A hydro company is laying down power lines in the ground. They have two sections of cable so far that are 325.0 m and 430.0 m long. These two sections of power lines meet at an angle of 23°. How much extra cable is necessary to connect the two pieces of existing cable? Round to the nearest metre.

Answer:

[tex]\frac{dy}{dy}=1[/tex], so I assume you mean "find [tex]\frac{dy}{dx}[/tex]".

We can rewrite this as an implicit equation to avoid using too much of the chain rule, namely

[tex]y = \sqrt[3]{\dfrac{e^x (x+1)}{x^2+1}} \implies (x^2+1) y^3 = e^x (x+1)[/tex]

Differentiate both sides with respect to [tex]x[/tex] using the product and chain rules.

[tex]2x y^3 + 3(x^2+1) y^2 \dfrac{dy}{dx} = e^x(x+1) + e^x[/tex]

[tex]\implies 3(x^2+1) y^2 \dfrac{dy}{dx} = e^x (x+2) - 2x y^3[/tex]

[tex]\implies \dfrac{dy}{dx} = \dfrac{e^x (x+2) - 2x y^3}{3(x^2+1) y^2}[/tex]

Now substitute the original expression for [tex]y[/tex].

[tex]\dfrac{dy}{dx} = \dfrac{e^x (x+2) - 2x \left(\sqrt[3]{\frac{e^x(x+1)}{x^2+1}}\right)^3}{3(x^2+1) \left(\sqrt[3]{\frac{e^x(x+1)}{x^2+1}}\right)^2}[/tex]

[tex]\implies \dfrac{dy}{dx} = \dfrac{e^x (x+2) - \frac{2e^x(x^2+x)}{x^2+1}}{3(x^2+1) \left(\frac{e^x(x+1)}{x^2+1}\right)^{2/3}}[/tex]

[tex]\implies \dfrac{dy}{dx} = e^x \dfrac{x^3-x+2}{3(x^2+1)^2 \frac{e^{2x/3}(x+1)^{2/3}}{(x^2+1)^{2/3}}}}[/tex]

[tex]\implies \dfrac{dy}{dx} = e^{x/3} \dfrac{x^3-x+2}{3(x^2+1)^{4/3} (x+1)^{2/3}}[/tex]

Now, since

[tex]y = \sqrt[3]{\dfrac{e^x (x+1)}{x^2+1}} = \dfrac{e^{x/3} (x+1)^{1/3}}{(x^2+1)^{1/3}}[/tex]

we can write

[tex]\dfrac{dy}{dx} = e^{x/3} \dfrac{x^3-x+2}{3(x^2+1)^{4/3} (x+1)^{2/3}} = \dfrac{e^{x/3} (x+1)^{1/3}}{(x^2+1)^{1/3}} \times \dfrac{x^3-x+2}{3(x^2+1)^{3/3} (x+1)^{3/3}}[/tex]

[tex]\implies \dfrac{dy}{dx} = y \dfrac{x^3-x+2}{3(x^2+1)(x+1)}[/tex]

Focusing on the rational expression in [tex]x[/tex], we have the partial fraction expansion

[tex]\dfrac{x^3 - x + 2}{(x^2 + 1) (x+1)} = a + \dfrac{bx+c}{x^2+1} + \dfrac d{x+1}[/tex]

where we have the constant term on the right side because both the numerator and denominator have degree 3.

Writing everything with a common denominator and equating the numerators leads to

[tex]x^3 - x + 2 = a (x^2+1) (x+1) + (bx+c)(x+1) + d(x^2+1) \\\\ = ax^3 + (a+b+d)x^2 + (a+b+c)x + a+c+d[/tex]

[tex]\implies \begin{cases} a = 1 \\ a+b+d=0 \\ a+b+c = -1 \\ a+c+d=2 \end{cases}[/tex]

[tex]\implies a=1, b=-2, c=0, d=1[/tex]

[tex]\implies \dfrac{x^3 - x + 2}{(x^2 + 1) (x+1)} = 1 - \dfrac{2x}{x^2+1} + \dfrac 1{x+1}[/tex]

and it follows that

[tex]\boxed{\dfrac{dy}{dx} = \dfrac y3 \left(1 - \dfrac{2x}{x^2+1} + \dfrac1{x+1}\right)}[/tex]

Answer:

The greatest common factor of 72 and 162 is 18.

