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4 years ago

PLZZz help me out this is due today and need correct answer plz!!!also no spam answers!!!

Mathematics

2 answers:

Salsk061 [2.6K]4 years ago

7 0

Answer: 1) organelles

2) membrane

3) cytoplasm

4) nucleus

Step-by-step explanation: I’ve studied cells.

musickatia [10]4 years ago

7 0

The nucleus is the leader basically or the control center. A membrane is the shield of the cell that keeps the bad stuff out or a selective barrier. The cytoplasm is a substance that moves things around. Organelles are structures within a cell that perform specific functions like controlling cell growth and producing energy. I hope this helps. I made sure that everything was correct.

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Equivalent expression for 5a+30

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Answer:

5a +30

take 5 common

5( a+6)

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3 years ago

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3 years ago

The volume of a cube is 343 cm³ what is the surface area of the cube

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3 years ago

Read 2 more answers

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 ...

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

_____

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.)

_____

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 ...

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

_____

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.

6 0

3 years ago

Toby can but T-shirts in packs of 5 for 13.00 or in packs of 2 for 7.22. How much money does he save by buying 30 T-shirts at th

Helen [10]

Toby saves 30.30 money by buying 30 T-shirts at better price.

Step-by-step explanation:

Cost of pack of 5 T-shirts = 13.00 money.

Cost of pack of 2 T-shirts = 7.22 money.

Toby wants to buy total T-shirts = 30 T-shirts.

Case 1 :

For 30 T-shirts, If Toby buys in packs of 5 T-shirts then the number of packs he should buy = 30/5 = 6 packs.

The cost for six packs of 5 T-shirts = 6*13.00 = 78.00 money.

Case 2 :

For 30 T-shirts, If Toby buys in packs of 2 T-shirts then the number of packs he should buy = 30/2 = 15 packs.

The cost for 15 packs of 2 T-shirts = 15*7.22 = 108.30 money.

The money saved = 108.30 -78.00 = 30.30 money.

Therefore, The better price to buy 30 T-shirts is 78.00 money.

3 0

3 years ago

PLZZz help me out this is due today and need correct answer plz!!!also no spam answers!!!​

PLZZz help me out this is due today and need correct answer plz!!!also no spam answers!!!