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Yak Travel Agency arranges trips for climbing Mount Everest. For each trip, they charge an initial fee in addition to $0.15 for each vertical meter climbed. For instance, the price for climbing all the way to the summit, which is 3500 meters above the base of the mountain, is $645. Let F represent the fee (in dollars) of a trip where they climbed ddd vertical meters. Complete the equation for the relationship between the fee and vertical distance.

Answer:

Three points are (0,-5.5), (-1,-3), (-2.2,0) and graph is shown below.

Step-by-step explanation:

The given equation is

[tex]-2y=5x+11[/tex]

We need to find three points that solve the equation.

Put x=0,

[tex]-2y=5(0)+11[/tex]

[tex]-2y=11[/tex]

[tex]y=-5.5[/tex]

Put x=-1,

[tex]-2y=5(-1)+11[/tex]

[tex]-2y=6[/tex]

[tex]y=-3[/tex]

Put y=0,

[tex]-2(0)=5x+11[/tex]

[tex]5x=-11[/tex]

[tex]x=-2.2[/tex]

So, three points (0,-5.5), (-1,-3) and (-2.2,0) are the solutions of the given equation.

Plot these points on a coordinate plane and connect them by a straight line as shown below.

Answer:

3* 10 ^12

Step-by-step explanation:

(5x10^7)(6x10^4)

Multiply the numbers together

5*6 =30

Add the exponents

10^7 * 10 ^ 4 = 10 ^(7+4) = 10 ^11

30 * 10 ^11

But this is not scientific notation since 30 >10

Move the decimal one place to the left and add 1 to the exponent

3* 10 ^12

Answer:

3* 10 ^12

Step-by-step explanation:

Answer: [tex]\frac{4}{3}[/tex]

Step-by-step explanation:

This is a multiplication problem. You are multiplying [tex]\frac{1}{3}[/tex] by 4. This also means 4 divided by 3. They are both the same.

Answer:

seeed

Step-by-step] explanation:

ddd~!`

Answer:

A: y= 1/4x - 7

if it is perpendicular, then it creates 4 right angles. so that new line would pass through (0,-7) and something else that isnt important. but the slope, or m, would be 1/4, and the y intercept would be -7. so the new equation is y=1/4x-7

Answer:

[tex] y = \frac{3}{4}x - 2 [/tex]

Step-by-step explanation:

Equation of a line is given as [tex] y = mx + b [/tex]

Where,

m = slope of the line = [tex] \frac{y_2 - y_1}{x_2 - x_1} [/tex]

b = y-intercept, which is the value at the point where the line intercepts the y-axis. At this point, x = 0.

Let's find m and b to derive the equation for the line.

[tex] m = \frac{y_2 - y_1}{x_2 - x_1} [/tex]

Use the coordinate pair of any two points on the line. Let's use the following,

[tex] (0, -2) = (x_1, y_1) [/tex] => on the line, when x = 0, y = -2

[tex] (4, 1) = (x_2, y_2) [/tex] => on the line, when x = 4, y = 1

Plug in the values and solve for m

[tex] m = \frac{1 - (-2)}{4 - 0} [/tex]

[tex] m = \frac{1 + 2}{4} [/tex]

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

b = -2 (the line intercepts the y-axis at this point)

Our equation would be =>

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

[tex] y = \frac{3}{4}x + (-2) [/tex]

[tex] y = \frac{3}{4}x - 2 [/tex]

Answer:

  no

Step-by-step explanation:

If any x-value is repeated, the relation is not a function. Both x=4 and x=9 are repeated values, so this relation is not a function.

Answer:

17/16 OR [tex]1\frac{1}{16}[/tex] minutes

Step-by-step explanation:

Since Jayla spent 1/16 of a minute AND one whole minute watching a millipede crawl, we'd need to first add the two numbers.

Since the given minute is out of 16, we can convert the one minute to 16/16.  This means we can add the other 1/16 of a minute.

This leaves us with Jayla watching the millipede for 17/16 OR [tex]1\frac{1}{16}[/tex] minutes.

