31.68 is 45% of what number?

The approximate length is 12 centimeters, of one side of the square which is the correct answer would be an option (C).

A square is a closed two-dimensional object with four equal-length sides and all interior angles of exactly 90 degrees.

The diagonal of the square is the distance between two vertices on opposing sides. A square has two diagonals of equal length.

We have been given that Justine draws a square with an area of 150 square centimeters.

The area of the square = (length of the side)²

150 square centimeters = (length of the side)²

Length of the side = √(150)

Length of the side = 12.24744

Rounded the nearest hundredth and we get

Length of the side = 12.24 centimeters

Therefore, the approximate length is 12 centimeters, of one side of the square.

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 1.1      Factoring:  x3-216 

Theory:

 A difference of two perfect cubes,  a3 - b3 can be factored into
              (a-b) • (a2 +ab +b2)

Proof :  (a-b)•(a2+ab+b2) =
            a3+a2b+ab2-ba2-b2a-b3 =
            a3+(a2b-ba2)+(ab2-b2a)-b3 =
            a3+0+0+b3 =
            a3+b3

Check:  216  is the cube of   6 
Check:  x3 is the cube of   x1

The solution to the equation x³ + 216 = 0 is x = -6, by subtracting 216 from both sides and taking the cube root.

To solve the equation x³ + 216 = 0, you must first isolate the variable on one side of the equation. To do this, we can subtract 216 from both sides, giving us the equation x³ = -216.

Next, to solve for x, we need to take the cube root of -216. The cube root of -216 is -6, so x = -6 is the root of the equation.

In conclusion, the solution to the equation x³ + 216 = 0 is x = -6.

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The subject of this question is Mathematics, specifically involving basic algebra and arithmetic. Students are required to apply the math principles given in the example boxes throughout the chapter.

The subject of this question is Mathematics. It involves basic algebra and arithmetic to solve mathematical problems and apply the principles given in the example boxes throughout the chapter. The question likely requires students to use their math skills to solve a specific problem or answer a mathematical question.

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Answer:

The value of x is 135.

Step-by-step explanation:

Given : Expression [tex]\frac{4}{9}x+\frac{1}{5}x=87[/tex]

To find : What is the value of x?

Solution :

Expression [tex]\frac{4}{9}x+\frac{1}{5}x=87[/tex]

Take LCM,

[tex]\frac{4x\times 5+x\times 9}{9\times 5}=87[/tex]

[tex]\frac{20x+9x}{45}=87[/tex]

[tex]\frac{29x}{45}=87[/tex]

[tex]x=\frac{87\times 45}{29}[/tex]

[tex]x=135[/tex]

Therefore, The value of x is 135.

Part a) Sam got a new job with a $7.00 per hour raise. He worked for 5 hours and got paid $85.00. How much did he make before the raise?

Let

x--------> amount that Sam earned before the raise

we know that

[tex] (x+7) *5 =85 [/tex]

Solve for x

[tex] 5x+35 =85 [/tex]

[tex] 5x =85-35 [/tex]

[tex] 5x =50 [/tex]

[tex] x=10\ dollars\ per\ hour [/tex]

therefore

the answer Part a) is the option

A. $10.00 per hour

Part b) Julia knows 19 Latin words. Each day, she learns 6 new words. How many days until she knows 145?

Let

x-------> the number of days

y-------> the total words Julia Knows

we know that

[tex] y=19+6x [/tex] ------> equation [tex] 1 [/tex]

For y=[tex] 145 [/tex]

Find the value of x

Substitute in the equation [tex] 1 [/tex] the value of y

[tex] 145=19+6x [/tex]

[tex] 6x=145-19 [/tex]

[tex] 6x=126 [/tex]

[tex] x=21\ days [/tex]

therefore

the answer Part b) is the option

B. 21

Part c) Claire has a part-time job walking dogs Monday through Friday. Business has really increased. She is walking 3 more dogs per day than she did last week. If she walked a total of 35 dogs this week, how many dogs did she walk per day last week?

Let

x--------> the number of dogs that Claire walked last week per day

we know that

[tex] x=\frac{35}{5} -3\\\\ x=7-3\\\\x=4\ dogs\ per\ day [/tex]

therefore

the answer Part c) is the option

C. 4

Part d) Kayla needs to have $560.00 to buy a new laptop. She already has $200.00 in the bank. Each month she will add $45.00 to her bank account. How many more months until she has enough money to buy her laptop?

Let

x--------> the number of months

y-------> amount in the bank account in dollars

we know that

[tex] y=45x+200 [/tex]

Find the value of x for y=[tex] 560 [/tex]

[tex] 560=45x+200 [/tex]

[tex] 45x=560-200 [/tex]

[tex] 45x=360 [/tex]

[tex] x=8\ days [/tex]

therefore

the answer part d) is the option

D. 8

Part e) Mark bought a notebook for $1.45 and some pencils that cost $0.42 each. If total cost was $6.91, how many pencils did he buy?

