what is the definition of an average?

Answer:

x=20

y and z=160

Step-by-step explanation:

the arc of the circle is congruent to the central angle therfore x is also 20

next for y and z we can tell they are congruent now to find the measure. we can tell A is 180 so we do 360-180-20=160

Check the picture below.

[tex]\textit{Sum and Difference Identities} \\\\ sin(\alpha + \beta)=sin(\alpha)cos(\beta) + cos(\alpha)sin(\beta) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ sin(255^o)\implies sin(135^o+120^o) \\\\\\ sin(135^o)cos(120^o)~~ + ~~cos(135^o)sin(120^o) \\\\\\ \left( \cfrac{\sqrt{2}}{2} \right)\left( -\cfrac{1}{2} \right)~~ + ~~\left( -\cfrac{\sqrt{2}}{2} \right)\left( \cfrac{\sqrt{3}}{2} \right)\implies -\cfrac{\sqrt{2}}{4}~~ - ~~\cfrac{\sqrt{6}}{4}\implies \cfrac{-2-\sqrt{6}}{4}[/tex]

Answer:

See below for answers and explanations

Step-by-step explanation:

Problem 1

Rewrite and add each vector:

[tex]p=\langle40cos70^\circ,40sin70^\circ\rangle\\q=\langle50cos270^\circ,50sin270^\circ\rangle\\r=\langle60cos235^\circ,60sin235^\circ\rangle\\p+q+r=\langle40cos70^\circ+50cos270^\circ+60cos235^\circ,40sin70^\circ+50sin270^\circ+60sin235^\circ\rangle\\p+q+r\approx\langle-20.73,-61.56\rangle[/tex]

Find the magnitude of the resulting vector:

[tex]||p+q+r||=\sqrt{x^2+y^2}\\||p+q+r||=\sqrt{(-20.73)^2+(-61.56)^2}\\||p+q+r||\approx64.96[/tex]

Therefore, the best answer is D) 64.959

Problem 2

Think of the vectors like this:

[tex]t=\langle7cos240^\circ,7sin240^\circ\rangle\\u=\langle10cos30^\circ,10sin30^\circ\rangle\\v=\langle15cos310^\circ,15sin310^\circ\rangle[/tex]

By adding the vectors, we have:

[tex]t+u+v=\langle7cos240^\circ+10cos30^\circ+15cos310^\circ,7sin240^\circ+10sin30^\circ+15sin310^\circ\rangle\\t+u+v\approx\langle14.8,-12.55\rangle[/tex]

Since the direction of a vector is [tex]\theta=tan^{-1}(\frac{y}{x})[/tex], we have:

[tex]\theta=tan^{-1}(\frac{y}{x})\\\theta=tan^{-1}(\frac{-12.55}{14.8})\\\theta\approx-40.3^\circ[/tex]

Don't forget to take into account that the resulting vector is in Quadrant IV since the horizontal component is positive and the vertical component is negative, so we will add 360° to our angle to get the result:

[tex]\theta=360^\circ+(-40.3)^\circ\\\theta=319.7^\circ[/tex]

Therefore, the best answer is D) 320°

Problem 3

Again, rewrite the vectors and add them:

[tex]u=\langle30cos70^\circ,30sin70^\circ\rangle\\v=\langle40cos220^\circ,40sin220^\circ\rangle\\u+v=\langle30cos70^\circ+40cos220^\circ,30sin70^\circ+40sin220^\circ\rangle\\u+v\approx\langle-20.38,2.48\rangle[/tex]

Using the direction formula:

[tex]\theta=tan^{-1}(\frac{y}{x})\\\theta=tan^{-1}(\frac{2.48}{-20.38})\\\theta\approx-6.94^\circ[/tex]

As the vector is located in Quadrant II, we need to add 180° to our angle:

[tex]\theta=180^\circ+(-6.94)^\circ\\\theta=173.06^\circ[/tex]

Therefore, the best answer is D) 173°

Problem 4

Using scalar multiplication:

[tex]-3u=-3\langle35,-12\rangle=\langle-105,36\rangle[/tex]