Hope this helps!! <3 :)

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▹ Answer

(-39/35, 9/7)

▹ Step-by-Step Explanation

5y + 2y = 9

-5x - 2y = 3

Solve the equation:

y = 9/7

-5x - 2y = 3

Substitute the value of y:

-5x - 2 * 9/7 = 3

x = -39/35

(x, y) = (-39/35, 9/7)

Hope this helps!

CloutAnswers ❁

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To solve this system by addition, we start by adding both of our equations together but notice that the x terms and the y terms cancel out.

This leaves us with 0 on the left side and on the right side,

9 + 13 = 12 so we are left with the equation 0 = 12.

Since 0 = 12 is a false statement, this means that

there is no solution to our system of equations.

Answer:

the first  partial sum  [tex]\mathbf{S_1= \dfrac{1}{2}}[/tex]

the second partial sum  [tex]\mathbf{S_2= \dfrac{3}{4} }[/tex]

the third partial sum [tex]\mathbf{S_3= \dfrac{7}{8}}[/tex]

the fourth  partial sum [tex]\mathbf{S_4= \dfrac{15}{16}}[/tex]

the fifth partial sum [tex]\mathbf{S_5= \dfrac{31}{32}}[/tex]

the sixth partial sum [tex]\mathbf{S_6= \dfrac{63}{64}}[/tex]

Step-by-step explanation:

The term of the sequence are given as : [tex]\dfrac{1}{2}[/tex], [tex]\dfrac{1}{2^2}[/tex], [tex]\dfrac{1}{2^3}[/tex], [tex]\dfrac{1}{2^4 }[/tex] ,  .  .  .  

The nth term for this sequence is , [tex]\mathtt{a_n =( \dfrac{1}{2})^n}[/tex]

The nth partial sum of the sequence for [tex]\mathtt{a_1,a_2,a_3.... a_n}[/tex] is [tex]\mathtt{S_n}[/tex]

where;

[tex]\mathtt{S_n = a_1 +a_2+a_3+ ...+a_n}[/tex]

The first partial sum  is:  [tex]\mathtt{S_1= a_1}[/tex]

[tex]\mathtt{S_1= (\dfrac{1}{2})^1}[/tex]

[tex]\mathbf{S_1= \dfrac{1}{2}}[/tex]

Therefore, the first  partial sum  [tex]\mathbf{S_1= \dfrac{1}{2}}[/tex]

The second partial sum is: [tex]\mathtt{S_2= a_1+a_2}[/tex]

[tex]\mathtt{S_2= (\dfrac{1}{2})^1 + (\dfrac{1}{2})^2}[/tex]

[tex]\mathtt{S_2= \dfrac{1}{2} + \dfrac{1}{4}}[/tex]

[tex]\mathtt{S_2= \dfrac{2+1}{4} }[/tex]

[tex]\mathbf{S_2= \dfrac{3}{4} }[/tex]

Therefore, the second partial sum  [tex]\mathbf{S_2= \dfrac{3}{4} }[/tex]

The third partial sum is : [tex]\mathtt{S_3= a_1+a_2+a_3}[/tex]

[tex]\mathtt{S_3= (\dfrac{1}{2})^1 + (\dfrac{1}{2})^2+(\dfrac{1}{2})^3 }[/tex]

[tex]\mathtt{S_3= \dfrac{1}{2} + \dfrac{1}{4}+\dfrac{1}{8}}[/tex]

[tex]\mathtt{S_3= \dfrac{4+2+1}{8}}[/tex]

[tex]\mathbf{S_3= \dfrac{7}{8}}[/tex]

Therefore, the third partial sum [tex]\mathbf{S_3= \dfrac{7}{8}}[/tex]

The fourth partial sum : [tex]\mathtt{S_4= a_1+a_2+a_3+a_4}[/tex]

[tex]\mathtt{S_4= (\dfrac{1}{2})^1 + (\dfrac{1}{2})^2+(\dfrac{1}{2})^3+(\dfrac{1}{2})^4 }[/tex]

[tex]\mathtt{S_4= \dfrac{1}{2} + \dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}}[/tex]

[tex]\mathtt{S_4= \dfrac{8+4+2+1}{16}}[/tex]