Step-by-step explanation:

In my opinion, one of the easier ways to find the greatest common factor of two numbers is to get the prime factors of both and multiply the common ones. The prime factorization of 72 is 2*2*2*3*3, and the prime factorization of 162 is 2*3*3*3*3 (you can make a factor tree for both of these numbers to verify this). The common primes factors for both of these numbers are 2, 3, and 3 (both 72 and 162 have one 2 and at least two 3s). 2*3*3 is 18, which is the greatest common factor.

Answer:

-10u^3 - 4

Step-by-step explanation:

-3u^3-7u^3 = -10u^3

then -4

Answer:

See below

Step-by-step explanation:

SAS says the two triangles are congruent

 so   side 6x-4 = 3x+8     then x = 4  

x= 4 true

x = 3 false ( because we just found it = 4)

Pythag theorem says QT

   (6x-4)^2 = QT^2 + 12^2

   ( 6(4)-4)^2 = 12 ^2 + QT^2

       400  -144 = QT^2

       QT = 16   true

Answer: A

Step-by-step explanation:

A) Correct. [tex]x^2 \geq 0 \implies -x^2 \leq 0[/tex]

B) Wrong. [tex]2^x > 0[/tex], so the range is [tex](4, \infty)[/tex]

C) Wrong. Linear functions have a range of all real numbers.

D) Wrong. Same logic as C.

Answer:

35 and 10

Step-by-step explanation:

Let m and n be the numbers

m  is 5 more than 3 times n can be written as

m = 3n + 5    ... (1)

Sum of numbers is 45 so m + n = 45 .or

m = 45 - n ... (2)

Equating RHS of (1) and (2) gives us

3n + 5 = 45 - n

Collecting like terms

3n + n = 45 -5

4n = 40

n = 10

Therefore m = 45-10 = 35

Quick Check

In (1) substitute for m and n,

LHS is m = 35

RHS = 3 * 10 + 5 = 35

Answer:

Sarah runs at 14.4km/h

Step-by-step explanation:

36km = 2.5 hours

/ = per or divide

36km/2.5hours = 36 divided by 2.5 = 14.4km/h

Answer:

$183.7

Step-by-step explanation:

To find the deposited amount you need to....

[tex]$532.20-$348.50=$183.70[/tex]

Hello!

Subtract the old balance from new balance to find deposit amount.

⇒ Deposit = $532.20 - $348.50

⇒ Deposit = $183.70

The value that remains under the radical is 2

The correct expression in the question is:

1250^4

This can be rewritten as:

[tex]\sqrt[4]{1250}[/tex]

Express 1250 as product

[tex]\sqrt[4]{(2 * 5^4)}[/tex]

Expand the expression

[tex]\sqrt[4]{2} * \sqrt[4]{5^4}[/tex]

Simplify

[tex]\sqrt[4]{2} * 5[/tex]

Evaluate the product

[tex]5\sqrt[4]{2}[/tex]

Hence, the value that remains under the radical is 2

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The distance between the two given points coordinates is; Option C: √85

Formula for the distance between two coordinates is;

d = √((x₂ - x₁)² + (y₂ - y₁)²)

We are given the coordinates as;

(-2, 3) and (5, -3). Thus;

d = √((-2 - 5)² + (-3 - 3)²)

d = √(49 + 36)

d = √85

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The valid exclusion of the algebraic fraction is (c) a =0, b =0, a =2b

The expression is given as:

8ab^2x/4a^2b - 8ab^2

Set the denominator to 0

4a^2b - 8ab^2 = 0

Divide through by 4ab

a - 2b = 0

Add 2b to both sides

a = 2b

Hence, the valid exclusion of the algebraic fraction is (c) a =0, b =0, a =2b

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Answer:

The Answer is x>0

A true statement about this transformation is: C. It can be a rigid or a nonrigid transformation depending on whether the corresponding side lengths have the same measures.

In Geometry, a transformation can be defined as the movement of a point from its initial position to a new location. Hence, when an object is transformed, all of the points would also be transformed.

In this scenario, we can logically deduce that triangle J'K'L' can either be a rigid or a nonrigid transformation based on the magnitude of the corresponding side lengths in both triangles, considering that their angles are equal in magnitude.

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Complete Question:

Triangle JKL is transformed to create triangle J'K'L'. The angles in both triangles are shown.

J = 90° J' = 90°

K = 65° K' = 65°

L = 25° L' = 25°

Which statement is true about this transformation?

