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2 years ago

Explain how you can use long division to see that the rational number is

1 answer:

8 0

To use division to find the decimal equivalent of a rational number, divide the top number by the bottom number. You will get a decimal number.

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What is 1 + 1? answer

Lady bird [3.3K]

Answer:

Step-by-step explanation:

+ 1

hope this helps...

3 0

3 years ago

FIND.......16 2/3 % of 96 lamps

r-ruslan [8.4K]

Convert percentage to fraction by dividing by 100
16 2/3 = 16 2/3 ÷ 100 = 1/6

16 2/3 % of 96 = 1/6 x 96 = 16

Ans: 16 2/3 % of 96 lamps = 16

4 0

3 years ago

Vladimir buys 1.20 pounds of Skittles, 312 pounds of M&Ms, and 5.2 pounds of pretzels. If the candy cost $1.20 a pound and t

BlackZzzverrR [31]

Answer: $10.90

Step-by-step explanation:

Skittles-1.44

M&M's-3.744

Pretzels-5.72

Not sure if its a typo, thats if theres 3.12lbs of of M&M's

The answer is theres 312 M&M's is 374.4

6 0

3 years ago

There are 3 star fighters and 10 aliens in the play space trek. Each alien costume takes 2 1/4 yards of material. How much mater

Leno4ka [110]

Answer:

22.5 yards

Step-by-step explanation:

10 costumes at 2.25 yards each is ...

10 · 2.25 = 22.5 . . . . yards of material

6 0

3 years ago

Use Euler's method with step size 0.2 to estimate y(1), where y(x) is the solution of the initial-value problem y' = x2y − 1 2 y

irina [24]

Answer:

Therefore the value of y(1)= 0.9152.

Step-by-step explanation:

According to the Euler's method

y(x+h)≈ y(x) + hy'(x) ....(1)

Given that y(0) =3 and step size (h) = 0.2.

$y'(x)= x^2y(x)-\frac12y^2(x)$

Putting the value of y'(x) in equation (1)

$y(x+h)\approx y(x) +h(x^2y(x)-\frac12y^2(x))$

Substituting x =0 and h= 0.2

$y(0+0.2)\approx y(0)+0.2[0\times y(0)-\frac12 (y(0))^2]$

$\Rightarrow y(0.2)\approx 3+0.2[-\frac12 \times3]$ [∵ y(0) =3 ]

$\Rightarrow y(0.2)\approx 2.7$

Substituting x =0.2 and h= 0.2

$y(0.2+0.2)\approx y(0.2)+0.2[(0.2)^2\times y(0.2)-\frac12 (y(0.2))^2]$

$\Rightarrow y(0.4)\approx 2.7+0.2[(0.2)^2\times 2.7- \frac12(2.7)^2]$

$\Rightarrow y(0.4)\approx 1.9926$

Substituting x =0.4 and h= 0.2

$y(0.4+0.2)\approx y(0.4)+0.2[(0.4)^2\times y(0.4)-\frac12 (y(0.4))^2]$

$\Rightarrow y(0.6)\approx 1.9926+0.2[(0.4)^2\times 1.9926- \frac12(1.9926)^2]$

$\Rightarrow y(0.6)\approx 1.6593$

Substituting x =0.6 and h= 0.2

$y(0.6+0.2)\approx y(0.6)+0.2[(0.6)^2\times y(0.6)-\frac12 (y(0.6))^2]$

$\Rightarrow y(0.8)\approx 1.6593+0.2[(0.6)^2\times 1.6593- \frac12(1.6593)^2]$

$\Rightarrow y(0.6)\approx 0.8800$

Substituting x =0.8 and h= 0.2

$y(0.8+0.2)\approx y(0.8)+0.2[(0.8)^2\times y(0.8)-\frac12 (y(0.8))^2]$

$\Rightarrow y(1.0)\approx 0.8800+0.2[(0.8)^2\times 0.8800- \frac12(0.8800)^2]$

$\Rightarrow y(1.0)\approx 0.9152$

Therefore the value of y(1)= 0.9152.

4 0

3 years ago