Let

x------> the number of pencil

we know that

[tex] 6.91= 1.45+0.42x\\\\ 0.42x=6.91-1.45\\\\ 0.42x= 5.46\\\\ x= 13\ pencils [/tex]

therefore

the answer part e) is the option

B. 13

-12 / 23 = -0.52

 so none of those are the correct answer.

if the 23 is suppose to be the fraction 2/3

 then you have -12  / 2/3 which =  -12 * 3/2 = -36/2 = -18

Answer:

hello your answer is -36  

Step-by-step explanation:

1. 12 = 12/1 so we keep the first number then2. we change the division sign to times 3. flip the second number 2/3 is turned to 3/2 then  multiply  so 12/1 x 3/2 is 36/2  hope this helps have a nice day !

Answer:

[tex]70\leq x\leq 80[/tex]

Step-by-step explanation:

Let x represent the test scores of 60% of his students.

We have been given that Mr. Venn knows that 60% of his students will score no more than 5 points above or below the class average,75% on the test.

We will use tolerance formula to solve our given problem.

[tex]|\text{Actual}-\text{Ideal}|\leq\text{Tolerance}[/tex]

[tex]|x-75|\leq5[/tex]

Using rule [tex]|u|\leq a[/tex], then [tex]-a\leq u\leq a[/tex], we will get:

[tex]-5\leq x-75\leq5[/tex]

[tex]-5+75\leq x-75+75\leq5+75[/tex]

[tex]70\leq x\leq 80[/tex]

Therefore, the inequality [tex]70\leq x\leq 80[/tex] represents the test scores of 60% of Mr. Venn's students.

The Greatest Common Factor of 22ab2c2 and 40a3c2 is 2ac2, found by identifying the highest power of each variable and number that divides the coefficients 22 and 40.

When finding the Greatest Common Factor (GCF) we need to look for the highest power of each variable and number that appears in all the terms. In the terms 22ab2c2 and 40a3c2, the GCF will consists of the highest power of each variable found in both terms, and the highest number that divides both 22 and 40.

To find the GCF you need to divide both 22 and 40 by common factor. The highest number that divides 22 and 40 is 2. For the variables, the lowest power of 'a' and 'c' in both terms is a and c2, respectively. 'b' is not in both terms, so we omit it. So, the GCF of 22ab2c2 and 40a3c2 is 2ac2.

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Answer:

what even is this???????????????

Answer: A

Step-by-step explanation:

Answer:

1800

Step-by-step explanation:

2x-3=7

add 3 to both sides

2x = 10

divide both sides by 2

x = 10/2

x = 5

 since the equation is between 2 lines, which means absolute value x can be both positive and negative,

 so the solution set is -5 and 5

By rigid transformations, the image of the parabola y = x² is equal to parabola y = x² + 4.

In this problem, we need to determine the image of a parabola by means of a rigid transformation known as vertical translation:

y = f(x) → y = f(x) + k, where k represents a translation upwards and f(x) is the original function.

If we know that f(x) = x² and k = 4, then the image of the function is:

f(x) = x² + 4

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The required pairs of triangles in A and D are congruent triangles.

Congruent triangles are two triangles that have the same shape and size. In other words, they are identical in every way, including their angles and side lengths. When two triangles are congruent, their corresponding sides and angles are equal.

Here,

Congruent triangles are important in geometry because they have many useful properties and can be used to solve a variety of problems. For example, if two triangles are congruent, then any corresponding parts (such as their altitudes, medians, or angle bisectors) will also be congruent. This can be helpful in finding missing side lengths or angles of a triangle.

Thus, from the above definitions Pairs of triangles in A and D are congruent triangles.

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There are 54 possible 4-digit numbers that can be created using only the digits 0, 1, and 2, considering that the first digit cannot be 0 and each position has a limited selection of digits.

To determine the number of 4-digit numbers that can be written using only the digits 0, 1, and 2, we must take into account that a 4-digit number cannot start with 0, which leaves us two choices for the first digit (either 1 or 2). For the subsequent three digits, we have three choices each (0, 1, or 2). Therefore, the total number of 4-digit combinations is obtained by multiplying the number of choices for each digit's position.

The explanation you provided is correct, and the calculation is accurate. There are indeed 54 possible 4-digit numbers that can be created using only the digits 0, 1, and 2, considering the given conditions.

To summarize:

- The first digit has 2 possible choices (1 or 2).

- Each of the next three digits has 3 choices (0, 1, or 2).

So, the total number of 4-digit combinations is obtained by multiplying the number of choices for each digit's position:

\[2 \times 3 \times 3 \times 3 = 54.\]

Therefore, there are 54 possible 4-digit numbers that can be created following the given conditions.

The first digit has 2 possible choices, and each of the other three positions has 3 choices, which gives us a total of:

2 × 3 × 3 × 3 = 54 possible 4-digit numbers using the digits 0, 1, and 2.