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brilliants [131]

2 years ago

14

At a basketball game,a team made 54 successful shots they were a combination of 1&2 point shot.the team scored 90 points in

all.

1 answer:

umka2103 [35]2 years ago

4 0

Answer:

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Step-by-step explanation:

4wr er0gi j 8jf8g r8gj v b gg r fdcth s dc67 858 568 5685 685

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How do you do this?<br> Please help x

kirill115 [55]

Answer:

Step-by-step explanation:

a). Given equation is y = 3x - 4

Table for the input-output values is,

x -1 0 1 2 3

y -7 -4 -1 2 5

Now we can plot these points on the graph (graph attached).

b). Equation is y = -2x + 3

Table for the input-output values will be,

x -1 0 1 2 3

y 5 3 1 -1 -3

3 0

2 years ago

If you're starting salary is 40,000 and you receive a 3% increase at the end of every year what is the total amount in dollars y

storchak [24]

$\bf \qquad \textit{Amount for Exponential Growth}\\\\
A=I(1 + r)^t\qquad 
\begin{cases}
A=\textit{accumulated amount}\\
I=\textit{initial amount}\to &40000\\
r=rate\to 3\%\to \frac{3}{100}\to &0.03\\
t=\textit{elapsed time}\to &15\\
\end{cases}
\\\\\\
A=40000(1+0.03)^{16}\implies A=40000(1.03)^{16}$

7 0

2 years ago

Tacoma's population in 2000 was about 200 thousand, and had been growing by about 9% each year. a. Write a recursive formula for

KIM [24]

Answer:

a) The recurrence formula is $P_n = \frac{109}{100}P_{n-1}$ .

b) The general formula for the population of Tacoma is

$P_n = \left(\frac{109}{100}\right)^nP_{0}$ .

c) In 2016 the approximate population of Tacoma will be 794062 people.

d) The population of Tacoma should exceed the 400000 people by the year 2009.

Step-by-step explanation:

a) We have the population in the year 2000, which is 200 000 people. Let us write $P_0 = 200 000$ . For the population in 2001 we will use $P_1$ , for the population in 2002 we will use $P_2$ , and so on.

In the following year, 2001, the population grow 9% with respect to the previous year. This means that $P_0$ is equal to $P_1$ plus 9% of the population of 2000. Notice that this can be written as

$P_1 = P_0 + (9/100)*P_0 = \left(1-\frac{9}{100}\right)P_0 = \frac{109}{100}P_0$ .

In 2002, we will have the population of 2001, $P_1$ , plus the 9% of $P_1$ . This is

$P_2 = P_1 + (9/100)*P_1 = \left(1-\frac{9}{100}\right)P_1 = \frac{109}{100}P_1$ .

So, it is not difficult to notice that the general recurrence is

$P_n = \frac{109}{100}P_{n-1}$ .

b) In the previous formula we only need to substitute the expression for $P_{n-1}$ :

$P_{n-1} = \frac{109}{100}P_{n-2}$ .

Then,

$P_n = \left(\frac{109}{100}\right)^2P_{n-2}$ .

Repeating the procedure for $P_{n-3}$ we get

$P_n = \left(\frac{109}{100}\right)^3P_{n-3}$ .

But we can do the same operation n times, so

$P_n = \left(\frac{109}{100}\right)^nP_{0}$ .

c) Recall the notation we have used:

$P_{0}$ for 2000, $P_{1}$ for 2001, $P_{2}$ for 2002, and so on. Then, 2016 is $P_{16}$ . So, in order to obtain the approximate population of Tacoma in 2016 is

$P_{16} = \left(\frac{109}{100}\right)^{16}P_{0} = (1.09)^{16}P_0 = 3.97\cdot 200000 \approx 794062$

d) In this case we want to know when $P_n>400000$ , which is equivalent to

$(1.09)^{n}P_0>400000$ .

Substituting the value of $P_0$ , we get

$(1.09)^{n}200000>400000$ .

Simplifying the expression:

$(1.09)^{n}>2$ .

So, we need to find the value of $n$ such that the above inequality holds.

The easiest way to do this is take logarithm in both hands. Then,

$n\ln(1.09)>\ln 2$ .

So, $n>\frac{\ln 2}{\ln(1.09)} = 8.04323172693$ .

So, the population of Tacoma should exceed the 400 000 by the year 2009.

8 0

3 years ago

Read 2 more answers

Simply 2 x b x b x b x 4

elena-s [515]

2×4=8
b×b×b= b^3
so putting them together would be
8×b^3

5 0

3 years ago

Read 2 more answers

Do the following lengths form a right triangle?

Dmitry_Shevchenko [17]

Answer:

yes, this forms a right angle

5 0

2 years ago