How do you find the surface area of a cube

Answer:

On a linear graph, the rate of change is called the slope of a line

Step-by-step explanation:

Hope this helps

The complete statement is, on a linear graph, the rate of change is called the slope of a line.

We know lines have various types of equations, the general type is

Ax + By + c = 0, and the equation of a line in slope-intercept form is

y = mx + b.

Where slope = m and b = y-intercept.

the slope is the rate of change of the y-axis with respect to the x-axis and the y-intercept is the (0,b) where the line intersects the y-axis at x = 0.

From the above concepts we now know that, on a linear graph, the rate of change is called the slope of a line.

It can also be defined as, (y₂ - y₁)/(x₂ - x₁).

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Answer:

14000

Step-by-step explanation:

The number in the thousands place is 4 so we look at the number after it which is 2:

Using the rounding rule, the number 2 is below 5 so we keep the 4 at 4:

So our answer is 14000.

Answer:

14000

Step-by-step explanation:

Since they are asking for us to round it to the nearest thousands, we have to look at the hundreds place to figure out whether we should round up or down. The number in the hundreds place is 2, and that means we round down to 14000 ("four or less, let it rest. five or more, raise the score").

Therefore, 14295.67 rounded to the nearest thousand is 14000.

I hope this helps! Have a lovely day!! :)

Answer:

it is c

Step-by-step explanation:

:)

The sine equation is  [tex]\rm y = 10sin(\frac{\pi}{15} t)+35[/tex]  an option (c) is correct.

It is given that the blades of a windmill turn on an axis that is 35 feet above the ground. the blades are 10 feet long and complete two rotations every minute.

It is required to model the sine equation.

Trigonometry is a branch of mathematics that deals with the relationship between sides and angles of a right-angle triangle.

We know the standard equation for the sine equation:

[tex]\rm y = Asin(wx)+h[/tex]

Where [tex]\rm T=\frac{2\pi}{w}[/tex]

[tex]\rm w = \frac{2\pi}{30}[/tex]   ( In one minute, they complete two rotations.)

[tex]\rm w = \frac{\pi}{15}[/tex]  ( T = 30)

At time t=0, the blade is horizontal. Hence, at t = 0, h = 35

And A = 10

Put all the values in the sine equation:

[tex]\rm y = 10sin(\frac{\pi}{15} t)+35[/tex]

Thus, the sine equation is  [tex]\rm y = 10sin(\frac{\pi}{15} t)+35[/tex] an option (c) is correct.

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The number of rotations is reduced to 542.501 when the diameter of wheel is increased by 20 %.

The quantity of rotations (n), no unit, done by a wheel is equal by total traveled distance (s), in inches, divided by its circumference (p), in inches:

[tex]n = \frac{s}{p}[/tex]  

[tex]n = \frac{s}{\pi\cdot D}[/tex]   (1)

Where D is the diameter of the wheel, in inches.

If we know that n = 651 and D = 31 in, then the travelled distance is:

s = (651) · π · (31 in)   (2)

s ≈ 63400.481 in

The travelled distance of the 31-in diameter wheel with 651 rotations is approximately 63400.481 inches.

By (2) we understand that the travelled distance is directly proportional to the diameter. Hence, if the diameter of the wheel is increased by 20 %, then we must multiply the diameter by 1.2 and divide the travelled distance by this result:

s ≈ 1.2 · π · (31 in)

s ≈ 116.867 in

[tex]n = \frac{63400.481\,in}{116.867\,in}[/tex]

[tex]n = 542.501[/tex]

The number of rotations is reduced to 542.501 when the diameter of wheel is increased by 20 %. [tex]\blacksquare[/tex]

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Answer:

6

Step-by-step explanation:

, , , , ,, , , , , , , , ,sssss

Answer:

the x is 538

Step-by-step explanation:

sorry if im not right

Answer:

<DFE

Step-by-step explanation:

Verticle angles are when they are directly apart and equal. So,

Answer:

-4

Step-by-step explanation:

<symbol means squared or increase and so does ( )

