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2 years ago

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A recipe calls for 3/4 cup of sugar for every 1 1/2 cups of flour. How much sugar would be needed if the bakers uses 4 1/2 cup

s of flour

1 answer:

Murrr4er [49]2 years ago

7 0

2 1/4 cups hope it helps

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3+ jk + k*3 when j=2 and k=6

Licemer1 [7]

The answer is 39, jk=18 because j multiplied by k is 18 and then i added 3 and i got 21, then k (6) multiplied by 3 is 18, so then i added 21 and 18 and i got 39

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3 years ago

Read 2 more answers

What's the area of a square with a radius of 16 root 2

omeli [17]

That's impossible to say. The radius of a square is meaningless.

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2 years ago

Which statement is true about the equations –3x + 4y = 12 and 1/4x –1/3 y = 1?The system of the equations has exactly one soluti

iren [92.7K]

Answer:

The system of the equations has no solution; the two lines are parallel.

Step-by-step explanation:

The equations have the same slope. In standard form, this can be seen as the same coefficients for x and y. We will multiply the second equation by -12 to reveal this.

$\frac{1}{4} x-\frac{1}{3}y=1 \\-12(\frac{1}{4} x-\frac{1}{3}y=1)\\\frac{-12}{4} x-\frac{-12}{3}y=-12\\-3x+4y=-12$

This means the equations are parallel and will never cross. There is no solution.

4 0

2 years ago

Solve the following equations x1/3=2 <br> Y-2=9<br> (2s)1/2=9<br> 2n-1=16

Nookie1986 [14]

$x \times \frac{1}{3} = 2 \\ = > x = 2 \times 3 \\ = > x = 6$

$y - 2 = 9 \\ = > y = 9 + 2 = 11$

$(2s) \frac{1}{2} = 9 \\ = > s = 9$

$2n - 1 = 16 \\ = > 2n = 16 + 1 \\ = > 2n = 17 \\ = > n = \frac{17}{2} = 8.5$

<h3>The answers are :</h3><h3>x = 6</h3><h3>y = 11</h3><h3>s = 9</h3><h3>n = 8.5</h3><h3>Hope it helps!</h3><h3 />

7 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