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Firlakuza [10]
3 years ago
5

Find the ninth term of the sequence sqrt 5, sqrt 10, ^2sqrt 5

Mathematics
1 answer:
hjlf3 years ago
5 0
The rule of geometric sequence is   ⇒⇒⇒ a * r^(n-1)
Where a is the first term and  r is the common ratio
for the given sequence  √5 , √10 , 2√5 , .......
a = √5
r = √10 / √5 = √2
The ninth term = √5 * (√2)^(9-1) = √5 * (√2)⁸ = 16 √5

the correct answer is the third option 16√5
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√5 percent of 5√5 =<br> a. 0.05<br> b. 0.25<br> c. 0.5<br> d. 2.5<br> e. 25
Stels [109]
\sqrt{5} \% \times 5\sqrt{5}=\frac{\sqrt{5}}{100} \times 5\sqrt{5}=\frac{5 \times \sqrt{25}}{100}=\frac{5 \times 5}{100}=\frac{25}{100}=0.25

The answer is B. 0.25.
8 0
3 years ago
If T (n) =6-1 what is the 1st term.
Mariulka [41]
T(n)=6-1=5
that means all terms are 5 because n does not appear on the right hand side.

However, if T(n)=6n-1, then
first term = T(1)=6(1)-1=5,
but second term T(2)=6(2)-1=11...
8 0
3 years ago
Simplify (5x2 + 3x + 4) + (5x2 + 5x - 1).
marishachu [46]
23+8X is the answer to this problem
5 0
3 years ago
A circular swimming pool has a diameter of 40 meters. The depth of the pool is constant along west-east lines and increases line
Nataliya [291]

Answer:

Volume is 2000\pi\ m^{3}

Solution:

As per the question:

Diameter, d = 40 m

Radius, r = 20 m

Now,

From north to south, we consider this vertical distance as 'y' and height, h varies linearly as a function of y:

iff

h(y) = cy + d

Then

when y = 1 m

h(- 20) = 1 m

1 = c.(- 20) + d = - 20c + d              (1)

when y = 9 m

h(20) = 9 m

9 = c.20 + d = 20c + d                  (2)

Adding eqn (1) and (2)

d = 5 m

Using d = 5 in eqn (2), we get:

c = \frac{1}{5}

Therefore,

h(y) = \frac{1}{5}y + 5

Now, the Volume of the pool is given by:

V = \int h(y)dA

where

A = r\theta

A = rdr\ d\theta

Thus

V = \int (\frac{1}{5}y + 5)dA

V = \int_{0}^{2\pi}\int_{0}^{20} (\frac{1}{5}rsin\theta + 5) rdr\ d\theta

V = \int_{0}^{2\pi}\int_{0}^{20} (\frac{1}{5}r^{2}sin\theta + 5r}) dr\ d\theta

V = \int_{0}^{2\pi} (\frac{1}{15}20^{3}sin\theta + 1000) d\theta

V = [- 533.33cos\theta + 1000\theta]_{0}^{2\pi}

V = 0 + 2\pi \times 1000 = 2000\pi\ m^{3}

7 0
3 years ago
For which of the following compound inequalities is there no solution? ​
Bess [88]

Answer:

(a)6m \le - 36 and m + 24 > 20

Step-by-step explanation:

Required

Which of a to d has no solution

(a)6m \le - 36 and m + 24 > 20

We have: m + 24 > 20

Sole for m

m >20 - 24

m >-4

6m \le - 36

Divide by 6

m \le -6

So, we have:

m \le -6 and m >-4

m \le -6 implies that: m = -6,-7,-8,-9.....

m >-4 implies that m = -3,-2,-1,0.....

<em>Hence, there is no solution</em>

8 0
2 years ago
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