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3 years ago

What is the answer to 7/8 times 2/9

Mathematics

2 answers:

deff fn [24]3 years ago

6 0

Answer:

7/36

Step-by-step explanation:

7/8 x 2/9

=> Multiply the numerators and the denominators

=> 7 x 2 / 8 x 9

=> 14 / 72

=> 7/ 36

aleksandrvk [35]3 years ago

4 0

Answer:

the answer would be 7/36

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Your friend earns $10.50 per hour. This is 125% of her hourly wage last year. How much did your friend earn per hour last year?

Mrac [35]

I believe it is 8.40

5 0

3 years ago

Ifteen bales of hale were loaded into a truck. Some of the bales weight 150 pounds, and some weighed 100 punds. Hpw many bales o

horrorfan [7]

Answer:

There are 7 bales of 100 pounds and 8 bales of 150 pounds hale

Step-by-step explanation:

Let the amount of bale of 100 pounds be x and the bale of hale of 150pounds be y

If fifteen bales of hale were loaded into a truck, then x+y = 15 ... 1

Also if some of the bales weight 150 pounds, and some weighed 100 pounds with total weight of pounds to be 1900 pounds, then mathematically;

150y + 100x = 1900

Dividing through by 100;

1.5y+x = 19 ..... 2

Solving equation 1 and 2 simultaneously;

x+y = 15 ... 1

1.5y+x = 19 .... 2

Subtracting 1 from 2

y-1.5y = 15-19

-0.5y = -4

Divide both sides by 0.5

-0.5y/-0.5 = -4/-0.5

y = 8

Substituting y = 8 into equation 1, we will have;

x+y = 15

x+8 = 15

x = 15-8

x = 7

Hence there are 7 bales of 100 pounds and 8 bales of 150 pounds hale

6 0

3 years ago

An=an-1+6an-2 n>2 a0=3 a1=6

Svetach [21]

$\begin{cases}a_0=3\\a_1=6\\a_n=a_{n-1}+6a_{n-2}&\text{for }n>2\end{cases}$

Let the generating function for $a_n$ be

$A(x)=\displaystyle\sum_{n\ge0}a_nx^n$

Multiplying both sides by $x^{n-2}$

$a_nx^{n-2}=a_{n-1}x^{n-2}+6a_{n-2}x^{n-2}$

Summing both sides over the non-negative integers greater than or equal to 2 gives

$\displaystyle\sum_{n\ge2}a_nx^{n-2}=\sum_{n\ge2}a_{n-1}x^{n-2}+6\sum_{n\ge2}a_{n-2}x^{n-2}$
$\displaystyle\frac1{x^2}\sum_{n\ge2}a_nx^n=\frac1x\sum_{n\ge1}a_nx^n+6\sum_{n\ge0}a_nx^n$
$\displaystyle\frac1{x^2}\left(\sum_{n\ge0}a_nx^n-a_0-a_1x\right)=\frac1x\left(\sum_{n\ge0}a_nx^n-a_0\right)+6\sum_{n\ge0}a_nx^n$
$\displaystyle\frac1{x^2}\left(A(x)-3-6x\right)=\frac1x\left(A(x)-3\right)+6A(x)$
$A(x)=\dfrac{3x+3}{1-x-6x^2}$
$A(x)=\dfrac3{5(1+2x)}+\dfrac{12}{5(1-3x)}$

For $|x|$ , the two series converge to

$\displaystyle A(x)=\frac35\sum_{n\ge0}(-2x)^n+\frac{12}5\sum_{n\ge0}(3x)^n$
$\implies A(x)=\displaystyle\sum_{n\ge0}\dfrac{3(-2)^n+12(3)^n}5x^n$
$a_n=\dfrac{3(-2)^n+12(3)^n}5=\dfrac{3(-2)^n+4(3)^{n+1}}5$