Find the magnitude of the resulting vector:

[tex]||-3u||=\sqrt{x^2+y^2}\\||-3u||=\sqrt{(-105)^2+(36)^2}\\||-3u||=111[/tex]

Find the direction of the resulting vector:

[tex]\theta=tan^{-1}(\frac{y}{x})\\\theta=tan^{-1}(\frac{36}{-105})\\\theta\approx-18.92^\circ[/tex]

As the vector is located in Quadrant II, we need to add 180° to our angle:

[tex]\theta=180^\circ+(-18.92)^\circ\\\theta=161.08^\circ[/tex]

Therefore, the best answer is C) 111; 161°

Problem 5

Using scalar multiplication:

[tex]5v=5\langle25cos40^\circ,25sin40^\circ\rangle=\langle125cos40^\circ,125sin40^\circ\rangle[/tex]

Since the direction of the angle doesn't change in scalar multiplication, it only affects the magnitude, so the magnitude of the resulting vector would be [tex]||5v||=125[/tex], making the correct answer C) 125; 200°

.................................................................................

the answer Is b and c

Step-by-step explanation:

because I did it

Answer:

40+20 = 72-12 is the simplified version.

Step-by-step explanation:

Your equations

4(10+5) = 6(12-2)

Multiply 4 by 10 and 5 for the "4(10+5)"
Multiply 6 by 12 and -2 for the "6(12-2)"

40+20 = 72-12 is the simplified version.

If you want to solve/verify:

40+20 = 72-12

60 = 60

The nanswer is correct.

Answer:

40 + 20 = 72 - 12

Step-by-step explanation:

The number outside the parenthesis will be multiplied to both numbers inside the parenthesis, making the equation

(4 * 10) + (4 * 5) = (6 * 12) + (6 * (-2))

simplifying the equation...

40 + 20 = 72 - 12

Answer:

the answers 1, 4, and 5.

m∠N = 59° and m∠O = 121°.

Option 3 and 4 are correct.

A parallelogram is a geometric object with sides that are parallel to one another in two dimensions. It is a form of polygon with four sides (sometimes known as a quadrilateral) in which each parallel pair of sides have the same length. A parallelogram has adjacent angles that add up to 180 degrees. You must have studied a variety of 2D shapes and sizes in geometry, including circles, squares, rectangles, rhombuses, etc. Each of these forms has a unique set of characteristics.

As per the given data:

LMNO is a parallelogram

The sum of adjacent angles in a parallelogram is always equal to 180°

11x + 6x - 7 = 180

17x = 187

x = 11

m∠L = m∠N {opposite angles in a parallelogram are equal}

m∠N = 6x - 7 = 6(11) - 7 m∠N

m∠L = 59°

m∠M = m∠O {opposite angles in a parallelogram are equal}

m∠M = 11x = 11(11) = 121°

m∠O = 121°

Hence, m∠N = 59° and m∠O = 121°.

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Answer:=8/17

Step-by-step explanation:

Answer:

concave up

Step-by-step explanation:

there is no negatives in the equation (in front of x-value)

[tex] \Large \bold{SOLUTION\ 1:} [/tex]

[tex] \small \begin{array}{l} \text{First, we need to check if the given differential} \\ \text{equation is exact.} \\ \\ (1-xy)^{-2} dx + \big[y^2 + x^2 (1-xy)^{-2}\big]dy = 0 \\ \\ \dfrac{dx}{(1-xy)^2} + \left[y^2 + \dfrac{x^2}{(1-xy)^2}\right]dy = 0 \\ \\ \quad M(x, y) dx + N(x, y) dy = 0 \end{array} [/tex]

