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2 years ago

What is the solution to the equation 1/4x+2=-5/8x-5? O X=-8 O X=-7 O X=7 O X=8

Mathematics

1 answer:

Elza [17]2 years ago

7 0

Answer:

The solution to the equation $\frac{1}{4}x+2=\frac{-5}{8}x-5$ is $\mathbf{x=-8}$

Option A is correct option.

Step-by-step explanation:

What is the solution to the equation $\frac{1}{4}x+2=\frac{-5}{8}x-5$

Solving the equation

$\frac{1}{4}x+2=\frac{-5}{8}x-5$

Subtracting 2 on both sides

$\frac{1}{4}x+2-2=\frac{-5}{8}x-5-2\\\frac{1}{4}x=\frac{-5}{8}x-7$

Adding 5/8x on both sides

$\frac{1}{4}x+\frac{5}{8}x=\frac{-5}{8}x-7+\frac{5}{8}x\\\frac{2x+5x}{8}=-7 \\\frac{7x}{8}=-7$

Multiply both sides by 8/7

$\frac{7x}{8}\times \frac{8}{7}=-7 \times \frac{8}{7}\\x=-8$

So, we get x = -8

The solution to the equation $\frac{1}{4}x+2=\frac{-5}{8}x-5$ is $\mathbf{x=-8}$

Option A is correct option.

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2 years ago

A scientist mixes water (containing no salt) with a solution that contains 35% salt. She wants to obtain 245 ounces of a mixture

sesenic [268]

Answer:

140 ounces of water and 105 ounces of 35% salt solution.

Explanation:

Let $x$ be the ounces of water that need to be added and $y$ be the ounces of 35% solution that need to to be added.

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35% of $x$ is the salt, and since no salt is added with addition of $y$ liters of water, this must be equal to the amount of salt in the final 15% solution:

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hichkok12 [17]

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Given

$\vec v = f(x,y,z)\,\vec\imath+g(x,y,z)\,\vec\jmath+h(x,y,z)\,\vec k \\\\ \vec v = \sin(x)\cos(y)\tan(z)\,\vec\imath + x^2y^2z^2\,\vec\jmath+x^4y^4z^4\,\vec k$

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$\displaystyle \nabla\times\vec v = \left(\frac{\partial h}{\partial y}-\frac{\partial g}{\partial z}\right)\,\vec\imath - \left(\frac{\partial h}{\partial x}-\frac{\partial f}{\partial z}\right)\,\vec\jmath + \left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right)\,\vec k$

$\nabla\times\vec v = \left(4x^4y^3z^4-2x^2y^2z\right)\,\vec\imath \\\\ - \left(4x^3y^4z^4-\sin(x)\cos(y)\sec^2(z)\right)\,\vec\jmath \\\\ + \left(2xy^2z^2+\sin(x)\sin(y)\tan(z)\right)\,\vec k$

$\nabla\times\vec v = \left(4x^4y^3z^4-2x^2y^2z\right)\,\vec\imath \\\\ + \left(\sin(x)\cos(y)\sec^2(z)-4x^3y^4z^4\right)\,\vec\jmath \\\\ + \left(2xy^2z^2+\sin(x)\sin(y)\tan(z)\right)\,\vec k$

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3 years ago