How to do this question plz answer me step by step plzz plz ​

Answer:

Scouts are on the third side in the sense they are running

Step-by-step explanation:

A regular hexagonal shape of perimeter 270 has each side of                 270/6 = 45  

Let´s call d  the run distance  then

d = 2/5 * 270      d = 108 m

We don´t have fig 9 available therefore if  X is a vertex in the hexagon or at the middle point of one side, scouts are 108 m from the starting point which means they had run 2,4 sides of the hexagon. If X is not either a vertex or a middle point of a side then, we have two solutions for the question depending on the sense the scouts took when began the run (clockwise or counterclockwise)

Answer:

1

Step-by-step explanation:

split the shape to triangle and a rectangle

the rectangle at the original trapezoid has 2 squares in width and 3 squares for height multiply those numbers by 1/2 you will get 1 square for width and 1.5 squares for the height which is showen in option 1

Answer:

C = 35x pence

Step-by-step explanation:

1 pen costs 15 , thus x will cost 15x

1 ruler costs 20, thus x will cost 20x

Total cost (C) will then be

C = 15x + 20x = 35x pence

The total cost of pens and rulers, C = 35x pence

Equations are mathematical statements with two algebraic expressions flanking the equals (=) sign on either side.

It demonstrates the equality of the relationship between the expressions printed on the left and right sides. LHS = RHS is a common mathematical formula.

Coefficients, variables, operators, constants, terms, expressions, and the equal to sign are some of the components of an equation. The "=" sign and terms on both sides must always be present when writing an equation.

Given:

Cost 1 pens is 15.

Then, cost for x pen is 15x

Cost of 1 ruler is 20

Then, cost of x ruler is 20x

So, the total cost is

= 15x + 20x

= 35x

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Answer:

25 + 10h = 50+5h

Step-by-step explanation:

Black Diamond Ski Resort

25 + 10h

Bunny Hill Ski Resort

50+5h

We want when they are equal

25 + 10h = 50+5h

Answer:

10x + 25 = 5x + 50

Step-by-step explanation:

Answer:

8

Step-by-step explanation:

x+37+x+37+90 = 180

2x + 74 = 90

2x = 16

x = 8

Answer:

x=8°

Step-by-step explanation:

JI is a diameter and K is on the circumference of a circle.

∴∠JKI=90°

also KJ=KI=x(say)

tan (x+37)=y/y=1=tan 45

so x+37=45

x=45-37=8°

Answer:

768 libras de fuerza

Step-by-step explanation:

Tenemos que encontrar la ecuación que los relacione.

F = Fuerza necesaria para evitar que el automóvil patine

r = radio de la curva

w = peso del coche

s = velocidad de los coches

En la pregunta se nos dice:

La fuerza requerida para evitar que un automóvil patine alrededor de una curva varía inversamente con el radio de la curva.

F ∝ 1 / r

Y luego con el peso del auto

F ∝ w

Y el cuadrado de la velocidad del coche

F ∝ s²

Combinando las tres variaciones juntas,

F ∝ 1 / r ∝ w ∝ s²

k = constante de proporcionalidad, por tanto:

F = k × w × s² / r

F = kws² / r

Paso 1

Encuentra k

En la pregunta, se nos dice:

Suponga que 400 libras de fuerza evitan que un automóvil de 1600 libras patine alrededor de una curva con un radio de 800 si viaja a 50 mph.

F = 400 libras

w = 1600 libras

r = 800

s = 50 mph

Tenga en cuenta que desde el

F = kws² / r

400 = k × 1600 × 50² / 800

400 = k × 5000

k = 400/5000

k = 2/25

Paso 2

¿Cuánta fuerza evitaría que el mismo automóvil patinara en una curva con un radio de 600 si viaja a 60 mph?

F = ?? libras

w = ya que es el mismo carro = 1600 libras

r = 600

s = 60 mph

F = kws² / r

k = 2/25

F = 2/25 × 1600 × 60² / 600

F = 768 libras

Por lo tanto, la cantidad de fuerza que evitaría que el mismo automóvil patine en una curva con un radio de 600 si viaja a 60 mph es de 768 libras.

Answer:

Step-by-step explanation:

The diagonals of the rhombus divide it into 4 congruent right triangles.

