a map of an amusement park is shown on a coordinate plane, where each square of the grid represents 1 square meter. The water ride is at (-17,12), the roller coaster is at (26,-8), and the Ferris wheel is at (2,20). Find each distance to the nearest tenth of a meter.
* what is the distance between the water ride and the roller coaster?
*What is the midpoint between the roller coaster and the ferris wheel? what is the distance from the midpoint to the ferris wheel

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Explanation:

The general cosine template is

y = A*cos(B(t - C)) + D

where in this case

We only really need to worry about the B value. To get the period T, we do the following

T = 2pi/B

T = (2pi)/(pi/4)

T = 2pi * (4/pi)

T = 8

Note how the pi terms canceled. The period is 8 seconds, which is the length of one full cycle. This is the time it takes for the pendulum to do one full swing (eg: start at the right, swing to the left all the way, then swing back to the right again).

The result of 8 we got has nothing to do with the D = 8 value (this D value could be any other number and T = 8 would still be the case as long as B doesn't change of course).

Answer:

false

Step-by-step explanation:

this is because there is no function of an unknown in this set

Answer:

Explanation:

Step 1

Multiply the denominator by the whole number

7 × 7 = 49

Step 2

Add the answer from Step 1 to the numerator

49 + 6 = 55

Step 3

Write answer from Step 2 over the denominator

Answer = 55/7

Answer:

55/7

Step-by-step explanation:

To convert 7 6/7 to an improper fraction

Step-by-step explanation:

cos(90) = 0

around this point the cos function "mirrors" with opposite signs. cos(<90) is positive and cos(>90) is negative.

but |cos(90-a)| = |cos(90+a)| for 0 <= a <= 90

75 = 90 - 15

105 = 90 + 15

so, a = 15

and because of

|cos(90-a)| = |cos(90+a)| for 0 <= a <= 90

cos (90-15) = cos(75) = -cos(90+15) = -cos(105)

we observe that this figure could be divided into a triangle and a rectangle.

for the area of the triangle, bxh/2, we have 15 x 12/2 = 90

and for the rectangle, we have 5 x 12 = 60

so, 90 + 60 = 150

hope it helps :)

Answer:

28125    

Step-by-step explanation:

Refer to the image below.

Answer: 7m⁵ -3m³ - 3

Working:

= (6m⁵ + 3 - m³ - 4m) -(-m⁵ + 2m³- 4m +6)

= 6m⁵ + 3 - m³ - 4m +m⁵-2m³+4m - 6

= 6m⁵ + m⁵-m³ -2m³ -4m + 4m - 6 +3

= 7m⁵ -3m³ - 3

Answered by Gauthmath must click thanks and mark brainliest

Answer:

37.15 pounds left

Step-by-step explanation:

Add the two pound values

1.3+1.75= 3.05

Subtract it from the total number of pounds in the bag

40.2-3.05

=37.15

Answer:

x = 10

Step-by-step explanation:

7(x+1+7)=6(x+5+6)

or, 7(x+8)=6(x+11)

or, 7x+56=6x+66

or, 7x-6x=66-56

or, x=10

Answered by GAUTHMATH

Answer:

Angelica's house is 11 miles away from the gas station.

Step-by-step explanation:

The most simple path from Angelica's house is 11 blocks away from the gas station and each unit/block is 1 mile. So 11 miles.

Answer:

C

Step-by-step explanation:

b=360-2(65)=360-130=230

Answer:

230

Step-by-step explanation:

b= 360-2(65)

=360-130

=230

Answer:

(x+8)(x+7)

Step-by-step explanation:

Find two numbers that add to

b

and mulitply to

a*c

so find two numbers that add to

15

and mulitply to

56

Those two numbers are 8 and 7

therefore the answer is

(x+8)(x+7)

Answer:

didn't shade all of them

Answer:

D

Step-by-step explanation:

x^2=y^3

Then x^6=y^9 but x^3z=z^9 so 3z=6, z=2

Answer:

z = 2

Step-by-step explanation:

[tex]x^{3z} = y^{9}\\\\x^{3z} =y^{3*3} \\\\x^{3z}= (y^{3})^{3}\\\\x^{3z}=(x^{2})^{3}\\\\x^{3z} = x^{6}[/tex]

