The perimeter of a rectangular playing field is 280 feet. The length of the field is 20 feet more than the width. What are the dimensions of the playing field? (You need to find the length and the width.)

Answer:

No

Step-by-step explanation:

4.5 IS NOT A WHOLE NUMBER!

Answer:

YES 4.5 IS NATURAL NUMBER

Step-by-step explanation:

Answer:

1620 peices

Step-by-step explanation:

9 ft = 9 x 12 (12 inches make 1 ft) = 108 x 15 rolls = 1620 inches

1620 inches divided by 4 inches = 405 peices

:)

Answer:

The inequality is x less than or equal to 6. So the amount of hours they can rent it is  6 hours or less.

Step-by-step explanation:

5.80x + 39 <_ 73.80 ----> subtract 39 on both side to get ---> 5.80x <_ 73.80 -----> then divide by 5.80 on both sides to get ---> x <_ 6.

Hope this helped!

Answer:

lolllll

Step-by-step explanation:

Answer:

12 inches

Step-by-step explanation:

The Pythagorean's theorem states that a triangle's hypotenuse's length is equal to a^2+b^2=c^2.

9 is a, b is the length, and c is the 15.

9^2+b^2=225

225-81=144

The square root of 144 is 12, therefore 12 inches is the length.

Answer:

A≈110.11

using formula 1/4[tex]\sqrt5(5+2\sqrt5)a2[/tex]

Step-by-step explanation:

Answer: Your answer will be [tex]\frac{-50}{5} =-10[/tex]

Step-by-step explanation: You will have

[tex](5)\frac{n}{5} =-10(5)[/tex] you will have to multiply 5 on both sides. So you will then simplify 5 and 5 to be left with n and multiply -10(5)=-50

So you get [tex]n=-50[/tex]

Answer:

Answer will be the 7/3<x>7/3

Answer: 6

Step-by-step explanation:

the biggest number that can go into both 24 and 18 evenly is 6. 24/6 is four, and 18/6 is three

Step-by-step explanation:

(A) squares are congruant .

(B) rectangle having equal length of same side is congruent.

(C)it is true because it have same side length are equal.

(D)it is true because same same side length are equal.

(E) it is truebecause same side length are equal.

The statements that are correct in the given question are: options A, B and C.

Quadrilaterals are figures or shape bounded by four straight sides. Thus each quadrilateral has its sum of interior angles to be [tex]360^{o}[/tex]. Examples include; rectangle, square, trapezium, rhombus, kite, parallelogram.

Considering the properties of each quadrilateral given in the question, it can be inferred that;

Therefore, the appropriate statements that are correct are: options A, B and C.

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Given:

The expression is -98 - 31.

To find:

The value of the given expression using the additive inverse.

Solution:

According to the  additive inverse, for any number a,

[tex]a+(-a)=0[/tex]

Using additive inverse, we have

[tex]98+(-98)=0[/tex]      ...(i)

[tex]31+(-31)=0[/tex]           ...(ii)

Now, using (i) and (ii), we get

[tex]98+(-98)+31+(-31)=0[/tex]

[tex](98+31)+(-98-31)=0[/tex]

[tex]129+(-98-31)=0[/tex]

Subtracting 129 from both sides, we get

[tex]-98-31=-129[/tex]

Therefore, the value of given expression is -129.

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:

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:

2.16

Step-by-step explanation:

c = -0.3

24c² =

plug in c

24(-0.3)²

solve for the exponent

24(0.09)

multiply

2.16

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]

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Answer:

4 out of the 6 possible numbers Josh wrote were greater than 480,000, while 2 out of the 6 possible numbers are less than 480,000

Step-by-step explanation:

The given parameters are;

The number with which Josh wrote the whole number = 8, 4, 7, 2, 9, and 0

The number in the thousands of the whole number that Josh wrote = 7

The number in the tens of the whole number that Josh wrote = 2

The value of the whole number that Josh wrote < 500,000

Therefore, we have;

The possible numbers in the hundreds of thousands place = 4, and 2

Which gives the following possible combinations;

The number of ways of selecting the first number = 1 way, the number is 4 to make the value less than 500,000

The number of ways of selecting the second number = 3 ways

The number of ways of selecting the third number = 1 way, the number is 7

The number of ways of selecting the fourth number = 2 ways as there are only two numbers left after filling the second place digit

The number of ways of selecting the fifth number = 1 way, the number is 2

The number of ways of selecting the sixth  number = 1 ways

The total number of possible number combinations = 3×2×1 = 6 possible numbers, which are;

487920 or 497820 or 407829 or 407928 or 487029 or 497028

Therefore, 2 out of the numbers are less than 480,000, the remaining 4 possible numbers are larger than 480,000.