[tex]\mathbf{S_4= \dfrac{15}{16}}[/tex]

Therefore, the fourth  partial sum [tex]\mathbf{S_4= \dfrac{15}{16}}[/tex]

The fifth partial sum : [tex]\mathtt{S_5= a_1+a_2+a_3+a_4+a_5}[/tex]

[tex]\mathtt{S_5= (\dfrac{1}{2})^1 + (\dfrac{1}{2})^2+(\dfrac{1}{2})^3+(\dfrac{1}{2})^4 +(\dfrac{1}{2})^5 }[/tex]

[tex]\mathtt{S_5= \dfrac{1}{2} + \dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}}[/tex]

[tex]\mathtt{S_5= \dfrac{16+8+4+2+1}{32}}[/tex]

[tex]\mathbf{S_5= \dfrac{31}{32}}[/tex]

Therefore, the fifth partial sum [tex]\mathbf{S_5= \dfrac{31}{32}}[/tex]

The sixth partial sum: [tex]\mathtt{S_5= a_1+a_2+a_3+a_4+a_5+a_6}[/tex]

[tex]\mathtt{S_6= (\dfrac{1}{2})^1 + (\dfrac{1}{2})^2+(\dfrac{1}{2})^3+(\dfrac{1}{2})^4 +(\dfrac{1}{2})^5 +(\dfrac{1}{2})^6 }[/tex]

[tex]\mathtt{S_6= \dfrac{1}{2} + \dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64} }[/tex]

[tex]\mathtt{S_6= \dfrac{32+16+8+4+2+1}{64}}[/tex]

[tex]\mathbf{S_6= \dfrac{63}{64}}[/tex]

Therefore, the sixth partial sum [tex]\mathbf{S_6= \dfrac{63}{64}}[/tex]

the graph has 12 segments so angle enclosed by each segment is [tex] {2\pi\over 12}=\frac{\pi}6[/tex]

anti-clockwise is taken as positive, so if you want positive q, you need to rotate 8 segments [tex] q=8\frac,{\pi}6=\frac{4\pi}3 [/tex] , and and 8 circles or units so r=8

and for a negative angle, you need to rotate clockwise

Which is 4 segments from the horizontal line. so [tex]q=-\frac{2\pi}3[/tex] and r will be same, 8 units.

[not sure about -r so I won't include it in answer]

Answer:

Points : ( 8, - 2π/3 ), ( - 8, π/3 ), ( - 8, - 5π/3 )

Step-by-step explanation:

For the first two cases, ( r, θ ) r would be > 0, where r is the directed distance from the pole, and theta is the directed angle from the positive x - axis.

So when r is positive, we can tell that this point is 8 units from the pole, so r is going to be 8 in either case,

( 8, 240° ) - because r is positive, theta would have to be an angle with which it's terminal side passes through this point. As you can see that would be 2 / 3rd of 90 degrees more than a 180 degree angle,or 60 + 180 = 240 degrees.

( 8, - 120° ) - now theta will be the negative side of 360 - 240, or in other words - 120

Now let's consider the second two cases, where r is < 0. Of course the point will still be 8 units from the pole. Again for r < 0 the point will lay on the ray pointing in the opposite direction of the terminal side of theta.

( - 8, 60° ) - theta will now be 2 / 3rd of 90 degrees, or 60 degrees, for - r. Respectively the remaining degrees will be negative, 360 - 60 = 300, - 300. Thus our second point for - r will be ( - 8, - 300° )

_________________________________

So we have the points ( 8, 240° ), ( 8, - 120° ), ( - 8, 60° ), and ( - 8, - 300° ). However we only want 3 cases, so we have points ( 8, - 120° ), ( - 8, 60° ), and ( - 8, - 300° ). Let's convert the degrees into radians,

Points : ( 8, - 2π/3 ), ( - 8, π/3 ), ( - 8, - 5π/3 )

Answer:

True

Step-by-step explanation:

We assume that is true for the nth case and prove it for the n+1 case

and show that it is true for the case when n=1

Answer:

6.02

six point zero two

Step-by-step explanation:

Answer:

602 / 100= 6,02

Step-by-step explanation:

