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3 years ago

A baby narwhal is born at 1 meter in length and grows / meter in length each year. Write an

Mathematics

2 answers:

Nana76 [90]3 years ago

6 0

Answer:

what subject is this for

Step-by-step explanation:

slamgirl [31]3 years ago

3 0

Stacy is doing a science project on narwhals. She needs to document the growth of narwhals over a period of time. If narwhals are born at 1 meter in length and grow 1 meter in length per year, how long will a narwhale be in 20 years?

hope this helps :)

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Help, please :)<br><br><br>I really appreciate if you help me with steps as well, thanks! <3

BlackZzzverrR [31]

Answer:

Step-by-step explanation:

All we have to do here is to find the measure of the central angle; the measure of the central angle of a circle is the same measure as the arc it intercepts. Angle DQE in degrees = arc DE in degrees. Use the arc length formula to find the measure of the central angle:

$AL=\frac{\theta}{360}*2 \pi r$ and filling in:

$8.73=\frac{\theta}{360}*(2)(3.14)(10)$ which simplifies to

$8.73=\frac{62.8\theta}{360}$ which simplifies a bit more to

$\frac{360(8.73)}{62.8}=\theta$ so

$\theta=50.04$ degrees. That means that arc DE also measures 50.04 degrees.

8 0

3 years ago

A pattern exists among the units digit of the posers of 8 as shown in the diagram.

guajiro [1.7K]

Answer:

B) 4

Step-by-step explanation:

We can solve this by observing some pattern.

The powers ending in 4 as unit digit are:

${8}^{2} , {8}^{6} , {8}^{10} , {8}^{14} , {8}^{20} ,...$

The exponents form the sequence:

2,6,10,14,20,...

We need to check if 62 belongs to this sequence.

This is an arithmetic sequence with a common difference of 4 and a first term of 2.

The explicit formula is

$2 + 4(n - 1)$

We equate this to 62 and solve for n.

$2 + 4(n - 1) = 62 \\ 4(n - 1) = 60 \\ n - 1 = 15 \\ n = 16$

Since n is a natural number, 62 belongs to the sequence.

Hence

${8}^{62}$

will have a unit digit of 4.

4 0

3 years ago

The difference between the roots of the equation 3x2+bx+10=0 is equal to 4 1 3 . find<br> b.

inna [77]

Given that the difference between the roots of the equation $3x^2+bx+10=0$ is $4 \frac{1}{3} = \frac{13}{3}$ .

Recall that the sum of roots of a quadratic equation is given by $- \frac{b}{a}$ .

Let the two roots of the equation be $\alpha$ and $\alpha + \frac{13}{3}$ , then

$\alpha + \alpha + \frac{13}{3} =2 \alpha + \frac{13}{3} =- \frac{b}{a} =- \frac{b}{3} \\ \\ i.e.\ \ 2 \alpha + \frac{13}{3}=- \frac{b}{3}$ . . . (1)

Also recall that the product of the two roots of a quadratic equation is given by $\frac{c}{a}$ , thus:

$\alpha \left( \alpha + \frac{13}{3} \right)= \alpha ^2+ \frac{13}{3} \alpha = \frac{c}{a} = \frac{10}{3} \\ \\ i.e.\ \ \alpha ^2+ \frac{13}{3} \alpha=\frac{10}{3}$ . . . (2)

From (1), we have:

$2 \alpha =- \frac{b}{3} - \frac{13}{3} \\ \\ \Rightarrow \alpha =- \frac{b}{6} - \frac{13}{6}$

Substituting for alpha into (2), gives:

$\left(- \frac{b}{6} - \frac{13}{6}\right)^2+ \frac{13}{3} \left(- \frac{b}{6} - \frac{13}{6}\right)= \frac{10}{3} \\ \\ \Rightarrow \frac{b^2}{36} + \frac{13b}{18} + \frac{169}{36} - \frac{13b}{18} - \frac{169}{18} = \frac{10}{3} \\ \\ \Rightarrow \frac{b^2}{36} - \frac{289}{36} =0 \\ \\ \Rightarrow \frac{b^2}{36} = \frac{289}{36} \\ \\ \Rightarrow b^2=289 \\ \\ \Rightarrow b=\pm\sqrt{289}=\pm17$

5 0

3 years ago

The table shows the number of books donated to a library each month. Suppose the growth continues exponentially.

telo118 [61]

$\bf \begin{array}{ccll}
\stackrel{t}{months}&\stackrel{A}{books}\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
0&80\\
1&100\\
2&125
\end{array}$ , we know that on Month 0, the Books were 80

$\bf \qquad \textit{Amount for Exponential Growth}\\\\
A=I(1 + r)^t\qquad 
\begin{cases}
A=\textit{accumulated amount}\to &80\\
I=\textit{initial amount}\\
r=rate\to r\%\to \frac{r}{100}\\
t=\textit{elapsed time}\to &0\\
\end{cases}
\\\\\\
80=I(1+r)^0\implies 80=I\qquad therefore\qquad \boxed{A=80(1+r)^t}$

we also know that on the first month there were 100 books,

$\bf \qquad \textit{Amount for Exponential Growth}\\\\
A=I(1 + r)^t\qquad 
\begin{cases}
A=\textit{accumulated amount}\to &100\\
I=\textit{initial amount}\\
r=rate\to r\%\to \frac{r}{100}\\
t=\textit{elapsed time}\to &1\\
\end{cases}
\\\\\\
100=80(1+r)^1\implies \cfrac{100}{80}=1+r\implies \cfrac{5}{4}=1+r
\implies 
\cfrac{5}{4}-1=r
\\\\\\
 \cfrac{1}{4}=r\implies 0.25=r\qquad therefore\qquad \boxed{A=80(1+0.25)^t}$

now, how many books when t = 8? A=80(1+0.25)⁸, or A=80(1.25)⁸.

7 0

3 years ago

Read 2 more answers

What is the average rate of change for this quadratic function for the interval fr x=2 to x=4?

myrzilka [38]

Answer:

0-4

derivate first

function is missing

substitute the value of X

8 0

2 years ago