The difference between twelve and 2/3 of a number is 18. What is the number?

Answer:

Step-by-step explanation:

x^2 + x + 1 - 2x + 5

x^2 - x + 6

Answer:

your answer will be 149/month

Step-by-step explanation:

you wanna ask urself how many months are in 5 years

The answer is: 60 months

then, you wanna divide the cost of the car over 60 months

So:

[tex]$8950/60=149.166[/tex]

round that to the nearest dollar AKA the ones

it'll be 149

and don't forget the unit of time

which is month

$149/month

Answer:

it should be 3/2  !

Answer:

Part A: The proposed route does not meet requirement because it is longer than the maximum required length of 3 miles

Part B: For the total distance is as close to 3 miles as possible, the start point of the parade should be at the point on Broadway with coordinates (9.941, 4.970)

Part C: The coordinates of the cameras stationed half way down each road are;

For central avenue; (4, 2)

For Broadway; (7.97, 2.49)

Step-by-step explanation:

Part A: The length of the given route can be found using the equation for the distance, l, between coordinate points as follows;

[tex]l = \sqrt{\left (y_{2}-y_{1} \right )^{2}+\left (x_{2}-x_{1} \right )^{2}}[/tex]

Where for the Broadway potion of the parade route, we have;

(x₁, y₁) = (12, 3)

(x₂, y₂) = (6, 0)

[tex]l_1 = \sqrt{\left (0 -3\right )^{2}+\left (6-12 \right )^{2}} = 3 \cdot \sqrt{5}[/tex]

For the Central Avenue potion of the parade route, we have;

(x₁, y₁) = (6, 0)

(x₂, y₂) = (2, 4)

[tex]l_2 = \sqrt{\left (4 -0\right )^{2}+\left (2-6 \right )^{2}} = 4 \cdot \sqrt{2}[/tex]

Therefore, the total length of the parade route =-3·√5 + 4·√2 = 12.265 unit

The scale of the drawing is 1 unit = 0.25 miles

Therefore;

The actual length of the initial parade =0.25×12.265 unit = 3.09 miles

The proposed route does not meet requirement because it is longer than the maximum required length of 3 miles

Part B:

For an actual length of 3 miles, the length on the scale drawing should be given as follows;

1 unit = 0.25 miles

0.25 miles = 1 unit

1 mile =  1 unit/(0.25) = 4 units

3 miles = 3 × 4 units = 12 units

With the same end point and route, we have;

[tex]l_1 = \sqrt{\left (0 -y\right )^{2}+\left (6-x \right )^{2}} = 12 - 4 \cdot \sqrt{2}[/tex]

y² + (6 - x)² = 176 - 96·√2

y² = 176 - 96·√2 - (6 - x)²............(1)

Also, the gradient of l₁ = (3 - 0)/(12 - 6) = 1/2

Which gives;

y/x = 1/2

y = x/2 ..............................(2)

Equating equation (1) to (2) gives;

176 - 96·√2 - (6 - x)² = (x/2)²

176 - 96·√2 - (6 - x)² - (x/2)²= 0

176 - 96·√2 - (1.25·x²- 12·x+36) = 0

Solving using a graphing calculator, gives;

(x - 9.941)(x + 0.341) = 0

Therefore;

x ≈ 9.941 or x = -0.341

Since l₁ is required to be 12 - 4·√2, we have and positive, we have;

x ≈ 9.941 and y = x/2 ≈ 9.941/2 = 4.97

Therefore, the start point of the parade should be the point (9.941, 4.970) on Broadway so that the total distance is as close to 3 miles as possible

Part C: The coordinates of the cameras stationed half way down each road are;

For central avenue;

Camera location = ((6 + 2)/2, (4 + 0)/2) = (4, 2)

For Broadway;

Camera location = ((6 + 9.941)/2, (0 + 4.970)/2) = (7.97, 2.49).

This exercise is necessary to use the given map information and then make the distance between points, in this way we find that:

A)The proposed route does not meet requirement because it is longer than the maximum required length of 3 miles.