602 to divide 100 = 6,02

Answer:

m = [tex]\frac{y-6}{x}[/tex]

Step-by-step explanation:

Given

y = mx + 6 ( subtract 6 from both sides )

y - 6 = mx ( divide both sides by x )

[tex]\frac{y-6}{x}[/tex] = m

Answer:

The answer is option B

Step-by-step explanation:

To find the coefficient of the third term in

[tex](x + 5)^{7} [/tex]

Rewrite the expansion in the form

[tex](a + x)^{n} [/tex]

where n is the index

So we have

[tex] ({5 + x})^{7} [/tex]

After that we use the formula

[tex]nCr( {a}^{n - r} ) {x}^{r} [/tex]

where r is the term we are looking for

For the third term we are looking for the term containing x²

that's

r + 1 = 3

r = 2

So to find the coefficient of the third term

We have

[tex]7C2[/tex]

Hope this helps you

first of all, the notation is wrong it should be [tex] {}^nC_r \text{ and more usual notation is } {n \choose k} [/tex]

second, the

[tex](r+1)^{\text{th}} \text{ term } T_{r+1} \text{ in the expansion of } (x+a)^n \text{ is } {n \choose r}x^{(n-r)}a^r[/tex]

here [tex] a=5 \text{ and } n=7 \text{ and for } 3^{\text{rd}} \text{ term } T_3, \quad r+1=3 \implies r=2 [/tex]

so the coefficient of third term is, [tex]{7 \choose 2}={7\choose 5}[/tex]

an important property of binomial coefficient you should know:

[tex] {n \choose k}={n \choose {n-k}}[/tex]

and if you interchange [tex] x \text{ and } a[/tex]

only the "order" will get reversed. i.e. the series will start from back.

another thing, the [tex] k^{\text{th}} \text{ term from beginning, is the } (n-k+2)^{\text{th}} \text{ term from behind}[/tex]

Answer:

see below

Step-by-step explanation:

Scientific notation is a * 10 ^b

a must be a number between 1 ( including 1 ) and less than 10

12 is greater than 10 so it is not scientific notation

Answer:

Although the question is not clear, It most likely looks like you were asking for the calculation of the savings for the month after increase.

savings for the month after increase = $1172.4

Step-by-step explanation:

First, let us calculate how much was saved before the increase in savings:

monthly income = $19,540

Percentage saved = 4% of monthly income

= 4/100 × 19,540 = 0.04 × 19,540 = $781.6

Next, we are given the ratio of increase in savings as 3:2

Let the new savings amount be x

3 : 2 = x : 781.6

[tex]\frac{3}{2} = \frac{x}{781.6} \\781.6\ \times 3\ =2x\\2344.8 = 2x\\x =\frac{2344.8}{2} \\x = \$1172.4[/tex]

therefore savings for the month after increase = $1172.4

Just incase you were looking for the savings before the increase, the answer is $781.6 (as calculated above)

Answer:

work pictured and shown

Answer:

Last one

Step-by-step explanation:

● [ ( 3^2 × 5^0) / 4 ]^2

5^0 is 1 since any number that has a null power is equal to 1.

●[ (3^2 ×1 ) / 4 ]^2

● (9/4)^2

● 81 / 16

Classica aplicação de regra de 3:

é dito que: 1 milha = 1,6km

Logo, eis a regra de 3:

  milha           km

     1   --------    1,6

     X  --------   240

1,6X = 240.1

X = 240/1,6

Logo 240km equivalem a 150milhas

Answer:

Step-by-step explanation:

Hello, please consider the following.

For x > 1, we can apply Pythagoras theorem as below.

[tex]\text{Let's estimate this sum of two squares.} \\\\(2x)^2+(x^2-1)^2=4x^2+x^4-2x^2+1=x^4+2x^2+1\\\\\text{Let's estimate this square, now.} \\\\(x^2+1)^2=x^4+2x^2+1\\\\\text{The two expressions are equal, meaning.} \\\\(2x)^2+(x^2-1)^2=(x^2+1)^2\\\\\text{Using Pythagoras' theorem, we can say that this is a right triangle.}[/tex]

Thank you

Step-by-step explanation:

Hi, there!!!