So we can use Pythagorean theorem to calculate side of a rhombus.

[tex](\frac e2)^2+(\frac f2)^2=s^2\\\\e=30\,cm\quad\implies\quad\frac e2=15\,cm\\\\f=16\,cm\quad\implies\quad\frac f2=8\,cm\\\\15^2+8^2=s^2\\\\s^2=225+64\\\\s^2=289\\\\s=17[/tex]

Perimeter:

P = 4s = 4•17 = 68 cm

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▹ Answer

Y = 1

▹ Step-by-Step Explanation

[tex]\frac{10}{2} = \frac{5}{y} \\\\10/5 = 2\\2/2 = 1\\\\Y =1[/tex]

Hope this helps!

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Answer:

Step-by-step explanation:

9 = m/3 + 4

5 = m/3

m = 15

Answer: m=15

Step-by-step explanation:

[tex]9=\frac{m}{3}+4[/tex]

subtract 4 on both sides

[tex]\frac{m}{3}+4-4=9-4[/tex]

[tex]\frac{m}{3}=5[/tex]

multiply 3 on both sides

[tex]\frac{3m}{3}=5\cdot \:3[/tex]

[tex]m=15[/tex]

Answer:

<4=<2

x+30=2x+15

x=15

therefore <4=(15)+30

=45°

Answer:

C is a function

Step-by-step explanation:

We can use the vertical line test.  If a vertical straight lines passes through the graph more than one, it is not a function

A and B are not functions

C is a function

Answer:

See below.

Step-by-step explanation:

[tex]\cos(20)-\sin(20)=\sqrt{2}\sin(25)[/tex]

First, use the co-function identity:

[tex]\sin(90-x)=\cos(x)[/tex]

We can turn the second term into cosine:

[tex]\sin(20)=\sin(90-70)=\cos(70)[/tex]

Substitute:

[tex]\cos(20)-\cos(70)=\sqrt{2}\sin(25)[/tex]

Now, use the sum to product formulas. We will use the following:

[tex]\cos(x)-\cos(y)=-2\sin(\frac{x+y}{2})\sin(\frac{x-y}{2})[/tex]

Substitute:

[tex]\cos(20)-\cos(70)=-2\sin(\frac{20+70}{2})\sin(\frac{20-70}{2})\\\cos(20)-\cos(70) =-2\sin(45)\sin(-25)\\\cos(20)-\cos(70)=-2(\frac{\sqrt{2}}{2})\sin(-25)\\ \cos(20)-\cos(70)=-\sqrt{2}\sin(-25)[/tex]

Use the even-odd identity:

[tex]\sin(-x)=-\sin(x)[/tex]

Therefore:

[tex]\cos(20)-\cos(70)=-\sqrt{2}\sin(-25)\\\cos(20)-\cos(70)=-\sqrt{2}\cdot-\sin(25)\\\cos(20)-\cos(70)=\sqrt{2}\sin(25)[/tex]

Replace the second term with the original term:

[tex]\cos(20)-\sin(20)=\sqrt{2}\sin(25)[/tex]

Proof complete.

Answer:

a) 2 h 45 min

b) 2 h 50 min

c) 2 h 20 min

d) 3 h 20 min

Step-by-step explanation:

Answer:

a) hours: 9 hours 15 minutes

minutes: 555

b) hours: 9hours 10 minutes

minutes: 550

c) hours: 2hours 20 minutes

minutes: 140

d) hours: 3 hours 20 minutes

minutes: 200

Step-by-step explanation:

Answer:

Daily pay= $18

5 days pay = $90

Step-by-step explanation:

Archer's daily pay =p

Pay for 5 days= 5p

Gas = 1/2 of 5p

= 1/2 × 5p

= 5p/2

Water = $5

Balance = $40

5p = 5/2p + 5 + 40

5p - 5/2p = 45

10p -5p /2 = 45

5/2p = 45

p= 45÷ 5/2

= 45 × 2/5

= 90/5

P= $18

5p= 5 × $18

=$90

The equation to determine Archer's daily pay is

5p = 5/2p + 5 + 40

Divide both sides by 5

p = 5/2p + 45 ÷ 5

= (5/2p + 45) / 5

p= (5/2p + 45) / 5

Answer:

Step-by-step explanation:

Let y(t) be the amount of salt in the tank after time t.

