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3 years ago

Mr. Chen spends $32 for tickets to a play. If the tickets cost $8 each how many tickets does Mr. Chen buy?

2 answers:

8 0

Hey!

You must divide the amount of money he spent by the amount of money each ticket costs:

32 ÷ 8 = 4

So, Mr. Chen bought 4 tickets.

leva [86]3 years ago

5 0

The answer is 4 because take 32 divided by 8 and it will give u 5

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A self-serve frozen yogurt shop has 8 candy toppings and 4 fruit toppings to choose from. How many ways are there to top a froze

Ilia_Sergeevich [38]

4096 ways : 8 candy toppings plus 4 fruit toppings is 12. Then 2^12 (2 different types of toppings and the 12 of total toppings) = 4096

4 0

3 years ago

PLEASE HELP! 40 POINTS

Anuta_ua [19.1K]

Answer:

-1 -1

Step-by-step explanation:

because thats were they touch

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2 years ago

Two similiar rectangles have corresponding sides in the ratio 10:3. What is the ratio of their areas

lianna [129]

Answer:

Step-by-step explanation:

The ratio of its areas is equal to

Step-by-step explanation:

we know that

If two figures are similar, then the ratio of its areas is equal to the scale factor squared

In this problem the scale factor is equal to the ratio 10:3

Let

z-------> the scale factor

z2=(10/3)2=100/9

7 0

3 years ago

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

3 years ago

Is it true that any set of data can be displayed using a line graph

fenix001 [56]

No, line graphs are used only to show change over time

6 0

3 years ago

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