[tex] \small \begin{array}{l l}\tt\: M(x,y) = \dfrac{1}{(1 - xy)^2}, & N(x,y) = y^2 + \dfrac{x^2}{(1-xy)^2}\\ \\\tt \dfrac{\partial M}{\partial y} = \dfrac{-2(-x)}{(1 - xy)^3}, & \dfrac{\partial N}{\partial x} = \dfrac{2x}{(1 - xy)^2} + \dfrac{-2(-y)x^2}{(1 - xy)^3} \\ \\\tt \dfrac{\partial M}{\partial y} = \dfrac{2x}{(1 - xy)^3}, & \dfrac{\partial N}{\partial x} = \dfrac{2x(1 - xy)+2x^2y}{(1 - xy)^3} \\ \\\tt \: & \dfrac{\partial N}{\partial x} = \dfrac{2x}{(1 - xy)^3} \end{array} [/tex]

[tex] \small \begin{array}{l} \tt\dfrac{\partial M}{\partial y} = \dfrac{\partial N}{\partial x} \implies \text{Differential equation is exact.} \\ \\\tt \dfrac{\partial F}{\partial x} = M(x, y) = \dfrac{1}{(1 - xy)^2} \\ \tt\displaystyle F(x, y) = \int \dfrac{1}{(1 - xy)^2} \partial x = -\dfrac{1}{y} \int \dfrac{1}{(1 - xy)^2}(-y)\partial x \\ \\ \tt\:F(x, y) = \dfrac{1}{y(1 - xy)} + h(y) \\ \\ \tt\dfrac{\partial F}{\partial y} = N(x, y) = y^2 + \dfrac{x^2}{(1-xy)^2} \\ \\\tt \dfrac{\partial}{\partial y}\left[\dfrac{1}{y(1 - xy)} + h(y)\right] = y^2 + \dfrac{x^2}{(1-xy)^2} \\ \\ \tt-\dfrac{1 - xy + y(-x)}{y^2(1 - xy)^2} + h'(y) = y^2 + \dfrac{x^2}{(1-xy)^2} \\ \\ \tt-\dfrac{1 - 2xy}{y^2(1 - xy)^2} + h'(y) = y^2 + \dfrac{x^2}{(1-xy)^2} \\ \\ h'(y) = y^2 + \dfrac{x^2}{(1-xy)^2} + \dfrac{1 - 2xy}{y^2(1 - xy)^2} \\ \\ \tt\:h'(y) = y^2 + \dfrac{x^2y^2 - 2xy + 1}{y^2(1-xy)^2} = y^2 + \dfrac{1}{y^2} \\ \\ h(y) = \dfrac{y^3}{3} - \dfrac{1}{y} + C \\ \\ \tt\text{Substituting to }F(x,y),\text{we get} \\ \\ \dfrac{1}{y(1 - xy)} + \dfrac{y^3}{3} - \dfrac{1}{y} = C \\ \\ \quad \quad \text{or} \\ \\ \tt\red{\boxed{\dfrac{x}{1 - xy} + \dfrac{y^3}{3} = C} \Longleftarrow \textit{Answer}} \end{array} [/tex]

[tex] \Large \bold{SOLUTION\ 2:} [/tex]

[tex] \small \begin{array}{l} \tt\text{Since we already know that the equation is exact,} \\ \text{we can then continue solving for the solution by} \\ \text{inspection method or by algebraic manipulation.} \\ \\ \tt(1-xy)^{-2} dx + \big[y^2 + x^2 (1-xy)^{-2}\big]dy = 0 \\ \\ \tt\dfrac{dx}{(1-xy)^2} + \left[y^2 + \dfrac{x^2}{(1-xy)^2}\right]dy = 0 \\ \\ \tt\dfrac{dx}{(1-xy)^2} + y^2 dy + \dfrac{x^2}{(1-xy)^2} dy = 0 \\ \\ \tt\dfrac{dx + x^2dy}{(1-xy)^2} + y^2 dy = 0 \\ \\ \tt\text{Divide both numerator and denominator of the} \\ \tt\text{fraction by }x^2. \end{array} [/tex]

