If 3 times a certain number is increased by 4 the result is 28. What is the number?

If x is "the number," which of the following equations could be used to solve for the number?

c.3x + 4 = 28

b.3(x + 4) = 28

a.3x = 4 + 28

Simply multiply the number outside the brackets with each one inside it..

3 [ 1 5 -5 6 0 0 ]
3 x 1 = 3
3 x 5 = 15
3 x -5 = -15
3 x 6 = 18
3 x 0 = 0
3 x 0 = 0
[ 3 15 -15 18 0 0 ]

[ 3 15 ]
[ -15 18 ]
[ 0 0]

The answer is B and I hope I explained this well for you.

Answer:  The answer is 0.5 and 5.

Step-by-step explanation: The given equation of the line is

[tex]2(y+1)=10x+3.[/tex]

We are to find the y-intercept and the slope of the given line.

We know that the slope-intercept form of a line is given by

y = mx + c, where, 'm' is the slope and 'c' is the y-intercept of the line.

We have

[tex]2(y+1)=10x+3\\\\\Rightarrow 2y+2=10x+3\\\\\Rightarrow 2y=10x+3-2\\\\\Rightarrow 2y=10x+1\\\\\Rightarrow y=5x+0.5.[/tex]

Therefore, c = 0.5 and m = 5.

Thus, the y-intercept of the line is 0.5 and the slope is 5.

Answer:

Option B is correct.

[tex]3x^2+4x+2[/tex].

Step-by-step explanation:

We are asked to find the quotient obtained by dividing the expression [tex](3x^3+10x^2+10x+4)[/tex] by the expression [tex](x+2)[/tex]

We can also write this expression as i.e. we are asked to find the value of the expression:

[tex]\dfrac{3x^3+10x^2+10x+4}{x+2}[/tex]

We can write the expression on the numerator as:

[tex]3x^3+10x^2+10x+4=(3x^2+4x+2)(x+2)[/tex]

Hence,

[tex]\dfrac{3x^3+10x^2+10x+4}{x+2}=\dfrac{(3x^2+4x+2)(x+2)}{x+2}[/tex].

Hence,

[tex]\dfrac{3x^3+10x^2+10x+4}{x+2}=3x^2+4x+2[/tex].

Hence, option B is correct.

Hence, the quotient is:

[tex]3x^2+4x+2[/tex].

I think it is c or D but it should be C

hope that help

[I don't think it did lol ]

BMK

Final answer:

To prove that the diagonals of a square are congruent and perpendicular, label the vertices of a square on a coordinate grid and calculate the slopes and lengths using the slope formula and distance formula respectively. The diagonals have slopes of +1 and -1, proving they are perpendicular, and they have equal lengths, proving they are congruent.

Explanation:

To prove that the diagonals of a square are congruent and perpendicular, we place a square with its vertices on a coordinate grid and label each vertex with variables.

Let's consider a unit square where c = 1 for simplicity, which means the length of each side is 1 unit. Place the square so that one vertex is at the origin (0,0), and label the vertices A(0,0), B(1,0), C(1,1), and D(0,1).

The diagonal AC will have endpoints at A(0,0) and C(1,1), and diagonal BD will have endpoints at B(1,0) and D(0,1). The slope of diagonal AC is (1 - 0)/(1 - 0) = 1, and the slope of diagonal BD is (1 - 0)/(0 - 1) = -1. Since the product of their slopes is -1 (1 * -1 = -1), this proves that they are perpendicular to each other.

To show they are congruent, we calculate their lengths using the distance formula: the distance between two points (x1,y1) and (x2,y2) is √[(x2 - x1)² + (y2 - y1)²]. Applying this to AC and BD reveals both lengths to be √[(1-0)² + (1-0)²] = √[1 + 1] = √2, proving the diagonals are congruent.

The velocity of the plane is 200 mph due north and the velocity of the wind is 23 mph due west. The velocity vector of the plane is 200 mph due north minus 23 mph due west. The ground speed of the plane can be found using the Pythagorean theorem.

Part I: The velocity of the plane can be represented as 200 mph due north, and the velocity of the wind can be represented as 23 mph due west.

Part II: To find the velocity vector of the plane, we subtract the velocity of the wind from the velocity of the plane. The resultant velocity vector of the plane is 200 mph due north minus 23 mph due west.

Part III: The ground speed of the plane is the magnitude of the resultant velocity vector of the plane. We can calculate it using the Pythagorean theorem: ground speed = square root of (200^2 + 23^2).

opposite angles are identical so if angle 1 = 130

 than angle 3 is also 130 degrees

Answer:

Angle 3

Step-by-step explanation:

we know that

[tex]m<1=m<3[/tex] -----> by vertical angles

we have

[tex]m<1=130\°[/tex]

therefore

[tex]m<3=130\°[/tex]

The angle of depression represents the angle from a horizontal layout to a lower surface. The distance between the two stadiums is 3115.1 meters

The given parameters have been illustrated using the attached image of triangles.