B) Broadway and Cedar Street

C) [tex]Central \ Avenue(4, 2)Broadway (7.97, 2.49)[/tex]

So from the distance between points and the informed map we have:

A) The equation for the distance, as follows:

[tex]l=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}[/tex]

Where for the Broadway potion of the parade route, we have:

[tex](x_1,y_1)=(12,3)\\(x_2,y_2)=(6,0)[/tex]

Applying the values ​​in the formula given above, we have:

[tex]l_1=\sqrt{(0-3)^2+(6-12)^2}=3\sqrt{5}[/tex]

For the Central Avenue potion of the parade route, we have:

[tex](x_1, y_1) = (6, 0)(x_2, y_2) = (2, 4)[/tex]

Applying the values ​​in the formula given above, we have:

[tex]l_2=\sqrt{(4-0)^2+(2-6)^2}=4\sqrt{2}[/tex]

Therefore, the total length of the parade route:

[tex]3\sqrt{5} + 4\sqrt{2} = 12.265 \\0.25*12.265 unit = 3.09\ miles[/tex]

B) The two streets that are the same distance apart are the streets: Broadway and Cedar Street.

C) The coordinates of the cameras stationed half way down each road are:

For Central Avenue:

[tex]Camera \ location = ((6 + 2)/2, (4 + 0)/2) = (4, 2)[/tex]

For Broadway:

[tex]Camera\ location = ((6 + 9.941)/2, (0 + 4.970)/2) = (7.97, 2.49)[/tex]

See more about distance at brainly.com/question/989117

Answer:

3 and 1/4

Step-by-step explanation:

Answer:

3.140

Step-by-step explanation:

Answer:

3.14

Step-by-step explanation:

-----

Answer:

1hr= 550 miles

0.35p

Because the price also indicates percentage

Answer:

p - 0.35, so C.

It's not A because that's the regular price times the percentage off.

y = 1

8-2y - 8 = 4y - 6

-2y = 4y - 6

-6y = -6

y = 1

Answer:

Step-by-step explanation:

8 - 2y - 8 = 4y - 6

-2y = 4y - 6

6y = -6

y = -1

Answer:

31 cm

Step-by-step explanation:

The complete question is attached.

A triangle is a polygon with three sides (three edges and three vertices). There are different types of triangles which are equilateral triangles, right triangles, scalene triangles, obtuse triangles, acute triangles, and isosceles triangles.

An equilateral triangle is a triangle in which all its sides are equal and all its angles are equal to 60°.

From the triangle, AB = BC = AC = 9.5 cm

Since the triangle is enlarged 3.25 times, hence:

new length of BC = 9.25 cm × 3.25 = 30.875

New length of BC = 31 cm to nearest cm

Answer:

Step-by-step explanation:

[tex]\frac{3}{2} + \frac{2m}{9} = \frac{37}{18}[/tex]

Multiply through by 18 to eliminate the fraction

That's

[tex]18 \times \frac{3}{2} + 18 \times \frac{2m}{9} = 18 \times \frac{37}{18} \\ 27 + 4m = 37[/tex]

Subtract 27 from both sides

That's

[tex]27 - 27 + 4m = 37 - 27 \\ 4m = 10[/tex]

Divide both sides by 4

That's

[tex] \frac{4m}{4} = \frac{10}{4} [/tex]

We have the final answer as

[tex]m = \frac{5}{2} [/tex]

Hope this helps you

[tex]\Huge{\boxed{\sf{\red{m = \frac{5}{2}}}}}[/tex]

[tex]\displaystyle{\sf{ \frac{3}{2} + \frac{2m}{9} = \frac{37}{18}}} [/tex]

| ∵ Transposing 3/2 to R.H.S

[tex]\implies[/tex] [tex]\displaystyle{\sf{ \frac{2m}{9} = \frac{37}{18} - \frac{3}{2}}} [/tex]

