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2 years ago

9

Write 2.04 × 10 ⁴ as an ordinary number

1 answer:

ivann1987 [24]2 years ago

8 0

Answer:

20400

Step-by-step explanation:

its 2.03x10x10x10x10 so first ten: 20.4 and ten: 204 3rd ten: 2040 last ten :20400

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On a coordinate plane, the orange K is shifted to form the purple K, reflected across the x-axis to form the green k, and rotate

PIT_PIT [208]

Answer:

the green kin quadrant IV

Step-by-step explanation:

i took the quiz and got it right

3 0

3 years ago

what is the value of 8/15÷(-0.35) A -75/14 B -32/21 C -21/32 D -14/75 I would really appreciate it if you could also explain how

barxatty [35]

Answer;-1.5238095238
First step; 8/15

8/15 = 0.53 and so on in a repeating form. (0.53333333333)

Divide; 0.53 / -0.35

= -1.5238095238

5 0

3 years ago

Read 2 more answers

Complete the table for the function y=x+3 <br><br> x - y <br> 0 -<br> 2 -<br> 4 - <br> 6 -

Rus_ich [418]

For each x value, add 3 to get y values.
0+3=3
2+3=5
4+3=7
6+3=9

Final answers, respectively: 3,5,7,9

7 0

3 years ago

Find the slope of two points: (-2,-1) and (4,3)

marta [7]

Answer: m = 2/3

$\frac{-1-3}{-2-4}$ = $\frac{-4}{-6}$ = $\frac{2}{3}$

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