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3 years ago

7

johnny saves all of his nickles and dimes in a jar. One day he counted them and found that there were 73 coins worth $4.55. How

many nickels and dimes were in the jar?

1 answer:

juin [17]3 years ago

7 0

Ummm I'm not sure but let me do the math and I'll tell u :)

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Jacob earned $80 babysitting and deposited the money into his savings account. The next week he spent $85 Lon video games. Use i

nexus9112 [7]

After one week, Jacob has earned $80 from babysitting, so his change is an increase of $80. The next week, he spends $85, so he experiences a decrease of $85. Over the two weeks, his overall change is $80-$85=-$5, or a decrease of $5.

5 0

3 years ago

2.6.5 A plant physiologist grew birch seedlings in the green-house and measured the ATP content of their roots. (See Example 1.1

Ilya [14]

Answer:

$(a)\ \bar x = 1.19$

$(b)\ \sigma_x = 0.18$

Step-by-step explanation:

Given

$n = 4$

$x: 1.45\ 1.19\ 1.05\ 1.07$

Solving (a): The mean

This is calculated as:

$\bar x = \frac{\sum x}{n}$

So, we have:

$\bar x = \frac{1.45 + 1.19 + 1.05 + 1.07}{4}$

$\bar x = \frac{4.76}{4}$

$\bar x = 1.19$

Solving (b): The standard deviation

This is calculated as:

$\sigma_x = \sqrt{\frac{\sum(x - \bar x)^2}{n - 1}}$

So, we have:

$\sigma_x = \sqrt{\frac{(1.45 -1.19)^2 + (1.19 -1.19)^2 + (1.05 -1.19)^2 + (1.07 -1.19)^2}{4 - 1}}$

$\sigma_x = \sqrt{\frac{(0.1016}{3}}$

$\sigma_x = \sqrt{0.033867}$

$\sigma_x = 0.18$

4 0

3 years ago

1. Let f(x, y) be a differentiable function in the variables x and y. Let r and θ the polar coordinates,and set g(r, θ) = f(r co

Olenka [21]

Answer:

$g_{r}(\sqrt{2},\frac{\pi}{4})=\frac{\sqrt{2}}{2}\\$

Step-by-step explanation:

First, notice that:

$g(\sqrt{2},\frac{\pi}{4})=f(\sqrt{2}cos(\frac{\pi}{4}),\sqrt{2}sin(\frac{\pi}{4}))\\$

$g(\sqrt{2},\frac{\pi}{4})=f(\sqrt{2}(\frac{1}{\sqrt{2}}),\sqrt{2}(\frac{1}{\sqrt{2}}))\\$

$g(\sqrt{2},\frac{\pi}{4})=f(1,1)\\$

We proceed to use the chain rule to find $g_{r}(\sqrt{2},\frac{\pi}{4})$ using the fact that $X(r,\theta)=rcos(\theta)\ and\ Y(r,\theta)=rsin(\theta)$ to find their derivatives:

$g_{r}(r,\theta)=f_{r}(rcos(\theta),rsin(\theta))=f_{x}( rcos(\theta),rsin(\theta))\frac{\delta x}{\delta r}(r,\theta)+f_{y}(rcos(\theta),rsin(\theta))\frac{\delta y}{\delta r}(r,\theta)\\$

Because we know $X(r,\theta)=rcos(\theta)\ and\ Y(r,\theta)=rsin(\theta)$ then:

$\frac{\delta x}{\delta r}=cos(\theta)\ and\ \frac{\delta y}{\delta r}=sin(\theta)$

We substitute in what we had:

$g_{r}(r,\theta)=f_{x}( rcos(\theta),rsin(\theta))cos(\theta)+f_{y}(rcos(\theta),rsin(\theta))sin(\theta)$

Now we put in the values $r=\sqrt{2}\ and\ \theta=\frac{\pi}{4}$ in the formula:

$g_{r}(\sqrt{2},\frac{\pi}{4})=f_{r}(1,1)=f_{x}(1,1)cos(\frac{\pi}{4})+f_{y}(1,1)sin(\frac{\pi}{4})$

Because of what we supposed:

$g_{r}(\sqrt{2},\frac{\pi}{4})=f_{r}(1,1)=-2cos(\frac{\pi}{4})+3sin(\frac{\pi}{4})$

And we operate to discover that:

$g_{r}(\sqrt{2},\frac{\pi}{4})=-2\frac{\sqrt{2}}{2}+3\frac{\sqrt{2}}{2}$

$g_{r}(\sqrt{2},\frac{\pi}{4})=\frac{\sqrt{2}}{2}$

and this will be our answer

3 0

3 years ago

HELPPP ILL GIVE EXTRA POINTS QUESTION ABOVE

MatroZZZ [7]

Answer:

The answer is the top option choice "both i and ii."

Step-by-step explanation:

Hopes this helps.

8 0

3 years ago

16+36= distributive property

lutik1710 [3]

Answer:

4(4 + 9)

Step-by-step explanation:

Well... you can divide each by 4 and you get 4(4 + 9)

That's about all you can do for the distributive property.

8 0

3 years ago

Read 2 more answers