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3 years ago

A bulletin board is 3ft high, it's area is 15 square ft what is the perimeter of the bulletin board

Mathematics

2 answers:

Zinaida [17]3 years ago

6 0

Yes it is indeed 16 feet

Molodets [167]3 years ago

3 0

16 Feet because 15 divided by 3 is 5 then you add 3+3+5+5

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HELPPP PLEASEEE !!! John draws AABD, with CD as the perpendicular bisector of AB such that AACD and ABCD are congruent to each o

laila [671]

Answer:

The points on the perpendicular bisector of a side of a triangle are equidistant from the vertices of the side it bisects.

Step-by-step explanation:

It is the last option. The perpendicular bisector theorem states that if a point lies on the bisector of a segment it is equidistant from the endpoints.

Meaning

If a perpendicular bisector is a line of the side of the triangle , it bisects the sides forming two right angles .

The first three choices are incorrect because

1) the figure shows a triangle bisected into two triangles and option 1 tells about 1 isosceles triangle.

2) The base angles of any triangle can be different or same .

3) the three perpendicular bisectors meet at a point called the circumeter. We have 1 perpendicual bisector which is dividing the triangle into two equal triangles.

4 0

2 years ago

Help me out last time on this app for this lat question big points and brainlest

Ahat [919]

Answer:

0.20m+650<775

Step-by-step explanation:

Less than 775=<775

650 is independent

8 0

3 years ago

Read 2 more answers

Rewrite the expression without the negative exponent. 2x^−4

romanna [79]

Answer:

$\frac{2}{x^4}$

Step-by-step explanation:

The expression is $2x^{-4}$

We use the rule of exponents shown below to change the original to "positive" exponent form:

$a^-b=\frac{1}{a^b}$

Now we can change the original expression given into positive exponent:

$2x^{-4}\\=\frac{2}{x^4}$

8 0

2 years ago

Read 2 more answers

Find the derivative of following function.

Aleks04 [339]

Answer:

$\displaystyle y' = \frac{\big( -2 \cos x \sin x - \frac{3}{2\sqrt{x}} \big) \big( \tan^2 x + 5x \big) + \big( \cos^2 x - 3\sqrt{x} + 6 \big) \big( 2 \sec^2 x \tan x + 5 \big)}{ \big( \csc^2 x + 3 \big) \big( \sin^2 x + 6 \big)} + \frac{2 \cot x \csc^2 x \big( \cos^2 x - 3\sqrt{x} + 6 \big) \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big)^2 \big( \sin^2x + 6 \big)} - \frac{2 \cos x \sin x \big( \cos^2 x - 3\sqrt{x} + 6 \big) \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big) \big( \sin^2 x + 6 \big)^2}$

General Formulas and Concepts:
Calculus

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]:
$\displaystyle (cu)' = cu'$

Derivative Property [Addition/Subtraction]:
$\displaystyle (u + v)' = u' + v'$

Derivative Rule [Basic Power Rule]:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Product Rule]:
$\displaystyle (uv)' = u'v + uv'$

Derivative Rule [Quotient Rule]:
$\displaystyle \bigg( \frac{u}{v} \bigg)' = \frac{vu' - uv'}{v^2}$

Derivative Rule [Chain Rule]:
$\displaystyle [u(v)]' = u'(v)v'$

Step-by-step explanation:

*Note:

Since the answering box is way too small for this problem, there will be limited explanation.

Step 1: Define

Identify.

$\displaystyle y = \frac{\cos^2 x - 3\sqrt{x} +6}{\sin^2 x + 6} \times \frac{\tan^2 x + 5x}{\csc^2 x + 3}$

Step 2: Differentiate

We can differentiate this function with the use of the given derivative rules and properties.

Applying Product Rule:

$\displaystyle y' = \bigg( \frac{\cos^2 x - 3\sqrt{x} + 6}{\sin^2 x + 6} \bigg)' \frac{\tan^2 x + 5x}{\csc^2 x + 3} + \frac{\cos^2 x - 3\sqrt{x} +6}{\sin^2 x + 6} \bigg( \frac{\tan^2 x + 5x}{\csc^2 x + 3} \bigg) '$

Differentiating the first portion using Quotient Rule:

$\displaystyle \bigg( \frac{\cos^2 x - 3\sqrt{x} + 6}{\sin^2 x + 6} \bigg)' = \frac{\big( \cos^2 x - 3\sqrt{x} + 6 \big)' \big( \sin^2 x + 6 \big) - \big( \sin^2 x + 6 \big)' \big( \cos^2 x - 3\sqrt{x} + 6 \big)}{\big( \sin^2 x + 6 \big)^2}$