As bases are same, compare the powers

3z = 6

z = 6/3

z = 2

Answer:

Step-by-step explanation:

This is a "regular" sin graph that's "taller" than the original. The amplitude is 3; other than that, its period is the same and it has not shifted to the right or left, so the equation, judging from the graph, is

[tex]y=3sin(x)[/tex]

Answer:

36 ft²

Step-by-step explanation:

equilateral means all sides are equally long.

a triangle has 3 sides.

in our case they are all equally long.

and their sum (all 3 sides together) is the perimeter

P = 18 = 3 × side length

side length = 18/3 = 6 ft

a square with the same side length (6) had then an area of

side length × side length = 6×6 = 6² = 36 ft²

it is important to remember : a length is measured e.g. in feet (ft). an area is then meshed in square-feet (ft²). it is important to add this to the calculated numbers to express the right dimension of these numbers.

Answer:

22.08 ft^2

Step-by-step explanation:

First find the area of the full circle

A = pi r^2

The diameter is 15 so the radius is 1/2 (15) = 7.5

A = (3.14) (7.5)^2

   =176.625

45 degrees is  a fraction of a circle which is 360 degrees

45/360 = 1/8

Multiply the area of the circle by this fraction

1/8 (176.625) =22.078125

Rounding to the nearest hundredth

22.08

Answer:

22.08 ft2

Step-by-step explanation:

Answer:

B = multiply both sides by 2y+1

Step-by-step explanation:

Answer:

B. multiply both sides by the equation 2y + 1

Answer:

Width = 4 cm

Length = 8 cm

Step-by-step explanation:

Hi there!

Let [tex]l[/tex] be equal to the length of the rectangle.

Let [tex]w[/tex] be equal to the width of the rectangle.

1) Determine equations to find the length and width

We're given that the length is two times the length of the width:

[tex]l=2w[/tex]

We're also given that the area of the rectangle is 32 cm². Recall that the area of a rectangle is [tex]A=lw[/tex]:

[tex]A=lw\\32=lw[/tex]

Now, we have our two equations:

[tex]\displaystyle \left \{ {{l=2w} \atop {32=lw}} \right.[/tex]

2) Solve for the width using substitution

[tex]\displaystyle \left \{ {{l=2w} \atop {32=lw}} \right.[/tex]

Replace [tex]l[/tex] in the second equation with [tex]2w[/tex] from the first equation:

[tex]32=(2w)w\\32=2w^2[/tex]

Divide both sides by 2 to isolate w²:

[tex]16=w^2[/tex]

Take the square root of both sides to isolate w:

[tex]\pm4=w[/tex]

Because width cannot be negative, w=4. Therefore, the width of the rectangle is 4 cm.

3) Solve for the length

[tex]\displaystyle \left \{ {{l=2w} \atop {32=lw}} \right.[/tex]

Now, that we have the width (4 cm), we can solve for the length by plugging it back into one of the equations. Either of the equations work, but we can use the first:

[tex]l=2w\\l=2(4)\\l=8[/tex]

Therefore, the length of the rectangle is 8 cm.

3) Draw the rectangle

We can use what you had before as a foundation. You drew a rectangle with width 4 cm and length 7 cm. To draw the correct rectangle, add another row on top to make it 4 cm by 8 cm.

I hope this helps!

Answer:

2/8 or 1/4 simplified

Step-by-step explanation:

If you count how many equal sections there are, you will see there are 8, and 2 or those 8 sections are red so that would easily give you your answer, 2/8.

And to simplify it divide both the top and the bottom by 2 and that gives you 1/4.

Hope the helps! :)

The probability of a red on this spinner would be 1/4.

Used the concept of probability that states,

The term probability refers to the likelihood of an event occurring. Probability means possibility.

Given that,

A spinner is shown in the image.

Since there are a total of 8 parts with different colors.

And, the number of red colors on this spinner is 2.

Hence, the probability of a red on this spinner would be,

P = 2 / 8

P = 1 / 4

Therefore, the probability is 1/4.