A. It is a rigid transformation because the pre-image and image have the same corresponding angle measures.

B. It is not a rigid transformation because the corresponding side lengths are not equal.

C. It can be a rigid or a nonrigid transformation depending on whether the corresponding side lengths have the same measures.

D. It can be a rigid or a nonrigid transformation because a pair of corresponding angles measures 90°.

It is a rigid transformation because the pre-image and image have the same corresponding angle measures.

The complete question is added as an attachment

From the image and the preimage triangles, we have that:

The corresponding sides of both triangles are equal

This is identified by the marks I, II and III on the side lengths

Equal corresponding sides represent a rigid transformation

Hence, the true statement about the dilation is (a)

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Answer:

The answer is B: x = 0 and x = 4

Step-by-step explanation:

The zeroes of a function are the x-intercepts, or the x values of the points that are on the x-axis. (The x-axis is the numbered horizontal line.)

One of the zeroes passes through the origin, which is (0,0). The other point is right before the number 5 on the x axis, which means that the point is (0, 4). If you look at the x-values of the ordered pairs of the zeroes, that's the answer.

(x, y)

Answer:

=========

We know the perpendicular lines have opposite reciprocal slopes, that is the product of their slopes is - 1.

Find the slope of line 1 by converting the equation into slope-intercept from standard form:

Info:

Its slope is 3/k.

Find the slope of line 2, using the slope formula:

We have both the slopes now. Find their product:

So when k is 18, the lines are perpendicular.

Answer:

450 people were at the match

Step-by-step explanation:

So let's make equation :

W = women

M = Men

C = Children

First equation :

M = W + 250

Second equation :

C = 2W

Third equation :

M = 2(W+C)

Let's substitute the second equation into the third :

M = 2(W + 2W)

M = 2W + 4W

M = 6W

Let's substitute this equation into the first one :

6W = W + 250

5W = 250

W = 50

Now we know that that there were 50 women at the match and can substitute this value for the first equation and then the second :

M = 50 + 250

M = 300

This means that there must have been 300 men at the match.  

C = 2(50)

C = 100

This means that there must have been 100 children at the match.

Adding all values together gives us :

50 + 300 + 100  =

450 people were at the match

Hope this helped and have a good day

Answer:

Step-by-step explanation:

Answer:

y = 40°

z = 140°

x = 100°

Information:

(i) Sum of interior angles of a triangle sum ups to 180°

(ii) On a straight line, the angles sum up to 180°

(iii) One exterior angle is equal to two opposite interior angles.

Here the exterior angle theorem applies.

∠z = 120° + 20°

∠z = 140°

Find the angle C. Here angles lie on a straight line.

∠? + 120° = 180°

∠? = 180° - 120° = 60°

80°, 60° and y are interior angles of a triangle.

y + 80°+ 60° = 180°

y = 180° - 140°

y = 40°

∠? = 40° (vertically opposite angle)

Now,

y + x + 40° = 180°

40° + x + 40° = 180°

x = 100°

Answer:

(- 2, [tex]\frac{5}{3}[/tex] )

Step-by-step explanation:

y = [tex]\frac{2}{3}[/tex] x + 3 → (1)

x = - 2

substitute x = - 2 into (1)

y = [tex]\frac{2}{3}[/tex] × - 2 + 3 = - [tex]\frac{4}{3}[/tex] + [tex]\frac{9}{3}[/tex] = [tex]\frac{5}{3}[/tex]

solution is (- 2, [tex]\frac{5}{3}[/tex] )

Answer:

[tex]\sf \boxed{\bf y =5\sqrt{22}}[/tex]

Step-by-step explanation:

       The ratio of sides of 45 - 45 - 90 triangle is  a : a : a√2.

a is the side opposite to 45°.

From the figure, a = 5√11

The side opposite to 90° is a√2.

y = a√2

  [tex]\sf = 5\sqrt{11} * \sqrt{2}\\\\ = 5*\sqrt{11*2}\\\\ = 5\sqrt{22}[/tex]

The proof is shown below:

A parallelogram is a quadrilateral with opposite sides parallel (and therefore opposite angles equal).

As,

<A= 104

<B= 76

As AB || CD,

<A + <D =180

104 + <D =180

<D = 76

and <B + <C =180

76 + <C = 180

<C = 104.

As, opposite angles are equals and  AB || CD.

Hence, ABCD is a parallelogram.

For 2 part,

Use m<A = 104° and m<B= 76° to show that <A and <B are same-side interior angles. Then, use AB || CD to show that <A and <D are supplementary angles.

