Find the number if distinguishable arrangements of the letters of the word.<br><br>HEEBIE-JEEBIES

A Student Who Is 60 Inches Tall Measured Shadows To Find The Height Of A Tree. The Students Shadow Was 24 Inches Long, And The S

True [87]

There are a many ways to solve this question, but this is the easier way:

60/x = 24/13
780 = 24x
32,5 = x or x = 32,5

7 0

2 years ago

Find an equation of the tangent line to the curve 2(x^2+y^2)2=25(x^2−y^2) (a lemniscate) at the point (3,1)

Sholpan [36]

$\bf 2[x^2+y^2]^2=25(x^2-y^2)\qquad \qquad 
\begin{array}{lllll}
&x_1&y_1\\
% (a,b)
&({{ 3}}\quad ,&{{ 1}})\quad 
\end{array}\\\\
-----------------------------\\\\
2\left[ x^4+2x^2y^2+y^4 \right]=25(x^2-y^2)\qquad thus
\\\\\\
2\left[ 4x^3+2\left[ 2xy^2+x^22y\frac{dy}{dx} \right]+4y^3\frac{dy}{dx} \right]=25\left[2x-2y\frac{dy}{dx} \right]
\\\\\\
2\left[ 4x^3+2\left[ 2xy^2+x^22y\frac{dy}{dx} \right]+4y^3\frac{dy}{dx} \right]=50\left[x-y\frac{dy}{dx} \right]
\\\\\\
$

$\bf \left[ 4x^3+2\left[ 2xy^2+x^22y\frac{dy}{dx} \right]+4y^3\frac{dy}{dx} \right]=25\left[x-y\frac{dy}{dx} \right]
\\\\\\
4x^3+4xy^2+4x^2y\frac{dy}{dx}+4y^3\frac{dy}{dx}+25y\frac{dy}{dx}=25x
\\\\\\
\cfrac{dy}{dx}[4x^2y+4y^3+25y]=25x-4x^3+4xy^2
\\\\\\
\cfrac{dy}{dx}=\cfrac{25x-4x^3+4xy^2}{4x^2y+4y^3+25y}\impliedby m=slope$

notice... a derivative is just the function for the slope

now, you're given the point 3,1, namely x = 3 and y = 1

to find the "m" or slope, use that derivative, namely $f'(3,1)=\cfrac{25x-4x^3+4xy^2}{4x^2y+4y^3+25y}$

that'd give you a value for the slope

to get the tangent line at that point, simply plug in the provided values
in the point-slope form

$\bf y-{{ y_1}}={{ m}}(x-{{ x_1}})\qquad
\begin{cases}
x_1=3\\
y_1=1\\
m=slope
\end{cases}\\ \qquad \uparrow\\
\textit{point-slope form}$

and then you solve it for "y", I gather you don't have to, but that'd be the equation of the tangent line at 3,1

6 0

3 years ago

Josephine can correct her students' test papers in 9 hours, but if her teacher's assistant helps, it would take them 7 hours. Ho

kirill [66]

Using the together rate, it is found that it would take the assistant 31 hours and 30 minutes to do it alone.

<h3>What is the together rate?</h3>

It is the sum of each separate rate.

In this problem, the rates are given as follows:

Josephine: 1/9.

Assistant: 1/x.

Together: 1/7.

Hence:

$\frac{1}{9} + \frac{1}{x} = \frac{1}{7}$

$\frac{x + 9}{9x} = \frac{1}{7}$

$9x = 7x + 63$

2x = 63

x = 31.5.

It would take the assistant 31 hours and 30 minutes to do it alone.

More can be learned about the together rate at brainly.com/question/25159431

7 0

1 year ago

Select the correct answer from each drop-down menu.

n200080 [17]

Answer:

= 62.68 + 4x

^^ hope this helps at all! :)

8 0

2 years ago

Read 2 more answers

Food allergies affect an estimated 7% of children under age 5 in the US. What is the probability that in a kindergarden class of

bezimeni [28]

Answer:

Probability of atleast one of 12 student has food allergies ≈ 0.58 ( approx)

Step-by-step explanation:

Given: Probability of a children under age 5 has food allergies = 7%

= $\frac{7}{100}$

To find : Probability of atleast one of 12 student has food allergies

Probability of a chindren under age 5 does not have food allergies = $1-\frac{7}{100}$

⇒ prob = $\frac{93}{100}$

now we find Probability of atleast one of 12 student has food allergies this means we have to find prob of 1 student, 2 student, 3 student, till 12 student have allergy out of 12 student of class then add all prob.

But instead of finding all these probability we find probability of student having no allergy.i.e., 0 student then subtract it from 1(total probability)

Probability of 0 student having allergy out of 12 student = $(\frac{93}{100})^{12}$

Therefore, Probability of atleast one of 12 student has food allergies

= $1-(\frac{93}{100})^{12}$

= $0.581403702521$

≈ 0.58 ( approx)

8 0

2 years ago

Read 2 more answers

Find the number if distinguishable arrangements of the letters of the word.HEEBIE-JEEBIES​

Find the number if distinguishable arrangements of the letters of the word.HEEBIE-JEEBIES