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2 years ago

Candice is researching a career as a electrician

Mathematics

2 answers:

kicyunya [14]2 years ago

8 0

More information lol???

Naddik [55]2 years ago

5 0

Answer:

ridiculous I'm trying got get this done.

Step-by-step explanation:

You might be interested in

I B.

Sliva [168]

Answer:

89 meters per second.

Step-by-step explanation:

That is quite fast!

First, we are given that in 9 sec our falcon has flown 801 meters, let's write an equation for that:

9s = 801

So in order to get s, we divide each side by 9.

Solve!

s = 801/9

s = 89

Happy learning! Please give a brainliest if you found this helpful

6 0

2 years ago

In a G.P the difference between the 1st and 5th term is 150, and the difference between the

liubo4ka [24]

Answer:

Either $\displaystyle \frac{-1522}{\sqrt{41}}$ (approximately $-238$ ) or $\displaystyle \frac{1522}{\sqrt{41}}$ (approximately $238$ .)

Step-by-step explanation:

Let $a$ denote the first term of this geometric series, and let $r$ denote the common ratio of this geometric series.

The first five terms of this series would be:

$a$ ,
$a\cdot r$ ,
$a \cdot r^2$ ,
$a \cdot r^3$ ,
$a \cdot r^4$ .

First equation:

$a\, r^4 - a = 150$ .

Second equation:

$a\, r^3 - a\, r = 48$ .

Rewrite and simplify the first equation.

$\begin{aligned}& a\, r^4 - a \\ &= a\, \left(r^4 - 1\right)\\ &= a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) \end{aligned}$ .

Therefore, the first equation becomes:

$a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) = 150$ ..

Similarly, rewrite and simplify the second equation:

$\begin{aligned}&a\, r^3 - a\, r\\ &= a\, \left( r^3 - r\right) \\ &= a\, r\, \left(r^2 - 1\right) \end{aligned}$ .

Therefore, the second equation becomes:

$a\, r\, \left(r^2 - 1\right) = 48$ .

Take the quotient between these two equations:

$\begin{aligned}\frac{a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right)}{a\cdot r\, \left(r^2 - 1\right)} = \frac{150}{48}\end{aligned}$ .

Simplify and solve for $r$ :

$\displaystyle \frac{r^2+ 1}{r} = \frac{25}{8}$ .

$8\, r^2 - 25\, r + 8 = 0$ .

Either $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ or $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ .

Assume that $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = -\frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= -\frac{1522\sqrt{41}}{41} \approx -238\end{aligned}$ .

Similarly, assume that $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = \frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= \frac{1522\sqrt{41}}{41} \approx 238\end{aligned}$ .

4 0

2 years ago

At noon, ship A is 170 km west of ship B. Ship A is sailing east at 35 km/h and ship B is sailing north at 20 km/h. How fast is

garik1379 [7]

Check the picture below.

now, keep in mind that ship B is going at 20kph, thus from noon to 4pm, is 4 hours, so it has travelled by then 20 * 4 or 80 kilometers, thus b = 80.

whilst the ship B is moving north, the distance "a" is not really changing, and thus is a constant, that matters because the derivative of a constant is 0.

$\bf c^2=a^2+b^2\implies \stackrel{chain~rule}{2c\cfrac{dc}{dt}}=0+2b\cfrac{db}{dt}\implies \cfrac{dc}{dt}=\cfrac{b\frac{db}{dt}}{c}
\\\\\\
\begin{cases}
\frac{db}{dt}=20\\
c=10\sqrt{353}\\
b=80
\end{cases}\implies \cfrac{dc}{dt}=\cfrac{80\cdot 20}{10\sqrt{353}}\implies \cfrac{dc}{dt}=\cfrac{160}{\sqrt{353}}
\\\\\\
\textit{and rationalizing the denominator}\implies \cfrac{160\sqrt{353}}{353}$

8 0

2 years ago

What is the arc measure of BAD on circle P in degrees?

lianna [129]

Answer:

244

Step-by-step explanation:

74+42=116

360-116=244.

5 0

2 years ago

If the Math Olympiad Club consists of 11 students, how many different teams of 3 students can be formed for competitions?

vivado [14]

Answer:

165 different teams of 3 students can be formed for competitions

Step-by-step explanation:

Combinations of m elements taken from n in n (m≥n) are called all possible groupings that can be made with the m elements so that:

Not all items fit
No matter the order
Elements are not repeated

That is, a combination is an arrangement of elements where the place or position they occupy within the arrangement does not matter. In a combination it is interesting to form groups and their content.

To calculate the number of combinations, the following expression is applied:

$C=\frac{m!}{n!*(m-n)!}$

It indicates the combinations of m objects taken from among n objects, where the term "n!" is called "factorial of n" and is the multiplication of all the numbers that go from "n" to 1.

In this case:

n: 3
m: 11

Replacing:

$C=\frac{11!}{3!*(11-3)!}$

Solving:

$C=\frac{11!}{3!*8!}$

being:

3!=3*2*1=6
8!=8*7*6*5*4*3*2*1=40,320
11!=39,916,800

So:

$C=\frac{39,916,800}{6*40,320}$

C= 165

<u><em>165 different teams of 3 students can be formed for competitions</em></u>

3 0

2 years ago