[tex]\implies[/tex] [tex]\displaystyle{\sf{ \frac{2m}{9} = \frac{37-27}{18}}} [/tex]

[tex]\implies[/tex] [tex]\displaystyle{\sf{ \frac{2m}{9} = \frac{10}{18} = \frac{5}{9}}} [/tex]

[tex]\implies[/tex] [tex]\displaystyle{\sf{ 2m = \frac{5}{9} \times 9}} [/tex]

[tex]\implies[/tex] [tex]\displaystyle{\sf{ 2m = 5}} [/tex]

[tex]\longrightarrow[/tex] [tex]\large{\boxed{\sf{\red{ m = \frac{5}{2}}}}}[/tex]

Answer:

yes they are

Step-by-step explanation:

Answer:

Her door should be 3cm on scale

The actual dimension of the kitchen table is

[tex]Length = 1.6m[/tex]

[tex]Width = 1m[/tex]

Step-by-step explanation:

Given

[tex]Scale= 1 : 40[/tex]

[tex]Kitchen\ Dimension= 6m\ by\ 2m[/tex]

[tex]Door\ Width = 1.2m[/tex]

Solving (a): The width of the door on the scale

Represent the scale width with x

So, we have:

[tex]1 : 40[/tex]

and

[tex]x : 1.2[/tex]

Equate both ratios

[tex]1 : 40 = x : 1.2[/tex]

Represent as fraction

[tex]\frac{1}{40} = \frac{x}{1.2}[/tex]

Solve for x

[tex]x = \frac{1.2}{40}[/tex]

[tex]x = 0.03m[/tex]

[tex]x = 3cm[/tex]

Hence, her door should be 3cm on scale

Solving (b): Actual Dimension of the kitchen table

To solve this, we simply multiply the scale dimensions by 40

[tex]Length = 4cm * 40[/tex]

[tex]Length = 160cm[/tex]

[tex]Length = 1.6m[/tex]

[tex]Width = 2.5cm * 40[/tex]

[tex]Width = 100cm[/tex]

[tex]Width = 1m[/tex]

Hence:

The actual dimension is

[tex]Length = 1.6m[/tex]

[tex]Width = 1m[/tex]

Answer:

$8

Step-by-step explanation:

If you forgot to add something I'll answer that

Answer:

f(x) = 2(x + 3)²

Step-by-step explanation:

Parent Graph: f(x) = a(bx - h)² + k

a is vertical shrink/stretch

b is horizontal shrink/stretch

h is horizontal movement left/right

k is vertical movement up/down

h, k is vertex

Step 1: Define variables

h = -3

k = 0

a = 2

Step 2: Write quadratic

f(x) = 2(x + 3)²

Answer:

5.70%

Step-by-step explanation:

It takes the time to perform on Ciara's song = 15 seconds

Total time given as 4 minutes 23 seconds ≈ (4×60) seconds + 23 seconds

                                                                      = 240 seconds + 23 seconds

                                                                      = 263 seconds

Percent of time taken to perform of the full time = [tex]\frac{\text{time taken to perform}}{\text{Total time}}\times 100[/tex]

                                                                                = [tex]\frac{15}{263}\times 100[/tex]

                                                                                = 5.70 %

Therefore, percent of the time taken for the performance on the full song will be 5.70%  

subtract 7 from -5

Step-by-step explanation:

Answer:

Step-by-step explanation:

I just took the test and it said that Subtract 7 from –5 was right

Answer:

y = 15

Step-by-step explanation:

if y+15=30 we know that we need a total of 30, since we already have 15 y will also be 15

Answer:

y+15=30

y+15-15=30-15

y=15

Step-by-step explanation:

Answer:

We know that in a parallelogram two opposite angles are equal -

[tex] \\ \implies \sf \: 2x + 2y = 360 {}^{ \circ} \\ \\ \\ \implies \sf \: x + y = 180 {}^{ \circ} \qquad \quad \: (i) \\ [/tex]