According to the question we must find the area of shaded region, but we must find area of circle and rectangle to find area of shaded region,

So, let's simply work with it,

Firstly, finding the area of rectangle,

length = 11cm.

breadth = 11cm.

now, area= length× breadth.

or, a = 11cm× 11cm.

a= 121cm^2

Now, let's work out the area of circle.

radius= 5cm

and pi. = 3.14 {using pi value as 3.14}

now,

area of a circle = pi× r^2

or, a= 3.14×5^2

or, a = 78.5 cm^2.

Therefore, The area of a circle is 78.5cm^2.

Now lastly finding the area of shadedregion,

area of shaded region = area of rectangle - area of circle.

or, area of shaded region = 121cm^2 - 78.5cm^2

Therefore, the area of shaded region is 42.5 cm^2.

Hope it helps...

Answer:

3780.  

Step-by-step explanation:

To solve this we will start by just considering the number of ways to arrange 9 objects. We can do this in  9!  ways.

However since we have 3 reoccurring letters in Tennessee namely n,s and e we need to remove the times these form the same arrangement. Let me give an example to show what this means. Lets say we have the arrangement:

ennetssee

Now what happens if we exchange the places of the letters n for example? Of course we get the same arrangement of letters. We don’t want to count these as 2 different arrangements since for our interests they are the same. We therefore divide  9!  by the number of times this type of double counting occurs.

Since the word has the letter n occurring twice we will start by diving by  2! .

The letter s occurs 2 times as well so we will have to divide by  2!  again.

Finally the letter e occurs 4 times and so we will have to divide by  4!  here.

Now we get the following result:

9/(2 x 2 x 4)=3780.  

So in conclusion there are 3780 different ways to arrange the letters in Tennessee.

Answer:

7.5 < x≤10.5

Step-by-step explanation:

14<2x−1≤20

Add 1 to all sides

14+1<2x−1+1≤20+1

15<2x≤21

Divide each side by 2

15/2 <2x/2 ≤21/2

7.5 < x≤10.5

Steps to solve:

14 < 2x - 1 <= 20

~Add 1 to everything

15 < 2x <= 21

~Divide 2 to everything

7.5 < x <= 10.5

Best of Luck!

Answer:

  [tex]\dfrac{3}{(y+2)^3}[/tex]

Step-by-step explanation:

Maybe you want this written using math symbols. It will be ...

  [tex]\boxed{\dfrac{3}{(y+2)^3}}[/tex]

Answer:

0.03 square meters

Step-by-step explanation:

1. convert 20 cm, 15 cm into m by dividing the number by 100 then you got 0.2 m and 0.15 m respectively

2. calculate the area by multiply 2 numbers together

0.2 × 0.15 = 0.03 square meter

Answer:

Solution : Option A

Step-by-step explanation:

What we want to do here is eliminate the parameter t. In order to do that, we can isolate t in our first equation x = t + 6 ----- ( 1 ) and then plug that value for t in the second equation y = 3t - 1. Our solution will be an equation that is not present with t.

( 1 ) x = t + 6, t = x - 6

( 2 ) y = 3( x - 6 ) - 1 ( Distribute the " 3 " in 3( x - 6 ) )

y = 3x - 18 - 1 ( Combine like terms )

y = 3x - 19

As you can see our result will be option a, y = 3x - 19.

We have 26 letters and 3 slots to fill. We can reuse a letter if it has been picked, so we have 26^3 = 26*26*26 = 17,576 different three letter "words". I put that in quotes because a lot of the words aren't actual words, but more just a sequence of letters.

Answer:

different

Step-by-step explanation:

reflection is basically like a mirror where it reflects you. rotation is when an object spins/rotates.

Answer:

If the width of the compass changed, it would make the arc smaller or larger. This would then make the line segment to be drawn smaller or larger, so it would not match the original.

Step-by-step explanation:

We assume your construction is trying to copy the length of a line segment. That length is "measured" by the width of the compass. If it is changed, it no longer matches the length of the segment you're trying to copy, so you will not get the copy you want.

Answer:

The correct answer is option B