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Answer:

5, 12, 13

Step-by-step explanation:

let x be the longer leg then x + 1 is the hypotenuse and x - 7 the shorter leg

Using Pythagoras' identity in the right triangle

x² + (x - 7)² = (x + 1)² ← expand using FOIL

x² + x² - 14x + 49 = x² + 2x + 1

2x² - 14x + 49 = x² + 2x + 1 ( subtract x² + 2x + 1 from both sides )

x² - 16x + 48 = 0 ← in standard form

(x - 4)(x - 12) = 0 ← in factored form

Equate each factor to zero and solve for x

x - 4 = 0 ⇒ x = 4

x - 12 = 0 ⇒ x = 12

x = 4 , then x - 7 = 4 - 7 = - 3 ← not possible

x = 12, then x - 7 = 12 - 7 = 5 and x + 1 = 12 + 1 = 13

The lengths of the 3 sides are

longer leg = 12 m , shorter leg = 5 m and hypotenuse = 13 m

The point [tex](-2,2)[/tex] is a solution of the system given by [tex]y<2x+8[/tex] and [tex]y \geq -x-3[/tex]

Given point: [tex](-2,2)[/tex]

Given systems:

To find: The system to which the given point is a solution

If a point is a solution of a system, then the coordinates of the point satisfies all the equation(s) or inequation(s) of the system. So, we can substitute the x & y coordinates of the given point into the inequalities of each of the given systems and check if the inequalities are satisfied by the coordinates of the point.

(1) [tex]y>-2x+2[/tex] and [tex]y>x+5[/tex]

Substitute the coordinates of the point [tex](-2,2)[/tex] into the inequalities of the system.

Put [tex]x=-2,y=2[/tex] in [tex]y>-2x+2[/tex] to get,

[tex]2>-2(-2)+2[/tex]

[tex]2>4+2[/tex]

[tex]2>6[/tex]

The above inequality is clearly impossible and thus, the coordinates of the given point does not satisfy this inequality.

This implies that the given point is not a solution of this system.

(2) [tex]y<x+2[/tex] and [tex]y>x-1[/tex]

Substitute the coordinates of the point [tex](-2,2)[/tex] into the inequalities of the system.

Put [tex]x=-2,y=2[/tex] in [tex]y<x+2[/tex] to get,

[tex]2<-2+2[/tex]

[tex]2<0[/tex]

The above inequality is clearly impossible and thus, the coordinates of the given point does not satisfy this inequality.

This implies that the given point is not a solution of this system.

(3) [tex]y<2x+8[/tex] and [tex]y \geq -x-3[/tex]

Substitute the coordinates of the point [tex](-2,2)[/tex] into the inequalities of the system.

Put [tex]x=-2,y=2[/tex] in [tex]y<2x+8[/tex] to get,

[tex]2<2(-2)+8[/tex]

[tex]2<-4+8[/tex]

[tex]2<4[/tex]

This is a true inequality. Then, the given point satisfies the first inequality of the system.

We will now check if the point satisfies the second inequality of the system.

Put [tex]x=-2,y=2[/tex] in [tex]y \geq -x-3[/tex] to get,

[tex]2 \geq -(-2)-3[/tex]

[tex]2 \geq 2-3[/tex]

[tex]2 \geq -1[/tex]

This is also a true inequality. Then, the given point also satisfies the second inequality of the system.

Thus, the given point is a solution of this system.

(4) [tex]y<2x+3[/tex] and [tex]y \geq -2x-5[/tex]

Substitute the coordinates of the point [tex](-2,2)[/tex] into the inequalities of the system.

Put [tex]x=-2,y=2[/tex] in [tex]y<2x+3[/tex] to get,

[tex]2<2(-2)+3[/tex]

[tex]2<-4+3[/tex]

[tex]2<-1[/tex]

The above inequality is clearly impossible and thus, the coordinates of the given point does not satisfy this inequality.

This implies that the given point is not a solution of this system.

Thus, we can see that the coordinates of the given point [tex](-2,2)[/tex] satisfies the inequalities of the third system only.

Then, the point [tex](-2,2)[/tex] is a solution of the system given by [tex]y<2x+8[/tex] and [tex]y \geq -x-3[/tex].

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To be honest, I don't think it has anything to do with the exponent part at all. Instead, I think it has to do with the fact that integers are inherently easier to grasp compared to fractions (which is exactly what rational numbers are).