Answer:

8x + 3

Step-by-step explanation:

In the expression -3x + 14 - 11 + 11x, we have two pairs of like terms that can be combined together:

-3x, 11x

14, -11

If we rewrite this expression with the like terms together, it would look like this:

-3x+11x + 14-11

To make the addition of -3x and 11x easier, we could switch their places in the expression to make it look like this:

11x-3x + 14-11

(Note: Even though we are switching the terms' places, the -3x still keeps its negative sign when you move it around.

11x-3x is 8x and 14-11 is 3.

Therefore, the expression -3x + 14 - 11 + 11x can be simplified by combining like terms into 8x + 3.

Answer:

Step-by-step explanation:

The mode in a given  set of data is the number that appears the most.

4 appears twice

5 appears twice

3 appears once

1 appears once

6 appears once

The value of X is 9.

Solve for x by simplifying both sides of the equation, then isolating the variable.

Answer:

yes they are

Step-by-step explanation:

Answer: my guess is 9

Answer:

there would be the 8 in the tens place and the 6 in the ones place

Step-by-step explanation:

you can see this in a place value chart

Answer:

Removing the outlier lowers the standard deviation by 1.09

Step-by-step explanation:

Find the standard deviation of the data WITHOUT the outlier first.

If you don't know how to do standard deviation, here are the steps:

1. Find the mean

2. Subtract the mean by the data set values

3. Square the differences

4. Add the squared numbers together and divide by the number of data set values

5. What you have now is the variance, and all you have to do now is square the variance and you have the standard deviation

Repeat this, but WITH the outlier this time.

Final step, just subtract the smaller number (answer you got from the data with the outlier) by the greater number (answer you got from the data without the outlier) and you have your answer: 1.09!

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:

Step-by-step explanation:

First equation:

          15 - x = - 3

           -x = - 3 - 15

          -x  =  -18  (now negative signs on b.s cut eachother)

            [x  =  18]

Second equation :

         15 - x = 3

          -x = 3 - 15

          - x = -12 ( now negative sign on b.s cut eachother)

            [x = 12]

so x = 18 and x = 12 are not same

Answer:

Fourth option

Step-by-step explanation:

We know that A and B are rational numbers because they can be written as the quotient of two integers. Therefore, that eliminates the first and third options because they state that (at least) one of the numbers is irrational but we know that both are rational, hence, they are incorrect. The second option is incorrect because A · B = [tex]\frac{mn}{pq}[/tex], not [tex]\frac{mp + nq}{pq}[/tex], therefore, the correct answer is the fourth option.

Answer:

1hr= 550 miles

Answer:

x + y are either 2 or 4

Step-by-step explanation:

conclusion x + y = 2 OR 4

Answer:

[tex]x=2i-2,\:x=-2i-2[/tex]

Step-by-step explanation:

[tex]x^2 + 4x = -8\\\\\mathrm{Solve\:by\:completing\:the\:square}\\\\\mathrm{Write\:equation\:in\:the\:form:\:\:}x^2+2ax+a^2=\left(x+a\right)^2\\\\\mathrm{Solve\:for\:}a,\:2ax=4x:\quad a=2\\\\\mathrm{Add\:}a^2=2^2\mathrm{\:to\:both\:sides}\\\\x^2+4x+2^2=-8+2^2\\\\\mathrm{Simplify\:}-8+2^2:\quad -4\\x^2+4x+2^2=-4\\\\\mathrm{Complete\:the\:square}\\\left(x+2\right)^2=-4\\\\\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}\\[/tex]

[tex]\mathrm{Solve\:}\:x+2=\sqrt{-4}:\quad x=2i-2\\\\\mathrm{Solve\:}\:x+2=-\sqrt{-4}:\quad x=-2i-2\\\\x=2i-2,\:x=-2i-2\\[/tex]