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Nimfa-mama [501]
4 years ago
5

Does anyone know the answer to this

Mathematics
1 answer:
iVinArrow [24]4 years ago
4 0
The answer is A hehe

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Jaime has a budget in which 1/4 of the money is for photocopying 3/7 of the money is for computer expenses and the rest of the m
Marat540 [252]
Let
x-------> <span>Jaime budget
</span>s------> <span>fraction of Jaime budget for student tutors
</span><span>
we know that
x=(1/4)x+(3/7)x+s

3/7 is equal to-----> (3/7)*(2/2)----> 6/14
substitute
</span>x=(1/4)x+(6/14)x+s<span>
x=(7/14)x+s
s=x-(7/14)x
s=(7/14)x-------> (1/2)x------> </span>half of the budget is for <span>student tutors.

the answer is
</span>half of the budget is for student tutors
6 0
3 years ago
ILL GIVE BRAINLIEST <br><br> Given the diagram, which of the following must be true?
Alex787 [66]

Answer:

Choice D

Step-by-step explanation:

QS ≈ TR

they have same direction, sense and magnitude

8 0
3 years ago
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Explain how you would simplify the<br> following:
Juliette [100K]

Answer:

Step-by-step explanation:

185,000\\\sqrt{x}

8 0
4 years ago
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How can i differentiate this equation?
Dmitry_Shevchenko [17]

\bf y=\cfrac{2x^2-10x}{\sqrt{x}}\implies y=\cfrac{2x^2-10x}{x^{\frac{1}{2}}} \\\\\\ \cfrac{dy}{dx}=\stackrel{\textit{quotient rule}}{\cfrac{(4x-10)(\sqrt{x})~~-~~(2x^2-10x)\left( \frac{1}{2}x^{-\frac{1}{2}} \right)}{\left( x^{\frac{1}{2}} \right)^2}} \\\\\\ \cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})~~-~~(2x^2-10x)\left( \frac{1}{2\sqrt{x}} \right)}{\left( x^{\frac{1}{2}} \right)^2} \\\\\\ \cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})~~-~~\left( \frac{2x^2-10x}{2\sqrt{x}} \right)}{x}


\bf\cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})~~-~~\left( \frac{2x^2-10x}{2\sqrt{x}} \right)}{x} \\\\\\ \cfrac{dy}{dx}=\cfrac{ \frac{(4x-10)(\sqrt{x})(2\sqrt{x})~~-~~(2x^2-10x)}{2\sqrt{x}}}{x} \\\\\\ \cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})(2\sqrt{x})~~-~~(2x^2-10x)}{2x\sqrt{x}}


\bf \cfrac{dy}{dx}=\cfrac{(4x-10)2x~~-~~(2x^2-10x)}{2x\sqrt{x}}\implies \cfrac{dy}{dx}=\cfrac{8x^2-20x~~-~~(2x^2-10x)}{2x\sqrt{x}} \\\\\\ \cfrac{dy}{dx}=\cfrac{8x^2-20x~~-~~2x^2+10x}{2x\sqrt{x}} \implies \cfrac{dy}{dx}=\cfrac{6x^2-10x}{2x\sqrt{x}} \\\\\\ \cfrac{dy}{dx}=\cfrac{2x(3x-5)}{2x\sqrt{x}}\implies \cfrac{dy}{dx}=\cfrac{3x-5}{\sqrt{x}}

8 0
3 years ago
Solve for w: -4(w+1)=-24 Please explain this like you would a 2 year old and please explain it so it is the easiest way to solve
Andrews [41]

Answer:

Step-by-step explanation:

The first step in solving an equation is to look at it. Identify where the variable is that you're interested in, and see what operations are being performed on that variable.

Here, the variable w has 1 added to it, and the sum is multiplied by -4. The variable is only on the left side of the equation, and is inside parentheses. Inside parentheses, the coefficient of the variable is 1.

These observations tell you that, in some order, you will need to undo the operations of multiplication and addition, and you will have to eliminate parentheses.

You also observe that the multiplication factor of the parentheses (-4) is a factor of the constant on the right side of the equation (-24). This means you have a couple of options for solving this equation:

  1. eliminate parentheses first, using the distributive property
  2. divide by -4 first, to undo the multiplication

The second method is actually faster. We'll illustrate that one first.

__

Divide both sides of the equation by -4 to undo the multiplication by -4.

  -4(w +1)/(-4) = -24/(-4) . . . . . we have indicated the division we want to do

  w +1 = 6 . . . . . . . . . . . . . . . . here is the result of that division

Now, we can add the opposite of +1 to eliminate the constant from the left side. As with all operations on equations, we must do the same thing to both sides. This means we add -1 to both sides of the equation:

  w +1 -1 = 6 -1 . . . . . . we have indicated the addition we want to do

  w = 5 . . . . . . . . . . .  here is the result of that addition

The solution of the equation is ...

  w = 5

__

Often, when parentheses are involved, the numbers are such that the solution method is not as simple as what we just did. In those cases, you would take the approach of eliminating parentheses first. For this equation, it looks like this:

  -4(w +1) = -24 . . . . . . . given

  -4w -4 = -24 . . . . . . . . use the distributive property to eliminate parentheses

We want the variable term by itself, so we need to eliminate the constant -4 that is added on the left. To do that, we add its opposite. This means we add +4 to both sides of the equation.

  -4w -4 +4 = -24 +4 . . . . . . indicating the addition of +4

  -4w = -20 . . . . . . . . . . . . . the result of adding +4

Here, w is multiplied by -4. We want w by itself, so we want to undo that multiplication. That means we want to divide both sides of the equation by -4.

  -4w/(-4) = -20/(-4) . . . . . . indicate the division

  w = 5 . . . . . . . . . . . . . . . . the result of doing the division

The solution of the equation is ...

  w = 5

_____

Our basic approach is to <em>undo the operations done to the variable, in the reverse of the order in which they are done</em>. In that process, we observe the rules of equality: <em>whatever operation you do to one side of the equation must also be done to the other side</em>. (My teacher called this last rule: "keep the equal sign sacred.")

You can always use the distributive property for factoring, collecting terms, or eliminating parentheses.

3 0
3 years ago
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