What is the least whole number that has exactly 7 factors including 1 and itself?

If two or more rectangles have the same dimensions is because they are congruent. In this way, they share two corresponding sides. Our first rectangle has the following sides:

L1 = x

L2 = 3x + 4

Therefore the second and third rectangles need to have the same sides. Therefore, we will have that the sides of the second one is also:

L1 = x

L2 = 3x + 4

and the sides of the third one is also:

L1 = x

L2 = 3x + 4

Therefore, the total amount of fencing needed is:

T = 3(3x + 4) + 4(x)

T = 9x + 12 + 4x

T = 13x + 12

Moreover, we can compute the area of the fence as follows:

A = [(3x + 4) + (3x + 4) + (3x + 4)]x = (3x + 3x + 3x + 4 + 4 + 4)x

A = (9x + 12)x

A = 9x² + 12x

The algebraic expression that expresses the total amount of fencing needed is 11X + 20.

Given that a piece of land is to be fenced and subdivided as shown so that each rectangle has the same dimensions, to express the total amount of fencing needed as an algebraic expression in X, the following calculation must be performed:

Therefore, the algebraic expression that expresses the total amount of fencing needed is 11X + 20.

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Answer:

Question 1 is B and Question 2 is D

The first passage has a hopeful and ambitious tone, using optimistic language to express excitement for the future. The second passage has a melancholic and isolated tone, using somber phrases to convey discomfort and disconnection. These contrasting tones reflect a range of emotions in the narrative.

To contrast the tone of two passages, we must first identify the attitude or mood present in each. In Tone ONE, the writer’s use of optimistic phrases such as "Envisioning paths," "grooming him for interviews," and "future was bright" convey a hopeful and ambitious tone. This tone suggests excitement for the future possibilities and success that lie ahead for Russell.

In contrast, Tone 2 uses phrases like "too awkward," "profoundly still," and "didn’t talk" which reveal a somber, isolated and possibly melancholic tone. It reflects the author’s feelings of discomfort, silence, and the lack of communication and connection with the family.

The two tones differ significantly, with the first being uplifted and forward-looking, while the second expresses a sense of disconnection and sadness, giving us a comprehensive view of the contrasting emotional landscapes in the same narrative context.

By using the formula for the area of a rectangle, we will see that there are 117 square units on the office.

We know that for a rectangle of width W and length L, the area is given by:

A = W*L

In this case, we have a rectangle of width of 13 units, and length of 9 units, replacing that in the area equation:

A = (13 units)*(9 units) = 117 square units.

So the correct option is the second one, there are 117 square units in the office.

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Answer:

D)15; 100°.

Step-by-step explanation:

Given : Verticle opposite angles (6a + 10) and (3a + 55).

To find :  Find the value of a. b. Find the value of the marked angles.

Solution : We have given  (6a + 10) and (3a + 55).

Verticle opposite angles are equal

Then

6a + 10 = 3a + 55

On subtracting both sides by 3a

6a - 3a + 10 = 55

3a + 10 = 55

On subtracting by 10 both sides.

3a = 55 -10.

3a = 45.

On dividing the 3 by both sides

a = 15°.

Then ,

6 (15) + 10.

90 + 10 = 100

b =100°.

Therefore, D)15; 100°.

A function is a rule that maps each preimage to a unique image.

A rule that maps each preimage to a unique image is called a function. In mathematics, a function is a relation between a set of inputs (preimages) and a set of outputs (images) in which each input is associated with exactly one output.

For example, consider the function f(x) = 2x, where the input x is a preimage and the output 2x is the corresponding image. If we plug in x = 3, the preimage, we get f(3) = 2(3) = 6, the image.

Functions are commonly represented using algebraic expressions, tables, graphs, or diagrams. They are fundamental to many branches of mathematics and have applications in various fields. The phrase 'rule that maps each preimage' refers to the function or transformation that defines how to manipulate a preimage to obtain an image, either by applying a mathematical formula or by executing a geometric operation.

Final answer:

To solve the equation 3ax + 2b / c = 4d for x, isolate x on one side of the equation by multiplying both sides by c, subtracting 2b, and dividing by 3a.

Explanation:

To solve the equation 3ax + 2b / c = 4d for x, we need to isolate x on one side of the equation. Here are the steps:

Therefore, the solution for x is x = (4cd - 2b) / (3a).

The values of x and y are 4.29 and 90.