[tex] \small \begin{array}{c}\tt \dfrac{\dfrac{1}{x^2}dx + dy}{\dfrac{(1-xy)^2}{x^2}} + y^2 dy = 0 \\ \tt\\ \tt\dfrac{\dfrac{1}{x^2}dx + dy}{\left(\dfrac{1}{x}-y\right)^2} + y^2 dy = 0 \\ \\ \tt-\dfrac{\left(-\dfrac{1}{x^2}dx - dy\right)}{\left(\dfrac{1}{x}-y\right)^2} + y^2 dy = 0 \\ \\ \tt\displaystyle {\large{\int}} -\frac{d\left(\dfrac{1}{x}-y\right)}{\left(\dfrac{1}{x}-y\right)^2} + \int y^2 dy = \int 0 \\ \\ \tt\implies\tt \dfrac{1}{\dfrac{1}{x}-y} + \dfrac{y^3}{3} = C \\ \\\text{or} \\ \\ \tt\red{\boxed{\dfrac{x}{1 - xy} + \dfrac{y^3}{3} = C} \Longleftarrow \textit{Answer}} \end{array} [/tex]

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#BrainlyMathKnower

4 * x = 12.44

4/4 * x = 12.44 / 4

x = 3.11

Answer:

x = 3.11

Step-by-step explanation:

To find the value of 'x', divide both sides of the equation by 4

12.44 = 4x

[tex]\dfrac{12.44}{4}=\dfrac{4x}{4}\\\\\\3.11=x[/tex]

Answer:

11 5/4

Step-by-step explanation:

make the denominator the same then add.the common factor of 2 and 7 is 14 so make all the denominator to 14 then u can add.thus having the answer

angle ksp+61=108(being straight angle)

or,angle ksp=180-61

angle ksp=119

Answer: 48.6°

Step-by-step explanation:

Angle of elevation = sin inverse (6/8)

= 48.6 degrees

Therefore, Option C is right answer

Answer:

x = 5

Step-by-step explanation:

3^2 + 4^2 = x^2

9 + 16 = x^2

25

Now we need to find the square root of 25 :)

The answer is 5

X = 5

Have a great day!!

Please rate and mark brainliest!!

Answer:

Step-by-step explanation:

aThe person and the hot air balloon form a right triangle, with the lift-off point as the vertex that form the right angle. If you were to try to solve for the height of the balloon, you would need to use tangent to do so:

tanθ = opp/adj = y/x

y = xtanθ

What this problem ultimately wants to know, is how the distance of the balloon is changing with respect to time: dy/dt. So we need to take the derivative of the above equation (you need to product rule to do so):

dy/dt = (dx/dt)tanθ + xsec2θ(dθ/dt)

We need to find out values for all of the variables:

The observer is not moving, therefore dx/dt = 0.

The observer is standing 4 miles away from the lift-off point, so x = 4.

The angle between the horizontal and the observer's line of sight is π/3, therefore θ = π/3.

The rate of change for the angle, dθ/dt, is 0.1.

Just plug all the numbers in:

dy/dt = (0)tan(π/3) + (4)sec2(π/3)(0.1)

dy/dt = 0 + 1.6

dy/dt = 1.6 mi/min

Just have a good day

The dimension of the unknown variable x of the deck is equal to 14 feet in length.

A rectangle is a closed 2-D shape, having 4 sides, 4 corners, and 4 right angles (90°).The opposite sides of a rectangle are equal and parallel.

Given in the question is a diagram showing a plan for deck. From the data given, we can write -

Area of Deck = 511 ft.

The value of x can be found out by subtracting the area of the deck from the  area of rectangle having dimensions of length 29 ft and width (16 + 4) = 20 ft. The remaining area will be equated to the sum of the area of rectangle at top right corner having dimensions of length 9 ft and width 4 ft and the area of triangle at lower right corner.

Firstly, we will find the dimensions of the triangle at the lower right. The height of the triangle will be = (20 - 9) ft = 11 ft. The base of the triangle will be = 29 - ( x + 9) ft = (20 - x) ft. Now -

[Area of Rectangle (29 x 20)] - [Area of Deck] = [Area of Rectangle (9 x 4) + Area of triangle (lower right)].

Mathematically -

580 - 511 = 9 x 4 + 1/2 × (20 - x) × 11 = 69

36 + 5.5(20 - x) = 69

36 + 110 - 5.5x = 69

146 - 69 = 5.5x

5.5x = 77

x = 14 ft

Therefore, the dimension of the unknown variable x of the deck is equal to 14 feet.

