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4 years ago

−7y−4x=1 7y−2x=53 Solve the system of equations

1 answer:

3 0

-7y-4x=17y-2x=53
-4x=24y-2x=53
-2x=24y=53
24y+2x=53

You would have to solve one at a time.
y=53/24
y=2.21
x=53/2
x=26.5

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Neil wants to join three pieces of metal to form a right triangle. which of the following lengths could be used? please use step

vichka [17]

Answer: G. 12.5, 7.5, 10

Step-by-step explanation:

To determine the lengths of metal that could be used to form the tight angle triangle, we would apply Pythagoras theorem which is expressed as

Hypotenuse² = opposite side² + adjacent side²

The hypotenuse is the longest side while the opposite and adjacent sides are the shorter side. Therefore,

8² = 4² + 4²

64 = 16 + 16 = 32

A right angle triangle cannot be formed because they are not equal.

12.5² = 7.5² + 10²

156.25 = 56.25 + 100 = 156.25

A right angle triangle can be formed because they are equal.

23² = 11² + 9²

529 = 121 + 81 = 202

A right angle triangle cannot be formed because they are not equal.

96² = 12.5² + 6²

9216 = 156.25 + 36 = 192.25

A right angle triangle cannot be formed because they are not equal.

4 0

3 years ago

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:
<u>Calculus</u>

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 <em>way</em> too small for this problem, there will be limited explanation.

<u>Step 1: Define</u>

<em>Identify.</em>

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

<u>Step 2: Differentiate</u>

We can differentiate this function with the use of the given <em>derivative rules and properties</em>.

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

3 years ago

Read 2 more answers

Simplify 3 over 2 x 3 over 2 exact answer

nikitadnepr [17]

ANSWER : 9 over 4

or 2 and 1/4

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

*3 over 2 x 3 over 2 is 3/2 x 3/2* << this is for me to remember

Okay so multiplying fractions means for us to multiply across. That means if we line the fractions, we multiply. Like this

3 x 3

----- ----- 3 times 3 is = 9

2 x 2 2 times 2 is = 4

9 over 4 is the same thing as 2 and 1/4 << FRACTION

WHOLE

Hope I helped (:

8 0

3 years ago

Read 2 more answers

I need a real answer to this problem and if you can't answer then please don't answer this. Just give me the answer for the blan

Harlamova29_29 [7]

To divide the triangles into these regions, you should construct the <u>perpendicular bisector</u> of each segment.
These perpendicular bisectors intersect and divide each triangle into three regions.
The points in each region are those closest to the vertex in that <u>region</u>.

<h3>What is a triangle?</h3>

A triangle can be defined as a two-dimensional geometric shape that comprises three (3) sides, three (3) vertices and three (3) angles only.

<h3>What is a line segment?</h3>

A line segment can be defined as the part of a line in a geometric figure such as a triangle, circle, quadrilateral, etc., that is bounded by two (2) distinct points and it typically has a fixed length.

<h3>What is a perpendicular bisector?</h3>

A perpendicular bisector can be defined as a type of line that bisects (divides) a line segment exactly into two (2) halves and forms an angle of 90 degrees at the point of intersection.

In this scenario, we can reasonably infer that to divide the triangles into these regions, you should construct the <u>perpendicular bisector</u> of each segment. These perpendicular bisectors intersect and divide each triangle into three regions. The points in each region are those closest to the vertex in that <u>region</u>.

Read more on perpendicular bisectors here: brainly.com/question/27948960

#SPJ1

5 0

2 years ago

I need help writing this as a decimal!

Vesna [10]

Write it as 0.35. Hope it helps.

6 0

3 years ago

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−7y−4x=1 ​7y−2x=53 Solve the system of equations​​

−7y−4x=1 7y−2x=53 Solve the system of equations