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3 years ago

11

What is 6=x over 8 for fractions

2 answers:

boyakko [2]3 years ago

6 0

Answer:

48

<h3><em><u>Could I please have BRAINLIEST.</u></em></h3>

9966 [12]3 years ago

4 0

Answer:

x = $\frac{3}{4}$

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Which fraction shows a correct way to set up the slope

belka [17]

(4, -6) e (-2, 1)
a = Δy / Δx
a = (y-yo) / (x-xo)
a = [1-(-6)] / (-2-4)

7 0

3 years ago

Read 2 more answers

Solve the system of equations by substitution.<br> 3/8x + 1/3y = 17/24<br> x + 7y = 8

bixtya [17]

Using the second equation, let’s find x

x + 7y = 8
x = 8 - 7y

Substitute x in the other equation

3/8(8 - 7y) + 1/3y = 17/24
3 - 21/8y + 1/3y = 17/24
-55/24y = -55/24
y = 1

Plug in y in any equation

x + 7(1) = 8
x + 7 = 8
x = 1

x = 1 and y = 1

3 0

2 years ago

I need help please get on z o o m

sergiy2304 [10]

Answer:

Ok

Step-by-step explanation:

Send me link I will help u

5 0

2 years ago

Problem page the sum of two numbers is 47 . the larger number is 17 more than the smaller number. what are the numbers?

user100 [1]

Smaller number = X
Larger Number = X + 17

47 = X + (X+17)
47 = 2x + 17
47 - 17 = 2x
30 = 2x
X = 15

3 0

2 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