2} - 4ac

In this question:

0.8t^{2} + 6t - 15200 = 0

So a = 0.8, b = 6, c = -15200

Then

\bigtriangleup = 6^{2} - 4*0.8*(-15100) = 48356

t_{1} = \frac{-6 + \sqrt{48356}}{2*0.8} = 134.14

t_{2} = \frac{-6 - \sqrt{48356}}{2*0.8} = -141.64

We only take the positive value.

134 years after 2012.

.14 of an year is 0.14*365 = 51.1. The 51st day of a year happens in February.

So the population will reach 34,200 in February of 2146.

6 0
2 years ago
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