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Using linear function concepts, it is found that consumption would increase by $800.

A linear function is modeled by:

[tex]y = mx + b[/tex]

In which:

In this problem, the function is given by:

C = 200 + 0.8Y.

Which means that the slope is of m = 0.8 per unit of Y.

For 1000 units:

0.8 x 1000 = 800.

Consumption would increase by $800.

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Answer:

97.16

Step-by-step explanation:

in total, you have $668.00.

To reserve enough for the last day, you need to subtract the last day's budget:

This means that, for the remaining 6 days in the week, you have $583.00.

Assuming an equal amount is being spent each day, you can calculate the daily budget by dividing the remaining total by the number of days left.

To avoid going over budget, you can spend $97.16 per day for the other six days.

Step-by-step explanation:

please mark me as brainlest

Answer:

3x + (2 - x) = 10

3x + 2 - x = 10

2x + 2 = 10

2x = 10 - 2

2x = 8

x = 8/2

x = 4

y = 2 - x

y = 2 - 4

y = -2

Answer:

15 :)

Step-by-step explanation:

brainliest please

Answer:

The answer is -125

Proof:

Answer: Is Square feet/minute if there is two of them the second one is Cubic feet/minute

Step-by-step explanation: feet/minute would cover anything linear {forward, backward, left, right}

square feet/minute would cover two-dimensional areas  {walls, floors, sides of buildings, etc.}

cubic feet/minute would cover three dimensional situations  {swimming pools, buckets, containers, etc.}

Answer:

Step-by-step explanation:

[tex]=5\displaystyle\int_{0}^{\pi/6}dx\displaystyle\int_{0}^{2}x\sec^2(xy)dy\\=5\displaystyle\int_{0}^{\pi/6}dx\displaystyle\int_{0}^{2}\sec^2(xy)d(xy)\\=5\displaystyle\int_{0}^{\pi/6}dx\tan(xy)|_{y=0}^{y=2}[/tex]

[tex]=5\displaystyle\int_{0}^{\pi/6}\tan(2x)dx\\=-\frac{5}{2}\ln\cos(2x)|_{0}^{\pi/6}\\=-\frac{5}{2}[\ln\cos(\pi/3) - \ln\cos(0)]\\[/tex]

[tex]=-\frac{5}{2}\ln{\frac{1}{2}[/tex]

Answer:

B and E

Step-by-step explanation:

[tex]\dfrac 6p = x+a\\\\\implies \dfrac 6p \times\dfrac 16 = (x+a) \times\dfrac 16 ~~~~~~;\left[\text{Multiply both sides by}~ \dfrac 16 \right]\\\\\implies \dfrac 1p = \dfrac{x+a}6\\\\\implies p= \dfrac{6}{x+a}~~~~~~~~~~~~~~~~~~~~;[\text{Cross multiply}][/tex]

Answer:

65.2 cm

Step-by-step explanation:

P = a + b + c + d

P = 16.7 cm + 12.9 cm + 16.7 cm + 18.9 cm

P = 65.2 cm

The perimeter of the trapezoid is 65.2 cm.

Answer:

To find the perimeter of this trapezoid, simply add all four sides (since there are only four sides to a trapezoid):-

16.7 + 12.9 + 16.7 + 18.9

= 65.2 centimetres is the perimeter.

You can picture it like this;

Step-by-step explanation:

63/37 = 1.702 = 1.7

#Ihopethishleps

C = 2 x pi x r

if r = 5 cm

C = 2 x pi x 5 = 10 x pi

≈ 10 x 3,14 ≈ 31,4 cm

C = 2 x pi x r = pi x d

if d = 2 m

C = pi x 2

≈ 3,14 x 2 ≈ 6,28 m

Hello!

a)

C  = 2πr                     C= 2*π     *5                     31,4cm    

                                   =  2*3,14* 5

d)

r =2/2 => 1m

C=  2πr                   C= 2*π*1                             6,28m

                               =  2 *3,14*1