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Fudgin [204]
3 years ago
11

What is the answer 2/4

Mathematics
1 answer:
Lena [83]3 years ago
6 0

Answer:

1/2

Step-by-step explanation:

Divide the numerator and denominator by their common factor, to simplify in this case the common factor is 2.

2/4

1/2

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Ginger wants to fill her new swimming pool. She has two pumps; the larger pump takes 40 minutes to fill the pool, and the smalle
Talja [164]

As per the problem,

The larger pump takes 40 minutes to fill the pool.

The smaller one takes 60 minutes.

When both the pumps are working , let the time taken be x minutes , then we can write

\frac{1}{x}=\frac{1}{40}+\frac{1}{60}\\ \\ \text{Make the denominator equal we get}\\ \\    \frac{1}{x}=\frac{1}{40}*\frac{60}{60}+\frac{1}{60}*\frac{40}{40}\\ \\   \frac{1}{x}=\frac{60}{2400}+\frac{40}{24000}\\ \\   \text{Simplify we get}\\ \\ \frac{1}{x}=\frac{60+40}{2400}\\ \\ \frac{1}{x}=\frac{100}{2400}\\ \\ \frac{1}{x}=\frac{1}{24}\\ \\ x=24\\

Hence, when both the pumps works together, Time taken =24 minutes.

5 0
3 years ago
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Which of the following lines best fits the data shown in the scatter plot
xenn [34]

Answer:

Im pretty sure its A

Step-by-step explanation:

4 0
4 years ago
HELP MEeeeeeeeee g: R² → R a differentiable function at (0, 0), with g (x, y) = 0 only at the point (x, y) = (0, 0). Consider<im
GrogVix [38]

(a) This follows from the definition for the partial derivative, with the help of some limit properties and a well-known limit.

• Recall that for f:\mathbb R^2\to\mathbb R, we have the partial derivative with respect to x defined as

\displaystyle \frac{\partial f}{\partial x} = \lim_{h\to0}\frac{f(x+h,y) - f(x,y)}h

The derivative at (0, 0) is then

\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{f(0+h,0) - f(0,0)}h

• By definition of f, f(0,0)=0, so

\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{f(h,0)}h = \lim_{h\to0}\frac{\tan^2(g(h,0))}{h\cdot g(h,0)}

• Expanding the tangent in terms of sine and cosine gives

\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{\sin^2(g(h,0))}{h\cdot g(h,0) \cdot \cos^2(g(h,0))}

• Introduce a factor of g(h,0) in the numerator, then distribute the limit over the resulting product as

\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{\sin^2(g(h,0))}{g(h,0)^2} \cdot \lim_{h\to0}\frac1{\cos^2(g(h,0))} \cdot \lim_{h\to0}\frac{g(h,0)}h

• The first limit is 1; recall that for a\neq0, we have

\displaystyle\lim_{x\to0}\frac{\sin(ax)}{ax}=1

The second limit is also 1, which should be obvious.

• In the remaining limit, we end up with

\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{g(h,0)}h = \lim_{h\to0}\frac{g(h,0)-g(0,0)}h

and this is exactly the partial derivative of g with respect to x.

\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{g(h,0)-g(0,0)}h = \frac{\partial g}{\partial x}(0,0)

For the same reasons shown above,

\displaystyle \frac{\partial f}{\partial y}(0,0) = \frac{\partial g}{\partial y}(0,0)

(b) To show that f is differentiable at (0, 0), we first need to show that f is continuous.

• By definition of continuity, we need to show that

\left|f(x,y)-f(0,0)\right|

is very small, and that as we move the point (x,y) closer to the origin, f(x,y) converges to f(0,0).

We have

\left|f(x,y)-f(0,0)\right| = \left|\dfrac{\tan^2(g(x,y))}{g(x,y)}\right| \\\\ = \left|\dfrac{\sin^2(g(x,y))}{g(x,y)^2}\cdot\dfrac{g(x,y)}{\cos^2(g(x,y))}\right| \\\\ = \left|\dfrac{\sin(g(x,y))}{g(x,y)}\right|^2 \cdot \dfrac{|g(x,y)|}{\cos^2(x,y)}

The first expression in the product is bounded above by 1, since |\sin(x)|\le|x| for all x. Then as (x,y) approaches the origin,

\displaystyle\lim_{(x,y)\to(0,0)}\frac{|g(x,y)|}{\cos^2(x,y)} = 0

So, f is continuous at the origin.

• Now that we have continuity established, we need to show that the derivative exists at (0, 0), which amounts to showing that the rate at which f(x,y) changes as we move the point (x,y) closer to the origin, given by

\left|\dfrac{f(x,y)-f(0,0)}{\sqrt{x^2+y^2}}\right|,

approaches 0.

Just like before,

\left|\dfrac{\tan^2(g(x,y))}{g(x,y)\sqrt{x^2+y^2}}\right| = \left|\dfrac{\sin^2(g(x,y))}{g(x,y)}\right|^2 \cdot \left|\dfrac{g(x,y)}{\cos^2(g(x,y))\sqrt{x^2+y^2}}\right| \\\\ \le \dfrac{|g(x,y)|}{\cos^2(g(x,y))\sqrt{x^2+y^2}}

and this converges to g(0,0)=0, since differentiability of g means

\displaystyle \lim_{(x,y)\to(0,0)}\frac{g(x,y)-g(0,0)}{\sqrt{x^2+y^2}}=0

So, f is differentiable at (0, 0).

3 0
3 years ago
10. What simple interest rate is needed for $15,000 to double in 8 years? Round to the nearest tenth.​
Mrrafil [7]

Answer:

The simples interest rate needed is 12.5%

Step-by-step explanation:

Explanation is in the attachments.

Hope this helps! :)

7 0
3 years ago
Can you please simplify c5 x c
Rzqust [24]

Answer:

5c^2

Step-by-step explanation:

i think c is a variable so,

5xcxc = 5c^2

5 0
3 years ago
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