(A) Incoming rate = 0 (due to Pure water having no salt)

(B) Mixed solution comes out at 150 L/min. Initially the tank has 15,000 L of brine with 24 kg of salt.

concentration of salt at time t  = y(t) / 15000 kg/L

Outgoing rate = y(t)/15000 * 150 = y(t) / 100

(C) we know that,

[tex]\frac{dy}{dx} =(incoming\ rate) - (outgoing\ rate)[/tex]

[tex]\frac{dy}{dx} =0-\frac{y(t)}{100} = \frac{-y(t)}{100}[/tex]

Separate variable and integrate

[tex]\int {\frac{dy}{y} } = - \int {\frac{1}{100} } \, dt[/tex]

[tex]ln|y|=-\frac{1}{100}t + D[/tex]

[tex]y=e^{D} e^{\frac{-t}{100} }[/tex]

[tex]y= Ce^{\frac{-t}{100} }\ [C=e^{D} ][/tex]

At t= 0 , y(0) = 24 kg

[tex]24=C\ e^{0}[/tex]

C= 24

(D) Therefore, the amount of salt in the tank after time t :

[tex]y(t)=24e^{\frac{-t}{100} }\ kg[/tex]

Answer:

3(6)

??

Step-by-step explanation:

=====================================================

Explanation:

The double tickmarks for quadrilateral MATH show that MA = TH. Since TH is 5 units long, this makes MA the same length as well.

For quadrilateral ROKS, we have RO = 15. For "MATH" and "ROKS" we have "MA" and "RO" as the first two letters of each four-letter sequence; meaning that MA and RO correspond together.

The ratio of the corresponding segments is RO/MA = 15/5 = 3.

The larger quadrilateral has each side length 3 times longer than the smaller quadrilateral's corresponding side lengths.

--------------

In short,

larger side = 3*(smaller side)

--------------

Using this scale factor of 3, we can find MH

larger side = 3*(smaller side)

RS = 3*(MH)

21 = 3*MH

3*MH = 21

MH = 21/3

MH = 7

Step-by-step explanation:

Example 1:

a+c=b+c then a=b

First let the value of a and b be different (not equal)

a=5

b=7

c=10

a+c=b+c

5+10=7+10

15≠17

Example 2:

Let the value a and b be equal (the same)

a=5

b=5

c=10

a+c=b+c

5+10=5+10

15=15

So when,

a+c and b+c is equal, a and b are always equal.

Hope this helps ;) ❤❤❤

Answer:

a=b

Step-by-step explanation:

Reason:

a+c=b+c

a-b=c-c

c-c would be 0

if a-b=c-c=0

a-b=0

Only if a=b can a-b=0

You can also take it as:

b-a=c-c (a+c=b+c)

b-a=0=c-c

Therefore b=a

By the way even I am a BTS army

Answer:

27500

Step-by-step explanation:

meters are 100 times more than kilometers hope this helps:)

Answer:

d. (9, 13)

Step-by-step explanation:

5x-3y=6  /*6

6x-4y=2  /*(-5)

30x - 18y = 36

-30x +20y = - 10

            2y = 26

              y = 13

5x-3y=6

5x - 3*13 = 6

5x - 39 = 6

5x = 45

x = 9

(9, 13)

Answer:

the answer is 3

Step-by-step explanation:

[tex]\displaystyle f(x)=\sin(3x)\\\\\\f'(x)=\lim_{h\to0}\dfrac{\sin(3(x+h))-\sin(3x)}{h}\\\\f'(x)=\lim_{h\to0}\dfrac{\sin(3x+3h)-\sin(3x)}{h}\\\\f'(x)=\lim_{h\to0}\dfrac{2\cos\left(\dfrac{3x+3h+3x}{2}\right)\sin\left(\dfrac{3x+3h-3x}{2}\right)}{h}\\\\f'(x)=\lim_{h\to0}\dfrac{2\cos\left(\dfrac{6x+3h}{2}\right)\sin\left(\dfrac{3h}{2}\right)}{h}\\\\f'(x)=\lim_{h\to0}\dfrac{2\cos\left(\dfrac{6x+3h}{2}\right)\sin\left(\dfrac{3h}{2}\right)}{\dfrac{3h}{2}}\cdot\dfrac{3}{2}[/tex]