Given that ∆ABC and ∆FDE are congruent by a certain criterion, let's denote the corresponding sides as follows:

AB = FD (corresponding side)

BC = DE (corresponding side)

AC = FE (corresponding side)

Let's also denote the angles of ∆ABC and ∆FDE as follows:

∠A = ∠F (corresponding angle)

∠B = ∠D (corresponding angle)

∠C = ∠E (corresponding angle)

Now, if we're given specific values for some sides or angles, we can use that information to find the missing values.

Let's assume we're given that AB = 5 cm, BC = 7 cm, AC = 6 cm, ∠A = 40°, and ∠B = 50°.

First, let's find the missing side FD using the side-length proportion:

[tex]\[ \frac{AB}{FD} = \frac{BC}{DE} = \frac{AC}{FE} \][/tex]

Substitute the given values:

[tex]\[ \frac{5}{FD} = \frac{7}{DE} = \frac{6}{FE} \][/tex]

We can solve for FD by setting up the proportion:

[tex]\[ \frac{5}{FD} = \frac{7}{6} \][/tex]

Cross-multiply to solve for FD:

[tex]\[ 5 \cdot 6 = FD \cdot 7 \][/tex]

[tex]\[ 30 = 7FD \][/tex]

[tex]\[ FD = \frac{30}{7} \][/tex]

[tex]\[ FD \approx 4.29 \text{ cm} \][/tex]

Next, we can find the missing angle measures. Since ∠A = ∠F = 40° and ∠B = ∠D = 50°, we can find ∠C = ∠E using the fact that the sum of angles in a triangle is 180°:

[tex]\[ ∠C = 180° - ∠A - ∠B \][/tex]

[tex]\[ ∠C = 180° - 40° - 50° \][/tex]

[tex]\[ ∠C = 90° \][/tex]

So, we have found that FD ≈ 4.29 cm and ∠C = 90°. Thus, the values of x and y are 4.29 and 90, respectively.

Complete question

∆ABC and ∆FDE are congruent by the

criterion. (Use the three-letter abbreviation without spaces.)

The value of x is

, and the value of y is

.

Answer:

Slope of the line passing through the points (1, −5) and (4, 1) is 2 .

Step-by-step explanation:

The equation of the slope is given by

[tex]m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]

As given

line passing through the points (1, −5) and (4, 1) .

[tex]m = \frac{1-(-5)}{4-1}[/tex]

[tex]m = \frac{1+5}{4-1}[/tex]

[tex]m = \frac{6}{3}[/tex]

m = 2

Therefore the slope of the line passing through the points (1, −5) and (4, 1) is 2 .

Answer:

-7/6  

Step-by-step explanation:

-2-5/3+3= -7/6

Answer:

2,225 years apart

Step-by-step explanation:

Answer:

option C  and D

Step-by-step explanation:

Thomas earns x dollars each week walking his neighbor’s dog. He also earns $10 allowance each week.

we need to find how much Thomas has earned after 4 weeks

in 1 week , x dollars earned

Also $10 allowance each week.

So amount earned in 1 week is [tex]x+10[/tex]

To find the Amount earned in 4 weeks , multiply the expression by 4

amount of money Thomas has earned after 4 weeks is [tex]4(x+10)[/tex]

[tex]4(x+10)=4x+40=x+10+x+10+x+10+x+10[/tex]

option C  and D

Answer:

Step 1:  Use the straightedge to draw a line k that passes through A and intersects line j. Label the point of intersection as point P.  This is because you must have a point of intersection with the line before you can move forward.

Step 2:  Place the point of the compass on point P and draw an arc that intersects lines j and k. Label the intersections as points Q and R.  This will give us a compass width to move forward.

Step 3:  Place the point of the compass on point Q and open it to width ​ QR.  

Step 4:   Without changing the width of the compass opening, place the point of the compass on point A and draw an arc that intersects line k. Label the intersection as point S.   This gives us the point to use to construct the second point of our parallel line.

Step 5:  With the compass opening set to width QR , place the point of the compass on point S and draw an arc that intersects the arc that was drawn from point A. Label the intersection of the arcs as point B.   This is the second point of our parallel line.

Step 6:  Use the straightedge to draw AB←→.  

The steps are shown below:

A line segment is a part of a straight line that is bounded by two distinct end points, and contains every point on the line that is between its endpoints. The length of a line segment is given by the Euclidean distance between its endpoints.

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