Given -

[tex] \\ \implies \sf \: x = \frac{2}{3} y \\ \\ \\ \implies \sf \frac{x}{2} = \frac{y}{3} = k \\ \\ \\ \qquad \sf \small \underline{ x = 2k \: \: \& \: \: y = 3k} \\ [/tex]

Now, by using equation (1) :

[tex] \\ \implies \sf \: 2k + 3k = 180 \\ \\ \\ \implies \sf \: 5k = 180 \\ \\ \\ \implies \sf \: k = \frac{180}{5} \\ \\ \\ \large{ \boxed{ \sf{k = {36}^{ \circ} }}} \\ [/tex]

Now, by putting the value of k in x and y.

Therefore, the right option and smallest angle is b) 72°.

Perpendicular lines have opposite reciprocal slopes.

y = 1/4x - 10 is our original equation.

If we flip the slope and change the sign, we get -4

The answer is -4

Answer:

13x + 3

Step-by-step explanation:

9x + 4x + 8 -5

13x + 3

Just multiply (x + 2) by 4

and add simply.

I hope now you got it.

512x³, I think, because 2x cubed (2x³) is 8x, and 8x cubed again ([8x]³) is 512x³.

Hope I helped!!!

Answer:

-2

Step-by-step explanation:

3x - 8

Let x=2

3*2 -8

6-8

-2

Answer:

6-8 = -2

Step-by-step explanation:

3 x if x equals 2 then 3 times 2 is 6. 6 minus 8 is -2!

Hope this helps! If it does, please mark me brainliest because it will help me. Thank you so much! ;) :)

Answer:

k= 7

Step-by-step explanation:

if m is the midpoint of fm, and fm= 2k-5. and fg=18. then you can do: 2k-5+2k-5=18 and solve for x

The measure of FM is 9 and MG is 9.

Given,

M is the midpoint of FG.

FM = 2k - 5 and FG = 18.

We need to find the measure of MG.

If there is a midpoint between two points the midpoint will divide the distance between the two points into two equal halves.

AB is the distance between two points A and B

If C is the midpoint then,

AC = CB = AB / 2

We have,

FG is a line.

M is the midpoint of this line FG.

Now,

F______M______G

Since M is the midpoint we have,

We need to get,

FM = MG = FG/2

FG = FM + MG

We need to find the k value.

FM = FG/2

2k - 5 = 18/2

Multiplying 2 on both sides

4k - 10 = 18

Adding 10 on both sides

4k - 10 + 10 = 18 + 10

4k = 28

Dividing both sides by 4

k = 28/4 = 7

k = 7.

So,

FM = 2k - 5 = 2 x 7 - 5 = 14 - 5 = 9

Since M is the midpoint

FM = MG

We have,

MG = 9

We can see that,

FG = FM + MG

18 = 9 + 9

18 = 18

Thus the measure of FM is 9 and MG is 9.

Learn more about finding measures of a line between two points here:

https://brainly.com/question/617690

#SPJ2

Answer:

x-4

Step-by-step explanation:

6(x+1)-5(x+2)

=6x+6-5x-10

=x-4

Answer:

[tex] \boxed{ \bold{ \huge{ \boxed{ \sf{x - 9}}}}}[/tex]

Step-by-step explanation:

[tex] \star \: \sf{6( x + 1) - 5(x + 2)}[/tex]

Distribute 6 through the parentheses

Similarly, distribute -5 through the parentheses

[tex] \dashrightarrow{ \sf{6x + 6 - 5x - 10}}[/tex]

Collect like terms

Like terms are those which have the same base

[tex] \dashrightarrow{ \sf{6x - 5x + 6 - 10}}[/tex]

[tex] \dashrightarrow{ \sf{x + 6 - 10}}[/tex]

The negative and positive integers are always subtracted but posses the sign of the bigger integer

[tex] \dashrightarrow{ \boxed{ \sf{x - 4}}}[/tex]

Hope I helped!

Best regards! :D