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jek_recluse [69]
3 years ago
7

A town has a population of 13,000 and grows at 5% every year. What will be the population after 11 years to the nearest whole nu

mber
Mathematics
2 answers:
timofeeve [1]3 years ago
6 0

Answer:

So all you need to do is find 5% of 13000 which is 650.

Now 650 times 11= 7150

13000+ 7150=

20,150

ozzi3 years ago
3 0

Answer:

20150

Step-by-step explanation:

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Simplify 24 − (23 + 22).<br> HELP FAST
RSB [31]

Answer:-21

Step-by-step explanation:

23+22=45

45-24=21

then flip 24 and 45 to get negative

8 0
3 years ago
Read 2 more answers
In the figure below, \overline{AD} AD start overline, A, D, end overline and \overline{BE} BE start overline, B, E, end overline
VLD [36.1K]

Answer:

<u>The measure of the arc CD = 64°</u>

Step-by-step explanation:

It is required to find the measure of the arc CD in degrees.

So, as shown at the graph

BE and AD are are diameters of circle P

And ∠APE is a right angle ⇒ ∠APE = 90°

So, BE⊥AD

And so, ∠BPE = 90° ⇒(1)

But it is given: ∠BPE = (33k-9)° ⇒(2)

From (1) and (2)

∴ 33k - 9 = 90

∴ 33k = 90 + 9 = 99

∴ k = 99/33 = 3

The measure of the arc CD = ∠CPD = 20k + 4

By substitution with k

<u>∴ The measure of the arc CD = 20*3 + 4 = 60 + 4 = 64°</u>

6 0
3 years ago
Mr. Howard paid $99 for 18 feet of chain. At this rate,
vekshin1

Answer:

12.3 feet of chain is your answer.

3 0
3 years ago
What is the value of x?
Elanso [62]
Since all the angles are congruent, the sides must be congruent too.


So, just have one equation on one side equal to another.

5x= 3x+6

Subtract 3x from both sides.

2x=6

Divide 2 into both sides.

x=3.

We can check this by seeing if all the sides are the same.

5(3)= 15

6(3)-5= 15

3(3)+6= 15

I hope this helps!
~cupcake
7 0
3 years ago
Which of the following graphs shows the solution set for the inequality below? 3|x + 1| &lt; 9
Bas_tet [7]

Step-by-step explanation:

The absolute value function is a well known piecewise function (a function defined by multiple subfunctions) that is described mathematically as

                                 f(x) \ = \ |x| \ = \ \left\{\left\begin{array}{ccc}x, \ \text{if} \ x \ \geq \ 0 \\ \\ -x, \ \text{if} \ x \ < \ 0\end{array}\right\}.

This definition of the absolute function can be explained geometrically to be similar to the straight line   \textbf{\textit{y}} \ = \ \textbf{\textit{x}}  , however, when the value of x is negative, the range of the function remains positive. In other words, the segment of the line  \textbf{\textit{y}} \ = \ \textbf{\textit{x}}  where \textbf{\textit{x}} \ < \ 0 (shown as the orange dotted line), the segment of the line is reflected across the <em>x</em>-axis.

First, we simplify the expression.

                                             3\left|x \ + \ 1 \right| \ < \ 9 \\ \\ \\\-\hspace{0.2cm} \left|x \ + \ 1 \right| \ < \ 3.

We, now, can simply visualise the straight line,  y \ = \ x \ + \ 1 , as a line having its y-intercept at the point  (0, \ 1) and its <em>x</em>-intercept at the point (-1, \ 0). Then, imagine that the segment of the line where x \ < \ 0 to be reflected along the <em>x</em>-axis, and you get the graph of the absolute function y \ = \ \left|x \ + \ 1 \right|.

Consider the inequality

                                                    \left|x \ + \ 1 \right| \ < \ 3,

this statement can actually be conceptualise as the question

            ``\text{For what \textbf{values of \textit{x}} will the absolute function \textbf{be less than 3}}".

Algebraically, we can solve this inequality by breaking the function into two different subfunctions (according to the definition above).

  • Case 1 (when x \ \geq \ 0)

                                                x \ + \ 1 \ < \ 3 \\ \\ \\ \-\hspace{0.9cm} x \ < \ 3 \ - \ 1 \\ \\ \\ \-\hspace{0.9cm} x \ < \ 2

  • Case 2 (when x \ < \ 0)

                                            -(x \ + \ 1) \ < \ 3 \\ \\ \\ \-\hspace{0.15cm} -x \ - \ 1 \ < \ 3 \\ \\ \\ \-\hspace{1cm} -x \ < \ 3 \ + \ 1 \\ \\ \\ \-\hspace{1cm} -x \ < \ 4 \\ \\ \\ \-\hspace{1.5cm} x \ > \ -4

           *remember to flip the inequality sign when multiplying or dividing by

            negative numbers on both sides of the statement.

Therefore, the values of <em>x</em> that satisfy this inequality lie within the interval

                                                     -4 \ < \ x \ < \ 2.

Similarly, on the real number line, the interval is shown below.

The use of open circles (as in the graph) indicates that the interval highlighted on the number line does not include its boundary value (-4 and 2) since the inequality is expressed as "less than", but not "less than or equal to". Contrastingly, close circles (circles that are coloured) show the inclusivity of the boundary values of the inequality.

3 0
2 years ago
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