Step-by-step explanation:

1) C

2) B

5) C

6)D

7) D

8) B

9) Figure not clear

10) C

11) D

12) B

13) D

14) A

15) D

P.S try to do ur own tests

The linear function f(x) = ax + b has a constant rate

The conclusion from the each interval is that the function values changes at a constant rate

The linear function is given as:

f(x) = ax + b

The interval is given as:

[-3,3] and [9,15]

Since the function is a linear function, the conclusion from the each interval is that the function values changes at a constant rate

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Answer:

1. Colored men

2. Not really sure, but maybe on a wall ??

3. To notify colored men what the president proclaimed

Step-by-step explanation:

Answer:

What are you trying to do you have not Explained to us

Step-by-step explanation:

Answer:

x=20

Step-by-step explanation:

(1/2)*10.5*(8.8+x)=151.2

8.8+x=151.2*2/10.5

8.8+x=28.8

x=28.8-8.8

x=20

Answer: 20 cm

Step-by-step explanation:

area of trapeziod = ( (base + base) / 2) x height

151.2 = ( (8.8 + base) / 2 ) x 10.5

14.4 = ( (8.8 + base) / 2)

14.4 x 2 = 8.8 + base

28.8 - 8.8 = base

base = 20 cm

Answer:

Mean = 8 gerbils

MAD = 4.29 gerbils

Step-by-step explanation:

Find the mean:

Mean= 0+3+6+9+11+13+14 ÷ 7

Mean = 56 ÷ 7

Mean = 8 gerbils

you have to find the distance of each value of gerbil from the mean

8,5,2,1,3,5,6

MAD = 9+6+3+0+2+4+5 ÷ 7

MAD = 30 ÷ 7

MAD = 4.29  gerbils

Answer:

The deck's width is 9 feet
The deck's area is 18 feet

Correct po yan

Hope it helps po

Answer:

Deck Width: 6 Feet

Deck Area: 72 Feet Squared

Step-by-step explanation:

[l = length, w = width, f = feet, p = perimeter, a = area]

A = l x w

P = 2l + 2w [l = 12f, p = 36f]

P = 2l + 2w

=> 36f = 2 (12f) + 2w

=> 36f = 24f + 2w

=> 36f - 24f = 2w

=> 12f = 2w

=> 2w/2 = 12f/2

=> W = 6f

Therefore: Width = 6 Feet

A = l × w

A = 12f × 6f

A = 72f^2

Therefore: Area = 72 Feet Squared

Answer:

69

Step-by-step explanation:

Find the area of the rectangle. 15 x 3 = 75.

Find the area of the cut-out triangle.
1/2bh

1/2(3(4))

1/2(12)

6

75 - 6 = 69.

Answer: C (0,5)

Step-by-step explanation:

So firstly you're going to try to solve the inequality, so what I did first is used the second equation because if I divide both sides by 3, I'm able to get the y by itself:

6x+3y>/=15, 6/3=2, 3/3=1, 15/3= 5

So now its: 2x+y>/=5

now I should get the y by itself by subtracting 2x from both sides to get

y>/=15-2x, now if we wanted to plug it in to the top equation, we could make it as, y=15-2x because the it is GREATER THAN OR EQUAL TO.

so to plug it in to the first equation we get:

6x-2(15-2x)<10,

x<4, to find y we just plug it in to any equation now we have x, so I did

6(3)-2y<10

18-2y<10

-18        -18

-2y<10

divide both sides by -2 (flips the equation sign)

y>4,

So take consider both X and Y values, C would be correct because X is less than four and 0 is less than four, and 5 is greater than four.

Answer

Force a perfect square trinomial on the left side. Take the square root of both sides and solve for

3√  3, − 5 − 3 √ 3

Step-by-step explanation:

The y-intercept will be where it crosses the y-axis. The y-axis is the line that goes up and down also know a vertically.