For instance, it's much easier to say 2+3 = 5 than it is to say 1/2 + 1/4 = 3/4

So going back to the exponent example, it's easier to say

x^2*x^3 = x^(2+3) = x^5

than it is to say

x^(1/2)*x^(1/4) = x^(1/2+1/4) = x^(3/4)

So that's my opinion as to why rational exponents are more tricky to grasp compared to integer exponents. Of course, everyone learns math differently so maybe some find fractions easier than others.

Step-by-step explanation:

there must be some typos here. but I try to address the right issues.

there is nothing under (a). so, really nothing to answer.

(b)

there are 20 sweets in her bag. 9 if the 20 are yellow.

so, the probability is the ratio of desired possibilities vs. total possibilities.

that is, tada! 9/20

that is the probabilty for Samira to pick a yellow sweet.

[1] (i)

again, 20 sweets to start with.

6 are green.

when she picks the first one, her probability to pick a green one is 6/20 = 3/10 = 0.3

and now, under the assumption that this came true, she picks another sweet.

this time she had only 19 left, and 5 of them are green.

so, this probabilty is 5/19

now both events need to happen for the case we are discussing. there is no overlapping, no ors, ifs and buts. it is just the product of both probabilities.

3/10 × 5/19 = 15/190 = 3/38

I think that is what the description asks for.

(ii)

that probability is

first selection is red and second is not red +

first selection is green and second is not green +

first selection is yellow and second is not yellow

so,

red and not red

5/20 = 1/4 red

15/19 not red (there are still 15 sweets of other colors in the bag, but again now only 19 total).

red and not red = 1/4 × 15/19 = 15/76 = 0.1974

green and not green

6/20 = 3/10 = green

14/19 = not green

green and not green = 3/10 × 14/19 = 52/190 = 26/95 =

= 0.2737

yellow and not yellow

9/20 = yellow

11/19 = not yellow

yellow and not yellow = 9/20 × 11/19 = 99/380 = 0.2606

so, now assuming up all 3 possibilities ("or" = sum) gives us the general possibility of selecting two different colors

= 0.7316

Answer: electronics

Hope that helps

The option that is not considered a sin tax is option B, gambling.

A sin tax is a tax that is applied to "harmful things" to society or individuals. An example of this is Tobacco.

From the given options, the harmful ones are alcohol, gambling, and tobacco. These 3 are considered harmful, so would be included under the sin taxes.

Then the option that is not considered a sin tax is option B, gambling.

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[tex]\\ \sf\longmapsto 2x+y=2[/tex]

[tex]\\ \sf\longmapsto 2x=2-y[/tex]

[tex]\\ \sf\longmapsto x=\dfrac{2-y}{2}\dots(1)[/tex]

And

[tex]\\ \sf\longmapsto x+1=y+2[/tex]

[tex]\\ \sf\longmapsto x=y+2-1[/tex]

[tex]\\ \sf\longmapsto x=y+1[/tex]

[tex]\\ \sf\longmapsto \dfrac{2-y}{2}=y+1[/tex]

[tex]\\ \sf\longmapsto 2-y=2(y+1)[/tex]

[tex]\\ \sf\longmapsto 2-y=2y+2[/tex]

[tex]\\ \sf\longmapsto 2-2=2y+y[/tex]

[tex]\\ \sf\longmapsto 3y=0[/tex]

[tex]\\ \sf\longmapsto y=\dfrac{0}{3}[/tex]

[tex]\\ \sf\longmapsto y=\infty[/tex]

[tex]\\ \sf\longmapsto x=\dfrac{2-y}{2}[/tex]

[tex]\\ \sf\longmapsto x=\dfrac{2-\infty}{2}[/tex]

[tex]\\ \sf\longmapsto x=\dfrac{2}{2}[/tex]

[tex]\\ \sf\longmapsto x=1[/tex]

Answer:

x=1 , y=0

Step-by-step explanation:

2x+y=2.......1

x+1=y+2

x-y=2-1

x-y=1...... 2

from equation 2 we get ,

x-y=1

x=1+y

putting the value of x in equation 1 we get,

2*(1+y) +y=2

2+2y+y=2

2+3y=2

3y=2-2

3y=0

y=0/3

y=0

putting the value of y in equation 2 we get,

x-0=1

x=1+0

x=1

hence x=1 , y=0

Step-by-step explanation:

I need help to...