[tex]\displaystyle f'(x)=\lim_{h\to0}2\cos\left(\dfrac{6x+3h}{2}\right)\cdot\dfrac{3}{2}\\\\f'(x)=\lim_{h\to0}3\cos\left(\dfrac{6x+3h}{2}\right)\\\\f'(x)=3\cos\left(\dfrac{6x+3\cdot 0}{2}\right)\\\\f'(x)=3\cos\left(\dfrac{6x}{2}\right)\\\\f'(x)=3\cos(3x)[/tex]

The derivative of sin 3x using first principles is; 3cos(3x)

We want to find the derivative of sin 3x using first principles.

Step 1;

f'(x) = [tex]\lim_{h \to \ 0} \frac{(sin3(x + h)) - sin(3x)}{h}[/tex]

Step 2; Expand the bracket to get;

f'(x) = [tex]\lim_{h \to \ 0} \frac{(sin(3x + 3h)) - sin(3x)}{h}[/tex]

Step 3: According to trigonometric identities, we know that;

sin A - sin B = [tex]2cos\frac{A + B}{2} sin\frac{A - B}{2}[/tex]

Applying that to the answer in step 2 gives;

f'(x) = [tex]\lim_{h \to \ 0} \frac{2(cos\frac{(3x + 3h + 3x)}{2}) sin\frac{(3x + 3h - 3x)}{2})}{h}[/tex]

Step 4; Simplify the brackets to obtain;

f'(x) = [tex]\lim_{h \to \ 0} \frac{2(cos\frac{(6x + 3h)}{2}) sin\frac{(3h)}{2})}{h}[/tex]

Step 5; Rewrite the denominator to get;

f'(x) = [tex]\lim_{h \to \ 0} \frac{2(cos\frac{(6x + 3h)}{2}) sin\frac{(3h)}{2})}{\frac{3h}{2}*\frac{2}{3}}[/tex]

Step 6; Input the limit of 0 for h to get;

f'(x) = [tex]\frac{2(cos\frac{(6x)}{2}) sin\frac{(0)}{2})}{\frac{0}{2}*\frac{2}{3}}[/tex]

⇒ f'(x) = [tex]\frac{2(cos3x)}{2/3} \frac{ 0}{{0}}[/tex]

0/0 = 1. Thus;

f'(x) = 3cos(3x)

Read more about differentiation from first principles at; https://brainly.com/question/5313449

Answer:

√137

Step-by-step explanation:

[tex](x_1, y_1) = (6, 0)\\(x_2, y_2) = (-5, 4)\\\\d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\d = \sqrt{(-5-6)^2+(4-0)^2}\\ d = \sqrt{(-11)^2+(4)^2}\\ d = \sqrt{121+16}\\ d = \sqrt{137}\: or \:11.7[/tex]

Answer:

16 feet.

Step-by-step explanation:

Please refer to the attached diagram for the clear understanding of the given question statement.

A is the position of tour guide on deck.

B is the sea level. (Can be considered as zero on the number line)

C is the position of Diver and

D is the point on ocean floor.

Below sea level dimensions are given as negative in the question statement.

As per given statement,

AB = 5 feet

BC = 11 feet (Ignoring the negative sign as negative sign only depicts that it is below sea level)

BD = 18 feet

To find:

Distance from tour guide to the diver = ?

Solution:

We have to actually find the value of AC here as per the image attached.

AC = AB + BC = 5 + 11 = 16 feet

Answer:

[tex]\large \boxed{\sf \bf \ \ x^2-6x+45 \ \ }[/tex]

Step-by-step explanation:

Hello, please consider the following.

You need to develop the expression.

[tex]y=(x-3)^2+36\\\\=x^2-2 * 3 * x +3^2+36\\\\=x^2-6x+9+36\\\\=x^2-6x+45[/tex]

Thank you.