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A trapezoid is a quadrilateral which is having a pair of opposite sides as parallel and the length of the parallel sides is not equal. The correct option is C.

A trapezoid is a quadrilateral which is having a pair of opposite sides as parallel and the length of the parallel sides is not equal.

Trapezoid CDEF was reflected across the x-axis followed by a 90° rotation about the origin to create the other trapezoid shown on the graph. The congruency statement that applies to the trapezoids is CDEF ≅ NMOP.

Hence, the correct option is C.

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Answer:

1413.7

Step-by-step explanation:

A(sphere) = 4 * π * r².

You can think about it like two times the cap surface area of a hemisphere. Therefore, the hemisphere cap area equals:

Ac = A(sphere) / 2,

Ac = 2 * π * r².

The correct statement is that the value of x for System A will be equal to the value of y for System B because the first equation of System B is obtained by adding −4 to the first equation of System A and the second equations are identical. Option D

6x + y = 2 2x − 3y = −10    System A

-x − y = −3 −x − y = −3 System B

Let;s solve for x and y in system A

6x + y = 2

Make 'y' the subject

y = 2-6x

Substitute in the other equation

-x -y = -3

-x - (2-6x) = -3

-x -2+6x = -3

Collect like terms

5x = -3+2

x = -1/5

Substitute in y = 2-6x to find 'y'

y = 2- 6(-1/5)

y = 2+ 6/5

y = [tex]\frac{10+ 6}{5}[/tex]

y = 16/5

For system B

-x-y = -3

Make y subject, we have

-x + 3 = y

y = -x + 3

Substitute in the other equation, we have

2x − 3y = −10

2x - 3(-x+3) = -10

2x + 3x -9 = -10

Collect like terms

5x -9 = -10

5x = -10 + 9

x = 1/5

Substitute into y = -x + 3 to find 'y'

y = -(1) + 3

y = 2

Thus, the correct statement is that the value of x for System A will be equal to the value of y for System B because the first equation of System B is obtained by adding −4 to the first equation of System A and the second equations are identical. Option D

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The equations of the perpendicular lines are: y = 1/2x + 6, y = 15, y = -x - 2, y = 6x + 3 and y = 1/3x - 4

When a linear equation is represented as:

Ax + By = C

The slope (m) is:

m = -A/B

When the linear equation is represented as:

y = mx + c

The slope is m

A line perpendicular to a linear equation that has a slope of m would have a slope of -1/m

Using the above highlights, the equations of the lines are:

6. y = -2x + 5; (2, 7)

The slope is:

m = -2

The perpendicular slope is:

n = 1/2

The equation of the perpendicular line is:

y = n(x - x1) + y1

This gives

y = 1/2(x - 2) + 7

Evaluate

y = 1/2x - 1 + 7

This gives

y = 1/2x + 6

7. y = -5; (11, 15)

The slope is:

m = 0

The perpendicular slope is:

n = 1/0 = undefined

The equation of the perpendicular line is:

y = n(x - x1) + y1

This gives

y = 15

8. Graph ; (-12, 10)

The slope is:

m = (y2 - y1)/(x2 - x1)

Using the points on the graph, we have:

m = (2 - 3)/(3 - 4)

m = 1

The perpendicular slope is:

n = -1

The equation of the perpendicular line is:

y = n(x - x1) + y1

This gives

y = -1(x + 12) + 10

y = -x - 12 + 10

Evaluate

y = -x - 2

9. y = -1/6x + 1; (-2, -9)

The slope is:

m = -1/6

The perpendicular slope is:

n = 6

The equation of the perpendicular line is:

y = n(x - x1) + y1

This gives

y = 6(x + 2) - 9

Evaluate

y = 6x + 12 - 9

This gives

y = 6x + 3

10. 6x + 2y = 14; (12, 0)

The slope is:

m = -6/2

m = -3

The perpendicular slope is:

n = 1/3

The equation of the perpendicular line is:

y = n(x - x1) + y1

This gives

y = 1/3(x - 12) + 0

Evaluate

y = 1/3x - 4

Hence, the equations of the perpendicular lines are: y = 1/2x + 6, y = 15, y = -x - 2, y = 6x + 3 and y = 1/3x - 4

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The order of the side lengths from least to greatest is; GH< IH< FG< GI< FH.

As with other closed geometric figures, the shortest side length is opposite the smallest angle measure while the longest side length, is opposite the greatest angle measure.