Answer:

median = 55.5

Step-by-step explanation:

Here we have 10 numbers which is an even number

then the median for the data set is the average of the two middle numbers

firstly ,we have to put the number of the set in order from least to greatest like this :

3 < 12 < 38 < 38 < 46 < 65 < 71 < 78 < 87 < 99

————————                     ————————-

As you can see the middle two numbers are : 46 and 65

Now, we calculate their average :

[tex]\large \text Median = \frac{46+65}{2}= 55.5[/tex]            

keeping in mind that Mt Everest is 8000 meters, or namely 8 Kilometers, Check the picture below.

Answer:

50.46???

Step-by-step explanation: im bad at  this so if it help ima be proud nut uhhh i tryed ...

Step-by-step explanation:

It is given that, the sum of two positive integers is 31 and the sum of the squares of these numbers is 625 and we are to find the smaller of the numbers.

So, let the two positive integers be x and y.

Therefore,

[tex] \\ {\longrightarrow \pmb{\sf {\qquad x + y = 31 \: \: ........ \: (i) }}} \\ \\[/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x {}^{2} + y {}^{2} = 625 \: ...... \: (ii)}}} \\ \\[/tex]

Now, From the first equation we have,

[tex]\\ {\longrightarrow \pmb{\sf {\qquad x + y = 31 }}} \\ \\ [/tex]

[tex] {\longrightarrow \pmb{\sf {\qquad y = 31 - x \: ...... \: (iii)}}} \\ \\ [/tex]

Now, substituting the value of y in equation (ii) we get :

[tex] \\ {\longrightarrow \pmb{\sf {\qquad x {}^{2} + (31 - x) {}^{2} = 625}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x {}^{2} + (31 {}^{2} - 2.x.31 + x{}^{2} ) = 625}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x {}^{2} + (961 - 62x + x{}^{2} ) = 625}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x {}^{2} + 961 - 62x + x{}^{2} - 625 = 0}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad 2 x {}^{2} + 336 - 62x = 0}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad 2 x {}^{2} - 62x + 336 = 0}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad 2( x {}^{2} - 31 + 168) = 0}}} \\ \\[/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad \frac{2}{2} ( x {}^{2} - 31 + 168) = \frac{0}{2} }}} \\ \\[/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x {}^{2} - 31 + 168 = 0}}} \\ \\[/tex]

Now using the quadratic formula :

[tex]{\longrightarrow \pmb{\sf {\qquad x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a} }}} \\ \\[/tex]

Where,

[tex] \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x = \frac{ - (- 31) \pm \sqrt{ {31}^{2} - 4(1)(168)} }{2(1)} }}} \\ \\[/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x = \frac{ 31 \pm \sqrt{ 961 -672} }{2(1)} }}} \\ \\[/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x = \frac{ 31 \pm \sqrt{ 289} }{2(1)} }}} \\ \\[/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x = \frac{ 31 \pm 17 }{2(1)} }}} \\ \\[/tex]

Now, we have two equations,

[tex]{\longrightarrow \pmb{\sf {\qquad x = \frac{ 31 + 17 }{2} \: ... .....\: (iv)}}} \\ \\[/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x = \frac{ 31 - 17 }{2} ... .....\: (v)}}} \\ \\[/tex]

So, Equation (iv) :

[tex] \\ {\longrightarrow \pmb{\sf {\qquad x = \frac{ 31 + 17 }{2} }}} \\ \\ [/tex]

[tex]\\ {\longrightarrow \pmb{\sf {\qquad x = \frac{ 48 }{2} }}} \\ \\ [/tex]

[tex]\\ {\longrightarrow \pmb{\sf {\qquad x = 24 }}} \\ \\ [/tex]

Now, Equation (v) :

[tex] \\ {\longrightarrow \pmb{\sf {\qquad x = \frac{ 31 - 17 }{2}}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x = \frac{ 14 }{2}}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x = \: 7}}}\\ \\ [/tex]

Now, we are to find the value of y.

