1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Digiron [165]
3 years ago
11

The length and width of a rectangle are 5 feet and 2 feet, respectively. A similar rectangle has a width of 8 feet. What is the

length of the second rectangle
Mathematics
2 answers:
shutvik [7]3 years ago
4 0

Answer:

20

Step-by-step explanation:

they are proportional

so, since width increases by 4 times, length also increases by 4times

Sav [38]3 years ago
4 0

Answer:

20

Step-by-step explanation:

New Rectangle width: 8 ft

divide by old rectangle w (2ft)

= 4

4 is the rate of change.

length of O.R: 5ft

New R: 5 * 4 = 20

You might be interested in
Can someone help me answer this?
mart [117]

Answer:

x = 40 , y = 70

Step-by-step explanation:

y = 70°

x° = 180° - 2 × 70° = 40°

5 0
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 ...

a)

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 ...

d)

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 ...

f)

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

_____

<em>Comment on problem g</em>

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

6 0
2 years ago
What is the slope of a line that is parallel to the x-axis? 0 1 undefined
Troyanec [42]
Its 0 because the slope would be one if it was a positive line and itd go up and across one unitl and it is undefined if it is parallel to the y axis
3 0
2 years ago
Read 2 more answers
Fill in the blank. Given O below, you can conclude that OD is congruent to ___________.
Debora [2.8K]

The circle with center O has two chords AC and EF which are of same length 9.07.

OD and OB are the two perpendiculars drawn from the center O to the two chords AC and EF .It represents the distance of the chords from the centre.

The circle theorem states: congruent chords are equidistant from the center.

OD is congruent to OB.

Option A is the right answer.

4 0
3 years ago
Read 2 more answers
Ifm_2 = 68°, what is the measure of angle 7? 1 2. 3 5 7​
Oksana_A [137]

Answer:

∠7=22°

Step-by-step explanation:

∠2 and ∠7 are supplementary angles, meaning they add to 90°.

∠2+∠7=90°

substitute ∠2

68°+∠7=90°

∠7=22°

7 0
3 years ago
Other questions:
  • A car costs £14000. What would the price be after an 8% discount
    12·1 answer
  • What is the degree measure of x in this triangle?
    13·1 answer
  • Which algebraic expression is a polynomial with a degree of 3? 4x3 –  2y3 + 5y2 – 5y 3y3 –  –xy
    7·2 answers
  • Round each number to the nearest million, thousand, and ten
    6·1 answer
  • 45^2 can be found out by 4 x ____ x 100 + ____ = ____
    6·1 answer
  • Helppppp pleaseee :)
    5·1 answer
  • PLEASE ANSWER!!!!
    5·1 answer
  • Compute P(X) using the binomial probability formula. Then determine whether the normal distribution can be used to estimate this
    15·1 answer
  • Martha compra 2 kilos de tomate y 1 kilo de cebolla pagando S/ 8,40. Si la siguiente semana compra 1 kilo de tomate y 2 kilos de
    15·1 answer
  • 32% of 250 = ?<br> I need help hurry!!
    10·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!