I keep putting it in and no one has ever answered it and I'm struggling and getting stressed.

Answer:

a(n)= 1/2 * (-8) n-1

Step-by-step explanation:

In a geometric sequence, the ratio between successive terms is constant. This means that we can move from any term to the next one by multiplying by a constant value. Let's calculate this ratio over the first few terms:

\dfrac{-256}{32}=\dfrac{32}{-4}=\dfrac{-4}{\frac12}=\blue{-8}  

32

−256

​

=  

−4

32

​

=  

2

1

​

−4

​

=−8start fraction, minus, 256, divided by, 32, end fraction, equals, start fraction, 32, divided by, minus, 4, end fraction, equals, start fraction, minus, 4, divided by, start fraction, 1, divided by, 2, end fraction, end fraction, equals, start color #6495ed, minus, 8, end color #6495ed

We see that the constant ratio between successive terms is \blue{-8}−8start color #6495ed, minus, 8, end color #6495ed. In other words, we can find any term by starting with the first term and multiplying by \blue{-8}−8start color #6495ed, minus, 8, end color #6495ed repeatedly until we get to the desired term.

Let's look at the first few terms expressed as products:

nn 111 222 333 444

h(n)\!\!\!\!\!h(n)h, left parenthesis, n, right parenthesis \red{\dfrac12}\cdot\!\!\!\left(\blue{-8}\right)^{\,\large0}\!\!\!\!\!\!  

2

1

​

⋅(−8)  

0

start color #df0030, start fraction, 1, divided by, 2, end fraction, end color #df0030, dot, left parenthesis, start color #6495ed, minus, 8, end color #6495ed, right parenthesis, start superscript, 0, end superscript \red{\dfrac12}\cdot\!\!\!\left(\blue{-8}\right)^{\,\large1}\!\!\!\!\!\!  

2

1

​

⋅(−8)  

1

start color #df0030, start fraction, 1, divided by, 2, end fraction, end color #df0030, dot, left parenthesis, start color #6495ed, minus, 8, end color #6495ed, right parenthesis, start superscript, 1, end superscript \red{\dfrac12}\cdot\!\!\!\left(\blue{-8}\right)^{\,\large2}\!\!\!\!\!\!  

2

1

​

⋅(−8)  

2

start color #df0030, start fraction, 1, divided by, 2, end fraction, end color #df0030, dot, left parenthesis, start color #6495ed, minus, 8, end color #6495ed, right parenthesis, squared \red{\dfrac12}\cdot\!\!\!\left(\blue{-8}\right)^{\,\large3}  

2

1

​

⋅(−8)  

3

start color #df0030, start fraction, 1, divided by, 2, end fraction, end color #df0030, dot, left parenthesis, start color #6495ed, minus, 8, end color #6495ed, right parenthesis, cubed

We can see that every term is the product of the first term, \red{\dfrac12}  

2

1

​

start color #df0030, start fraction, 1, divided by, 2, end fraction, end color #df0030, and a power of the constant ratio, \blue{-8}−8start color #6495ed, minus, 8, end color #6495ed. Note that this power is always one less than the term number nnn. This is because the first term is the product of itself and plainly 111, which is like taking the constant ratio to the zeroth power.

Thus, we arrive at the following explicit formula (Note that \red{\dfrac12}  

2

1

​

start color #df0030, start fraction, 1, divided by, 2, end fraction, end color #df0030 is the first term and \blue{-8}−8start color #6495ed, minus, 8, end color #6495ed is the constant ratio):

a(n)=\red{\dfrac12}\cdot\left(\blue{-8}\right)^{\large{\,n-1}}a(n)=  

2

1

​

⋅(−8)  

n−1

a, left parenthesis, n, right parenthesis, equals, start color #df0030, start fraction, 1, divided by, 2, end fraction, end color #df0030, dot, left parenthesis, start color #6495ed, minus, 8, end color #6495ed, right parenthesis, start superscript, n, minus, 1, end superscript

Note that this solution strategy results in this formula; however, an equally correct solution can be written in other equivalent forms as well.