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3 years ago

Whats -1 1/10 divided by -11/10

Mathematics

2 answers:

Bogdan [553]3 years ago

4 0

Answer:

Step-by-step explanation:

Any number divided by itself is 1

Dafna11 [192]3 years ago

3 0

Answer:

$-1 \frac{1}{10} \div -\frac{11}{10} = \frac{1}{100}$

Decimal:

0.01

Step-by-step explanation:

$-1\frac{1}{10}\div -\frac{11}{10}$

Convert mixed numbers to improper fractions: $1 \frac{1}{10} = \frac{11}{10}$

$-\frac{11}{10}\div -\frac{11}{10}$

$\frac{\frac{11}{10} }{-11} = -\frac{11}{110}$

$=-\frac{-\frac{11}{110}}{10}$

Simplify $\frac{-\frac{11}{110}}{10} : -\frac{11}{1100}$

$=-\left(-\frac{11}{1100}\right)$

Cancel $\frac{11}{1100}: \frac{1}{100}$

$=-\left(-\frac{1}{100}\right)$

Apply rule $-\left(-a\right)=a$ :

$=\frac{1}{100}$

Hope I helped. If so, may I get brainliest and a thanks?

Thank you, have a good day! =)

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Solve these recurrence relations together with the initial conditions given. a) an= an−1+6an−2 for n ≥ 2, a0= 3, a1= 6 b) an= 7a

8_murik_8 [283]

Answer:

a) 3/5·((-2)^n + 4·3^n)
b) 3·2^n - 5^n
c) 3·2^n + 4^n
d) 4 - 3 n
e) 2 + 3·(-1)^n
f) (-3)^n·(3 - 2n)
g) ((-2 - √19)^n·(-6 + √19) + (-2 + √19)^n·(6 + √19))/√19

Step-by-step explanation:

These homogeneous recurrence relations of degree 2 have one of two solutions. Problems a, b, c, e, g have one solution; problems d and f have a slightly different solution. The solution method is similar, up to a point.

If there is a solution of the form $a[n]=r^n$ , then it will satisfy ...

$r^n=c_1\cdot r^{n-1}+c_2\cdot r^{n-2}$

Rearranging and dividing by $r^{n-2}$ , we get the quadratic ...

$r^2-c_1r-c_2=0$

The quadratic formula tells us values of r that satisfy this are ...

$r=\dfrac{c_1\pm\sqrt{c_1^2+4c_2}}{2}$

We can call these values of r by the names r₁ and r₂.

Then, for some coefficients p and q, the solution to the recurrence relation is ...

$a[n]=pr_1^n+qr_2^n$

We can find p and q by solving the initial condition equations:

$\left[\begin{array}{cc}1&1\\r_1&r_2\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]$

These have the solution ...

$p=\dfrac{a[0]r_2-a[1]}{r_2-r_1}\\\\q=\dfrac{a[1]-a[0]r_1}{r_2-r_1}$

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Using these formulas on the first recurrence relation, we get ...

$c_1=1,\ c_2=6,\ a[0]=3,\ a[1]=6\\\\r_1=\dfrac{1+\sqrt{1^2+4\cdot 6}}{2}=3,\ r_2=\dfrac{1-\sqrt{1^2+4\cdot 6}}{2}=-2\\\\p=\dfrac{3(-2)-6}{-5}=\dfrac{12}{5},\ q=\dfrac{6-3(3)}{-5}=\dfrac{3}{5}\\\\a[n]=\dfrac{3}{5}(-2)^n+\dfrac{12}{5}3^n$

The rest of (b), (c), (e), (g) are solved in exactly the same way. A spreadsheet or graphing calculator can ease the process of finding the roots and coefficients for the given recurrence constants. (It's a matter of plugging in the numbers and doing the arithmetic.)

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For problems (d) and (f), the quadratic has one root with multiplicity 2. So, the formulas for p and q don't work and we must do something different. The generic solution in this case is ...

$a[n]=(p+qn)r^n$

The initial condition equations are now ...

$\left[\begin{array}{cc}1&0\\r&r\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]$

and the solutions for p and q are ...

$p=a[0]\\\\q=\dfrac{a[1]-a[0]r}{r}$

Using these formulas on problem (d), we get ...

$c_1=2,\ c_2=-1,\ a[0]=4,\ a[1]=1\\\\r=\dfrac{2+\sqrt{2^2+4(-1)}}{2}=1\\\\p=4,\ q=\dfrac{1-4(1)}{1}=-3\\\\a[n]=4-3n$

And for problem (f), we get ...

$c_1=-6,\ c_2=-9,\ a[0]=3,\ a[1]=-3\\\\r=\dfrac{-6+\sqrt{6^2+4(-9)}}{2}=-3\\\\p=3,\ q=\dfrac{-3-3(-3)}{-3}=-2\\\\a[n]=(3-2n)(-3)^n$

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Comment on problem g

Yes, the bases of the exponential terms are conjugate irrational numbers. When the terms are evaluated, they do resolve to rational numbers.

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2 years ago

The rule for being the nth item in a certain sequence is n(n+1)/2. what is the 4th term in this sequence ?

Lorico [155]

Answer:

a.) 4

Step-by-step explanation:

a.)4

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2 years ago

There is 1000cm3 of aluminum available to cast a trophy that will be in the shape of a right square pyramid. Is this enough alum

stealth61 [152]

The 1000 cubic centimeters of aluminium is enough for aluminium a trophy that will be in the shape of a right square pyramid and has a base edge of 10 cm and a slant height of 13 cm.

Step-by-step explanation:

The given is,

Volume of aluminium available is 1000 cubic centimeters

Shape of trophy is right square pyramid

Trophy has a base edge of 10 cm and slant height of 13 cm

Step:1

Formula for volume of right square pyramid,

$Volume, V = a^{2}\frac{h}{3}$ .....................................(1)

Where, a - Base edge value

h - Height of pyramid

From given,

a = 10 cm

h = 13 cm

Equation (1) becomes,

$= 10^{2}(\frac{13}{3} )$

$= (100)(4.333)$

$= 433.33 cm^{3}$

Volume of trophy = 433.33 cubic centimeters

Compare with the volume of available aluminium and volume of right square pyramid,

$Volume of available aluminium > Volume of right square pyramid$

$1000 cm^{3} > 433.33 cm^{3}$

So, the given volume of aluminium is enough to make right square pyramid shaped trophy.

Result:

The 1000 cubic centimeters of aluminium is enough for aluminium a trophy that will be in the shape of a right square pyramid and has a base edge of 10 cm and a slant height of 13 cm.

6 0

3 years ago

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answers

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Step-by-step explanation:

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