Apply Derivative Rules and Properties, namely the Chain Rule:

$\displaystyle \bigg( \frac{\cos^2 x - 3\sqrt{x} + 6}{\sin^2 x + 6} \bigg)' = \frac{\big( -2 \cos x \sin x - \frac{3}{2\sqrt{x}} \big) \big( \sin^2 x + 6 \big) - \big( 2 \sin x \cos x \big) \big( \cos^2 x - 3\sqrt{x} + 6 \big)}{\big( \sin^2 x + 6 \big)^2}$

Differentiating the second portion using Quotient Rule again:

$\displaystyle \bigg( \frac{\tan^2 x + 5x}{\csc^2 x + 3} \bigg) ' = \frac{\big( \tan^2 x + 5x \big)' \big( \csc^2 x + 3 \big) - \big( \csc^2 x + 3 \big)' \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big)^2}$

Apply Derivative Rules and Properties, namely the Chain Rule again:
$\displaystyle \bigg( \frac{\tan^2 x + 5x}{\csc^2 x + 3} \bigg) ' = \frac{\big( 2 \tan x \sec^2 x + 5 \big) \big( \csc^2 x + 3 \big) - \big( -2 \csc^2 x \cot x \big) \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big)^2}$

Substitute in derivatives:

$\displaystyle y' = \frac{\big( -2 \cos x \sin x - \frac{3}{2\sqrt{x}} \big) \big( \sin^2 x + 6 \big) - \big( 2 \sin x \cos x \big) \big( \cos^2 x - 3\sqrt{x} + 6 \big)}{\big( \sin^2 x + 6 \big)^2} \frac{\tan^2 x + 5x}{\csc^2 x + 3} + \frac{\cos^2 x - 3\sqrt{x} +6}{\sin^2 x + 6} \frac{\big( 2 \tan x \sec^2 x + 5 \big) \big( \csc^2 x + 3 \big) - \big( -2 \csc^2 x \cot x \big) \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big)^2}$

Simplify:

$\displaystyle y' = \frac{\big( \tan^2 x + 5x \big) \bigg[ \big( -2 \cos x \sin x - \frac{3}{2\sqrt{x}} \big) \big( \sin^2 x + 6 \big) - 2 \sin x \cos x \big( \cos^2 x - 3\sqrt{x} + 6 \big) \bigg]}{\big( \sin^2 x + 6 \big)^2 \big( \csc^2 x + 3 \big)} + \frac{\big( \cos^2 x - 3\sqrt{x} +6 \big) \bigg[ \big( 2 \tan x \sec^2 x + 5 \big) \big( \csc^2 x + 3 \big) + 2 \csc^2 x \cot x \big( \tan^2 x + 5x \big) \bigg] }{\big( \csc^2 x + 3 \big)^2 \big( \sin^2 x + 6 \big)}$

We can rewrite the differential by factoring and common mathematical properties to obtain our final answer:

∴ we have found our derivative.

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Learn more about derivatives: brainly.com/question/26836290

Learn more about calculus: brainly.com/question/23558817

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Topic: Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

8 0

2 years ago

Read 2 more answers

Hi, teacher I was absent these days and I didn’t understand anything about this lesson and I need help this is not count as a te

motikmotik

Given:

There are given that the cos function:

$cos210^{\circ}=-\frac{\sqrt{3}}{2}$

Explanation:

To find the value, first, we need to use the half-angle formula:

So,

From the half-angle formula:

$cos(\frac{\theta}{2})=\pm\sqrt{\frac{1+cos\theta}{2}}$

Then,

Since 105 degrees is the 2nd quadrant so cosine is negative

Then,

By the formula:

$\begin{gathered} cos(105^{\circ})=cos(\frac{210^{\circ}}{2}) \\ =-\sqrt{\frac{1+cos(210)}{2}} \end{gathered}$

Then,

Put the value of cos210 degrees into the above function:

So,

$\begin{gathered} cos(105^{\circ})=-\sqrt{\frac{1+cos(210)}{2}} \\ cos(105^{\operatorname{\circ}})=-\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}} \\ cos(105^{\circ})=-\sqrt{\frac{2-\sqrt{3}}{4}} \\ cos(105^{\circ})=-\frac{\sqrt{2-\sqrt{3}}}{2} \end{gathered}$

Final answer:

Hence, the value of the cos(105) is shown below:

$cos(105^{\operatorname{\circ}})=-\frac{\sqrt{2-\sqrt{3}}}{2}$

4 0

1 year ago