Substituting the value of x (24) in equation (iii) :

[tex] \\ {\longrightarrow \pmb{\sf {\qquad y = 31 - x \:}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad y = 31 - 24 \:}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad y = 7 \:}}} \\ \\ [/tex]

Again, Substituting the value of x (7) in equation (iii) :

[tex]\\ {\longrightarrow \pmb{\sf {\qquad y = 31 - x \:}}} \\ \\ [/tex]

[tex] {\longrightarrow \pmb{\sf {\qquad y = 31 - 7 \:}}} \\ \\ [/tex]

[tex]\\ {\longrightarrow \pmb{\sf {\qquad y = 24 \:}}} \\ \\ [/tex]

Therefore,

[tex] \\ [/tex]

So, The smaller of the numbers is 7 .

The sum of two positive integers is 31. If the sum of the squares of these numbers is 625, find the smaller of the numbers.

This question belongs to quadratic equations so we have to find the answer by making equation and solving it .

According to question , sum of two positive integers is 31 . So ,

According to question , sum of square

of these numbers is 625 . So ,

From equation 1 ( x + y = 31 ) , Value

of x :

Now , putting value of x in eq. 2 :

Solving it by using middle term

splitting :

First number ,

Second Number ,

According to question ,

Sum of numbers is equal to 31 :

Sum of squares of these numbers is equal to 625 :

As shown, to obtain the blue quilt block pattern, we have to rotate the parallelogram.

A parallelogram is a quadrilateral consisting of pairs of parallel sides.

To obtain the blue quilt block pattern, we have to rotate the parallelogram. This rotation allows the parallelogram to take on a new orientation, and when it is positioned correctly, it creates the blue quilt block pattern. The process of rotating the parallelogram may seem simple, but it is a critical step in achieving the desired quilt design. It requires attention to detail and precision to ensure that the parallelogram is correctly rotated, and the resulting blue quilt block pattern is aligned correctly.

Moreover, the correct rotation of the parallelogram can significantly impact the overall appearance of the quilt, and even a slight misalignment can make a noticeable difference. Therefore, it is important to carefully follow the instructions to rotate the parallelogram to achieve the intended blue quilt block pattern.

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Answer:

for number 6 the width is 24 and the blank after the length of 9 is 12 and after 48 width it is 60. your welcome.

The linear relationship on the table has a constant slope

The linear relationship is (c) y is 3 times x

The table of values is given as:

x l y

3 l 9

0 l 0

For both x and y values on the table, we have:

y = 3x

i.e.

3 * 3 = 9 and 3 * 0 = 0

This means that the linear relationship is (c) y is 3 times x

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Answer:

See below ↓↓↓

Step-by-step explanation:

Part A

Total length = Side 1 + Side 2

Part B

Length of 3rd side = Perimeter - [Side 1 + Side 2]

Part C

Yes, as the resulting polynomial has a finite value we can conclude that polynomials are closed under addition and subtraction.

Part A: The total length of the two sides of the triangle is [tex]5x^2+7x-4[/tex]. This is obtained by adding the given two sides.

Part B: The length of the third side is [tex]5x^3-x^2-8x+1[/tex]. This is obtained by subtracting the sum of two sides from the perimeter of the triangle.

Part C: Yes, the Part A and Part B answers show that the polynomials are closed under addition and subtraction. This is because the expressions have like terms.

Given that, a triangle has two sides of length [tex]3x^2-2x-1[/tex] and [tex]9x+2x^2-3[/tex]

The perimeter of the triangle is [tex]5x^3+4x^2-x-3[/tex]

Part A:

To find the total length of two sides, adding the two side

⇒ [tex](3x^2-2x-1)+9x+2x^2-3\\[/tex]

⇒ [tex]5x^2+7x-4[/tex]

(adding the like terms w.r.t their sign)

Part B:

To find the length of the third side, subtract the sum of two sides from the perimeter.

⇒ [tex](5x^3+4x^2-x-3)-(5x^2+7x-4)[/tex]

⇒ [tex]5x^3-x^2-8x+1[/tex]

Part C:

From Part A and Part B, it is proved that the polynomials undergo addition and subtraction. Hence, it is justified.

Therefore, Part A, Part B, and Part C were obtained.

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