Answer:

D

Step-by-step explanation:

Answer: [tex]y-1=\dfrac32(x+3)[/tex]

Step-by-step explanation:

Slope of a line passes through (a,b) and (c,d) = [tex]\dfrac{d-b}{c-a}[/tex]

In graph(below) given line is passing through (-2,-4) and (2,2) .

Slope of the given line passing through (-2,-4) and (2,2) =[tex]\dfrac{-4-2}{-2-2}=\dfrac{-6}{-4}=\dfrac{3}{2}[/tex]

Since parallel lines have equal slope . That means slope of the required line would be .

Equation of a line passing through (a,b) and has slope m is given by :_

(y-b)=m(x-a)

Then, Equation of a line passing through(-3, 1) and has slope =  is given by

[tex](y-1)=\dfrac32(x-(-3))\\\\\Rightarrow\ y-1=\dfrac32(x+3)[/tex]

Required equation: [tex]y-1=\dfrac32(x+3)[/tex]

Answer:

25[tex]\sqrt{3}[/tex] +60

Step-by-step explanation: The first thing you need to do is realize that, this figure is a isosceles trapezoid due to the markings on each side.

So now we know both sides are 10.

We also know the the top two angles are congruent to each other and so are the bottom two angles due to the trapezoid being isosceles.

So the top two angles are 120 degrees and bottom two angles are 60 degrees.

It seems like we can't find the sides, let's try drawing two lines from each top angle all the way down to form two right triangles.

Wow, these two triangles are special right triangles in the form of

30 - 60 - 90 degrees.

shorter side = n

longer side = n[tex]\sqrt{3}[/tex]

hypotenuse = 2n

So, 2n = 10

n = 5 for the short side

The bottom base is 4[tex]\sqrt{3}[/tex] + 5 + 5 = 10 + 4[tex]\sqrt{3}[/tex]

The longer side is  5[tex]\sqrt{3}[/tex].

The area of trapezoid = (base1 + base2)/2 * height

= (4[tex]\sqrt{3}[/tex] + 10 + 4[tex]\sqrt{3}[/tex])/2 * 5[tex]\sqrt{3}[/tex] = (10 + 8[tex]\sqrt{3}[/tex])/2 * 5[tex]\sqrt{3}[/tex] = (5+4[tex]\sqrt{3}[/tex])*5[tex]\sqrt{3}[/tex] = 25[tex]\sqrt{3}[/tex] +60

So, 25[tex]\sqrt{3}[/tex] + 60 is our answer.

Answer:

  60 +25√3

Step-by-step explanation:

In the figure of the isosceles trapezoid below, the angles at C and D are supplementary to the given angle, so are 60°. That makes triangle BDE a 30°-60°-90° right triangle, which has side length ratios ...

  DE : BE : BD = 1 : √3 : 2 = 5 : 5√3 : 10

Triangle BDE can be relocated to the other end of the figure to become triangle CAD'. Then the area of concern is that of the rectangle with height 5√3 and length 5+4√3. The area is then ...

  Area = lh = (5√3)(5 +4√3) = 5·5√3 +5·4·3

  Area = 60 +25√3 . . . square units

_____

In the figure, 6.93 = 4√3, and 8.66 = 5√3, 16.93 = 10+4√3.

Answer:

3.03 • 10⁻³ is scientific notation

0.00303 is decimal form

Answer:

x = 20

Step-by-step explanation:

Intersecting Chords Theorem: ab = cd

Step 1: Label our variables

a = x

b = x - 11

c = x - 8

d = x - 5

Step 2: Plug into theorem

x(x - 11) = (x - 5)(x - 8)

Step 3: Solve for x

x² - 11x = x² - 8x - 5x + 40

x² - 11x = x² - 13x + 40

-11x = -13x + 40

2x = 40

x = 20

Answer: x=20

Step-by-step explanation:

[tex]ab=cd[/tex]

[tex]x(x - 11) = (x - 5)(x - 8)[/tex]

[tex]x^2 - 11x = x^2 - 13x + 40[/tex]

[tex]x^2 - 11x = x^2 - 8x - 5x + 40[/tex]

[tex]-11x = -13x + 40\\2x = 40\\x = 20[/tex]

Greetings from Brasil...

In this case, we can say:

Domain = [0; 6]

Image = [3; 192]

see attachment