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3 years ago

10 number for the binary 1000

Mathematics

1 answer:

puteri [66]3 years ago

5 0

The expanded form of 1000 base 2 is ...

... 1×2³ + 0×2² + 0×2¹ + 0×2⁰

... = 8+0+0+0 = 8 (base 10)

The only thing that changes is the number base. The rest of the behavior of the place-value number system is the same as in base ten.

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The sum of first three terms of a finite geometric series is -7/10 and their product is -1/125. [Hint: Use a/r, a, and ar to rep

Alchen [17]

Ooh, fun

geometric sequences can be represented as
$a_n=a(r)^{n-1}$
so the first 3 terms are
$a_1=a$
$a_2=a(r)$
$a_2=a(r)^2$

the sum is -7/10
$\frac{-7}{10}=a+ar+ar^2$
and their product is -1/125
$\frac{-1}{125}=(a)(ar)(ar^2)=a^3r^3=(ar)^3$

from the 2nd equation we can take the cube root of both sides to get
$\frac{-1}{5}=ar$
note that a=ar/r and ar²=(ar)r
so now rewrite 1st equation as
$\frac{-7}{10}=\frac{ar}{r}+ar+(ar)r$
subsituting -1/5 for ar
$\frac{-7}{10}=\frac{\frac{-1}{5}}{r}+\frac{-1}{5}+(\frac{-1}{5})r$
which simplifies to
$\frac{-7}{10}=\frac{-1}{5r}+\frac{-1}{5}+\frac{-r}{5}$
multiply both sides by 10r
-7r=-2-2r-2r²
add (2r²+2r+2) to both sides
2r²-5r+2=0
solve using quadratic formula
for $ax^2+bx+c=0$
$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$
so
for 2r²-5r+2=0
a=2
b=-5
c=2

$r=\frac{-(-5) \pm \sqrt{(-5)^2-4(2)(2)}}{2(2)}$
$r=\frac{5 \pm \sqrt{25-16}}{4}$
$r=\frac{5 \pm \sqrt{9}}{4}$
$r=\frac{5 \pm 3}{4}$
so
$r=\frac{5+3}{4}=\frac{8}{4}=2$ or $r=\frac{5-3}{4}=\frac{2}{4}=\frac{1}{2}$

use them to solve for the value of a
$\frac{-1}{5}=ar$
$\frac{-1}{5r}=a$
try for r=2 and 1/2
$a=\frac{-1}{10}$ or $a=\frac{-2}{5}$

test each
for a=-1/10 and r=2
a+ar+ar²= $\frac{-1}{10}+\frac{-2}{10}+\frac{-4}{10}=\frac{-7}{10}$
it works

for a=-2/5 and r=1/2
a+ar+ar²= $\frac{-2}{5}+\frac{-1}{5}+\frac{-1}{10}=\frac{-7}{10}$
it works

both have the same terms but one is simplified

the 3 numbers are $\frac{-2}{5}$ , $\frac{-1}{5}$ , and $\frac{-1}{10}$

6 0

3 years ago

1.9(2.3n+6)+10.45>43.7

Ulleksa [173]

Answer:

4.27n+21.4

Step-by-step explanation:

8 0

2 years ago

The surface area of a typical classroom floor is closest to______.

avanturin [10]

Answer would be C. Why A would be incorrect: 100m^2 is 100m × 100 m. That would be as big as a football field. Why B would be incorrect: 1cm^2 is 1cm × 1cm. You can use a regular 15 cm to measure a piece of paper with the length of 1cm each side. Why C would be correct: 1m^2 = 1m × 1m and it is also equals to 60cm × 60 cm. A 1 meter ruler = 60 cm = 4 times of your average 15 cm ruler. So, it is reasonable that a classroom is 10m by 10m in length and breadth. Why D would be incorrect: The same explanation applies to here as well. Things that can be measured with 1m × 1m is a square table so your classroom can't be that small to fit an average of 30 to 40 students in there. I hope my explanations were detailed and easy to understand. :)

4 0

2 years ago

Read 2 more answers

Trigonometry

Alla [95]

90° is equal to $\frac{\pi}{2}$ or 1.5707 radians.