Hence, it follows from the statement above that the order of side lengths is; GH< IH< FG< GI< FH.

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Answer:

390 m^2

Step-by-step explanation:

So let's start by finding the surface area of the two triangles on the side. Generally the area of a triangle can be calculated as: [tex]\frac{1}{2}bh[/tex]. but since there is two of them, we can calculate the area of both of them by simply canceling out the 1/2 to a 1. So the area of both triangles are 12 m * 5m as given in the diagram. This gives you 60m^2 area for both triangles on the side

Now let's calculate the rectangle on the top. It has a width and length of 13 and 11 m. So multiply these together to get: 143 m^2

Now let's calculate the rectangle that's on the bottom, it has the dimensions 11 and 12m and multiplying these together gets you: 132 m^2\

Now calculate the rectangle all the way in the back, it has the dimensions 5 and 11m, multiplying these together gets you 55 m^2.

Now add all these together: 60m^2 + 143m^2 + 132m^2 + 55m^2 = 390m^2

The system of equations represented by the attached graph is y = (x + 3)^2 - 2 and y = -x + 1

This question will be answered using the attached graph

The curve

The curve is a quadratic function, and it has the following features:

Vertex, (h, k) = (-3, -2)

Point (x, y) = (-1, 2)

A quadratic function is represented as:

y = a(x - h)^2 + k

So, we have:

y = a(x + 3)^2 - 2

Substitute (x, y) = (-1, 2)

2 = a(-1 + 3)^2 - 2

This gives

2 = 4a-2

Solve for a

a = 1

Substitute a = 1 in y = a(x + 3)^2 - 2

y = (x + 3)^2 - 2

The line

The line is a linear function, and it has the following features:

Point (x1, y1) = (-1, 2)

Point (x2, y2) = (-6, 7)

The linear function is calculated as:

y = (y2 - y1)/(x2 - x1) *(x- x1) + y1

So, we have:

y = (7 -2)/(-6 +1) *(x + 1) + 2

Evaluate the quotient

y =  -1(x + 1) + 2

Expand

y = -x -1 + 2

y = -x + 1

Hence, the system of equations represented by the attached graph is y = (x + 3)^2 - 2 and y = -x + 1

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Answer:

Step-by-step explanation:

Domain is the set of all possible inputs (x) for the function f(x)

The range is the set of all possible values for that domain

This function is 4x + y = 3 which can be re-written as y = 3-4x

Plug in each of the domain values and you will get the range

For -2 : y = 3-4(-2) = 3 + 8 = 11

For 1:  y = 3-4(1) = -1

For 5: y = 3-4(5) = -17

So range is {-17, -1, 11}  

The values of the trigonometric functions are

[tex]sin \theta = \frac{12}{13}[/tex]

[tex]cos \theta = \frac{5}{13}[/tex]

[tex]tan \theta = \frac{12}{5}[/tex]

 [tex]cot\theta = \frac{5}{12}[/tex]

[tex]csc\theta = \frac{13}{12}[/tex]

From the question, we are to determine the values of the given trigonometric functions

From the given information,

[tex]sec \theta =\frac{13}{5}[/tex]

∴ [tex]\frac{1}{cos \theta} =\frac{13}{5}[/tex]

[tex]cos \theta = \frac{5}{13}[/tex]

Thus,

Adjacent = 5

Hypotenuse = 13

Opposite = ?

Using the Pythagorean theorem

|Opp|² = |Hyp|² - |Adj|²

|Opp|² = 13² - 5²

|Opp|² = 169 - 25

|Opp|² = 144

|Opp| = √144

|Opp| = 12

∴ Opposite = 12

Thus,

By using SOH CAH TOA

[tex]sin \theta = \frac{12}{13}[/tex]

[tex]tan \theta = \frac{12}{5}[/tex]

[tex]cot\theta = \frac{1}{tan\theta}[/tex]

∴ [tex]cot\theta = \frac{5}{12}[/tex]

[tex]csc\theta = \frac{1}{sin\theta}[/tex]

∴ [tex]csc\theta = \frac{13}{12}[/tex]

Hence, the values of the trigonometric functions are

[tex]sin \theta = \frac{12}{13}[/tex]

[tex]cos \theta = \frac{5}{13}[/tex]

[tex]tan \theta = \frac{12}{5}[/tex]

 [tex]cot\theta = \frac{5}{12}[/tex]

[tex]csc\theta = \frac{13}{12}[/tex]

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