All categories

2 years ago

Help plz due in 10min

Mathematics

2 answers:

JulsSmile [24]2 years ago

4 0

Answer:

1.15

Step-by-step explanation:

NemiM [27]2 years ago

3 0

Answer:

1.15

Step-by-step explanation:

So, first it would be helpful to change this from an improper fraction to a mixed number. Because we have an improper fraction now, we need to make this a mixed number by seeing how many times 23 can go into 20. Twenty-three can't go into 20, but if we take away 3 from 23 it can.

So, now we have 20/20 but with a remainder of 3/20. Because 20/20 = 1, it becomes a whole number of 1 and 3/20. Because there is a whole number of 1, we can put this as the whole number for the decimal. The fraction 3/20 is equal to the decimal 0.15, so we can now put the decimal together now that we have all the parts for the answer needed, which is 1.15.

I hope that this helps.

You might be interested in

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

Pooppooppooppooppooppoop

Andrei [34K]

Answer:

ok?

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

The weight of 72 books is 9 kg.what is the weight of 80 such books? How many such books would weigh 6kg?

Alex17521 [72]

9 divided by 72 = 0.125
0.125 is the weight of one book
0.125 times 80 = 10kg.
So 80 books weight 10kg.
72 divided by 9 = 8 books make up 1 kg.
8 times 6 =48

4 0

3 years ago

Ryan earns $20 for every lawn that he mows.<br> How much will Ryan earn if he mows 15 lawns?

Fiesta28 [93]

20*15=300

It's "too short", but there's not much else to say here.

6 0

3 years ago

Read 2 more answers

What is the coefficient of x2y3 in the expansion of (2x + y)5?

Zigmanuir [339]

Option C:

The coefficient of $x^{2} y^{3}$ is 40.

Solution:

Given expression:

$(2 x+y)^{5}$

Using binomial theorem:

$(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}$

Here $a=2 x, b=y$

Substitute in the binomial formula, we get

$(2x+y)^5=\sum_{i=0}^{5}\left(\begin{array}{l}5 \\i\end{array}\right)(2 x)^{(5-i)} y^{i}$

Now to expand the summation, substitute i = 0, 1, 2, 3, 4 and 5.

$$=\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}+\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1}+\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}+\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}$