I need help please.......

The recursive formula for an arithmetic sequence is an = an - 1 + d.

LekaFEV [45]

If an is a5
and
an-1 is a4

then using your recursive formula for an arithmetic sequence
an=an-1 +d
then
a5=a4+d

now, a4 =6 and common difference "d" is d=-11
hence
a5=6 -11
6-11= -5

any questions?

3 0

3 years ago

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Can anybody explain how to subtract mixed fractions?

Ilya [14]

Answer:

<em>convert </em>the mixed fractions to improper fractions (where the numerator is greater than or equal to the denominator): multiply the whole number part by the fraction's denominator, add that to the numerator, write the result on top of the denominator.
if the denominators are not the same, work out the common denominator and <em>rewrite </em>the fractions with the same denominators
subtract by subtracting the numerators and writing the result over the denominator
convert back to mixed fractions by dividing the numerator by the denominator, write down the whole number answer, write down the remainder above the denominator.

Example

$3\frac23-1\frac45$

convert to improper fractions:

$\dfrac{3 \times 3+2}{3}-\dfrac{1 \times 5+4}{5}=\dfrac{11}{3}-\dfrac{9}{5}$

common denominator = 3 × 5 = 15, so:

$\dfrac{11}{3}-\dfrac{9}{5}=\dfrac{11\times 5}{3\times 5}-\dfrac{9\times 3}{5\times 3}=\dfrac{55}{15}-\dfrac{27}{15}$

subtract:

$\dfrac{55}{15}-\dfrac{27}{15}=\dfrac{55-27}{15}=\dfrac{28}{15}$

convert back to mixed fractions:

$28 \div 15=1 \textsf{ remainder }13=1 \frac{13}{15}$

7 0

2 years ago

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Одна зі сторін прямокутника в 11 разів менша від другої. Знайдіть

Ratling [72]

Answer:

for the plug-in settings to determine

Step-by-step explanation:

GG from Shahs of sunset Blvd suite is the best way to the answer to the answer to the answer to the plug-in the plug-in the whole thing is I don't have a email the e I m not sure if you have any questions or concerns please visit the plug-in settings to determine

5 0

1 year ago

What would be the image of Point B(3, 2) after the following transformations have been applied:

Ulleksa [173]

Answer:

(9,6)

Step-by-step explanation:

(6,9)

Because the scale factor is 3 so u r suppose to multiply by 3

6 0

2 years ago

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Enter the coefficients of the fifth Taylor polynomial T5(x) for the function f(x) = x5−3x4+2x2+5x−2 based at b=1. T5(x)= + (x−1)

DENIUS [597]

Compute the necessary values/derivatives of $f(x)$ at $x=1$ :

$f(1)=3$

$f'(1)=2$

$f''(1)=-12$

$f'''(1)=-12$

$f^{(4)}(1)=48$

$f^{(5)}(1)=120$

Taylor's theorem then says we can "approximate" (in quotes because the Taylor polynomial for a polynomial is another, exact polynomial) $f(x)$ at $x=1$ by

$T_5(x)=\dfrac3{0!}+\dfrac2{1!}(x-1)-\dfrac{12}{2!}(x-1)^2-\dfrac{12}{3!}(x-1)^3+\dfrac{48}{4!}(x-1)^4+\dfrac{120}{5!}(x-1)^5$

$T_5(x)=3+2(x-1)-6(x-1)^2-2(x-1)^3+2(x-1)^4+(x-1)^5$

###

Another way of doing this would be to solve for the coefficients $a,b,c,d,e,g$ in

$f(x)=a+b(x-1)+c(x-1)^2+d(x-1)^3+e(x-1)^4+g(x-1)^5$

by expanding the right hand side and matching up terms with the same power of $x$ .

5 0

2 years ago