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Svetlanka [38]
3 years ago
7

What is the equation for the nth term of the arithmetic sequence. 28, 25, 22, 19, ....

Mathematics
1 answer:
natka813 [3]3 years ago
4 0

Answer:

31-3n

Step-by-step explanation:

d=-3

a1=28

an= 28+(n-1)(-3)=28-3n+3

an=31-3n

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C

Step-by-step explanation:

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Need help ASAP!!! Tysm if you answer!! IF YOU ANSWER RIGHT AND GET BRAINLIEST!!
Delvig [45]

Answer:

a_n = 3 ({5})^{n - 1}

Step-by-step explanation:

The given sequence is 3,15,75,375,...

The first term of this geometric sequence is

a_1=3

The common ratio is

r =  \frac{15}{3}  = 5

The explicit formula is given by:

a_n = a_1 {r}^{n - 1}

We plug the first term and common ratio into the formula to get:

a_n = 3 ({5})^{n - 1}

8 0
3 years ago
Write 5 as a ratio of two integers
tino4ka555 [31]

The number 5 is a rational number, so you must be able to express it as a quotient, and you can. Dividing any number by 1 gives you the original number, so to express an integer like 5 as a quotient, you simply write 5/1. The same is true for negative numbers: -5 = -5/1

4 0
3 years ago
Read 2 more answers
Physics! Someone help me! Its due today please!
icang [17]

Step-by-step explanation:

We have,

Mass of ball is 9 kg

Initial speed, u = 5 m/s

Final speed, v = -2 m/s (negative as it bounces off)

(a) The change in velocity of the bowling ball is :

\Delta v=v-u\\\\\Delta v=-2-5\\\\\Delta v=-7\ m/s

(b) Change of momentum of the ball is :

p=m\Delta v\\\\p=90\times (-7)\\\\p=-630\ kg-m/s

|p| = 630 kg-m/s

(c) Impulse momentum theorem states that the change in momentum of the ball is equal to the impulse exerted on the ball. So, impulse is 630 kg-m/s.

(d) Impulse is also given by :

J=F\times t\\\\F=\dfrac{J}{t}\\\\F=\dfrac{630}{0.3}\\\\F=2100\ N

6 0
3 years ago
For question 25 please pick 1,2,3 or 4
Juli2301 [7.4K]

Answer:

4.

$x=\frac{2}{3} \pm  \frac{1}{6}i \sqrt{158}$

Step-by-step explanation:

18x^2-24x+87=0

Using Quadratic Formula

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-\left(-24\right)\pm \sqrt{\left(-24\right)^2-4\cdot \:18\cdot \:87}}{2\cdot \:18}$

Solving the discriminant:

\Delta = \left(-24\right)^2-4\cdot \:18\cdot \:87\\\Delta = 576-6264\\ \Delta=-5688

$x= \frac{24 \pm \sqrt{-5688} }{36} $

$x= \frac{24 \pm \sqrt{5688i} }{36} $

Once \sqrt{5688} =\sqrt{2^3\cdot \:3^2\cdot \:79}=6\sqrt{158}

$x=\frac{24\pm6\sqrt{158}i}{36}$

Dividing the denominator and numerator by 6

$x=\frac{4\pm \sqrt{158}i}{6}$

Now rewrite it:

$x=\frac{4}{6} \pm  i\frac{\sqrt{158}}{6}$

$x=\frac{2}{3} \pm i \frac{\sqrt{158}}{6}$

or

$x=\frac{2}{3} \pm  \frac{1}{6}i \sqrt{158}$

5 0
3 years ago
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