All categories

3 years ago

Find the next two terms of the sequence: 80, ?40, 20, ?10, , . A) 20, 40 B) 1 2 , 1 4 C) 5, ? 5 2 D) ?30, ?40

Mathematics

2 answers:

denis23 [38]3 years ago

6 0

Answer:

The correct option is C) 5, 5/2

The sequence given is:

80, 40, 20, 10, _____, ______

If we look at the given numbers, it appears that every next number is obtained by dividing the previous number of 2.

Like 40 is obtained by dividing 80 by 2

$\frac{80}{2}=40$

Similarly:

$\frac{40}{2}=20$

$\frac{20}{2}=10$

$\frac{10}{2} = 5$

$\frac{5}{2}$

Therefore, the complete sequence is:

80, 40, 20, 10, 5, 5/2

Hence the correct option is C) 5, 5/2

sergeinik [125]3 years ago

4 0

Answer:c

Step-by-step explanation:

You might be interested in

What is the missing polynomial?

german

Answer:

C. 40-4x-12x^2

Step-by-step explanation:

very simple

? – (20 – 4x – 5x2) = 20 – 7x2, so ? = 20 – 7x2 + (20 – 4x – 5x2)

? = 20 – 7x2 + (20 – 4x – 5x2)=20-7x²+20-4x-5x²=40-12x²-4x

7 0

3 years ago

Read 2 more answers

Integrala x la a treia ori ln la a doua dx va rog

Studentka2010 [4]

I don't speak Romanian, but the closest translation for this suggests you're trying to compute

$\displaystyle \int x^3 \ln(x)^2 \, dx$

Integrate by parts:

$\displaystyle \int x^3 \ln(x)^2 \, dx = uv - \int v \, du$

where

u = ln(x)² ⇒ du = 2 ln(x)/x dx

dv = x³ dx ⇒ v = 1/4 x⁴

$\implies \displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac12 \int x^3 \ln(x) \, dx$

Integrate by parts again:

$\displaystyle \int x^3 \ln(x) \, dx = u'v' - \int v' du'$

where

u' = ln(x) ⇒ du' = dx/x

dv' = x³ dx ⇒ v' = 1/4 x⁴

$\implies \displaystyle \int x^3 \ln(x) \, dx = \frac14 x^4 \ln(x) - \frac14 \int x^3 \, dx$

So, we have

$\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac12 \left(\frac14 x^4 \ln(x) - \frac14 \int x^3 \, dx \right)$

$\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac18 x^4 \ln(x) + \frac18 \int x^3 \, dx$

$\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac18 x^4 \ln(x) + \frac18 \left(\frac14 x^4\right) + C$

$\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac18 x^4 \ln(x) + \frac1{32} x^4 + C$

$\boxed{\displaystyle \int x^3 \ln(x)^2 \, dx = \frac1{32} x^4 \left(8\ln(x)^2 - 4\ln(x) + 1\right) + C}$

3 0

2 years ago

Write 6.75 as a mixed number.

KonstantinChe [14]

Answer:

Question 1 6 1/2

Step-by-step explanation:

Question 2 289/50

4 0

3 years ago

Read 2 more answers

Hello people ~<br>factorise↷<br> x^2 + 4x –21

EleoNora [17]

Let's see

$\\ \rm\rightarrowtail x^2+4x-21$

$\\ \rm\rightarrowtail x^2+7x-3x-21$

$\\ \rm\rightarrowtail x(x+7)-3(x+7)$

$\\ \rm\rightarrowtail (x-3)(x+7)$

8 0

2 years ago

Any one know the answer it has to be turned in tomorrow morning

Veseljchak [2.6K]

17: 900 because 3,000 + 900 + 40 +7 is 3,947 or 3,947-3,000-40-7= 900

6 0

3 years ago

Read 2 more answers

Find the next two terms of the sequence: 80, ?40, 20, ?10, ______, ______. A) 20, 40 B) 1 2 , 1 4 C) 5, ? 5 2 D) ?30, ?40

Find the next two terms of the sequence: 80, ?40, 20, ?10, , . A) 20, 40 B) 1 2 , 1 4 C) 5, ? 5 2 D) ?30, ?40