The stadiums are represented with A and B.

First, calculate distance BO using:

[tex]\tan T =\frac{BO}{TO}[/tex]

Where:

[tex]\angle T = 90 -75.2 = 14.8[/tex]

[tex]TO = 1100[/tex]

So, we have:

[tex]\tan(14.8^o) = \frac{BO}{1100}[/tex]

Make BO the subject

[tex]BO = 1100 * \tan(14.8^o)[/tex]

[tex]BO = 1100 * 0.2642[/tex]

[tex]BO = 290.62[/tex]

Next, calculate distance AO using:

[tex]\tan T =\frac{AO}{TO}[/tex]

But in this case:

[tex]\angle T = 90 -17.9 = 72.1[/tex]

[tex]TO = 1100[/tex]

So, we have:

[tex]\tan(72.1^o) = \frac{AO}{1100}[/tex]

Make AO the subject

[tex]AO = 1100 * \tan(72.1^o)[/tex]

[tex]AO = 1100 * 3.0961[/tex]

[tex]AO = 3405.71[/tex]

The distance AB between the 2 stadiums is:

[tex]AB = AO - BO[/tex]

[tex]AB = 3405.71-290.61[/tex]

[tex]AB = 3115.1[/tex]

Hence, the distance between the 2 stadiums is 3115.1 meters.

Read more about angles of depression at:

https://brainly.com/question/13697260

Answer: 12.496

Step-by-step explanation:

The formula to find the sum of geometric progression is given by :-

[tex]S_n=\dfrac{a(1-r^n)}{1-r}[/tex]

Given : The first term : [tex]a_1=10[/tex]

Common ratio = [tex]r=\dfrac{1}{5}=0.2[/tex]

Then , the sum of first five terms of a geometric series is given by :-

[tex]S_5=\dfrac{10(1-(0.2)^5)}{1-0.2}=12.496[/tex]

Hence, the sum of the first five terms of given geometric series =12.496

Answer:

9 square inches.

Step-by-step explanation:

We have been given that a rectangle has a length of the [tex]\sqrt[3]{81}[/tex] inches and a width of [tex]3^{\frac{2}{3}}[/tex] power inches. We are asked to find the area of given rectangle.

We know that area of rectangle in length times width of rectangle.

[tex]\text{Area of rectangle}=\sqrt[3]{81}\times 3^{\frac{2}{3}}[/tex]

We can write 81 as [tex]3^4[/tex] as:

[tex]\text{Area of rectangle}=\sqrt[3]{3^4}\times 3^{\frac{2}{3}}[/tex]

Using exponent rule [tex]\sqrt[n]{a^m}=a^{\frac{m}{n}}[/tex], we can write [tex]\sqrt[3]{3^4}=3^{\frac{4}{3}}[/tex].

[tex]\text{Area of rectangle}=3^{\frac{4}{3}}\times 3^{\frac{2}{3}}[/tex]

Using exponent rule [tex]a^b\cdot a^c=a^{b+c}[/tex], we will get:

[tex]\text{Area of rectangle}=3^{\frac{4}{3}+\frac{2}{3}}[/tex]

[tex]\text{Area of rectangle}=3^{\frac{4+2}{3}}[/tex]

[tex]\text{Area of rectangle}=3^{\frac{6}{3}}[/tex]

[tex]\text{Area of rectangle}=3^{2}[/tex]

[tex]\text{Area of rectangle}=9[/tex]

Therefore, the area of given rectangle is 9 square inches.

Answer:  Second option is correct.

Step-by-step explanation:

Since we have given that

ΔABC has measures a=2, b=2, m∠A=30⁰

As we know the "Law of sines " i.e.

[tex]\frac{a}{\sin A}=\frac{b}{\sin B}\\[/tex]

so, we put the given values in above formula:

[tex]\frac{2}{\sin 30\textdegree}=\frac{2}{\sin B}\\\\\implies \sin 30\textdegree=\sin B\\\\\implies B=30\textdegreee[/tex]

Hence, Second option is correct.

An odd function, by definition, is a function that is symmetric about the origin.

An even function, by definition, is a function that is symmetric with respect to the y-axis.

Since the question says that f(x) is an odd function, it has rotational symmetry about the origin. First option is correct.

ANSWER: symmetric about the origin.

Answer:It has rotational symmetry about the origin.

Step-by-step explanation:

An odd function : is a function that is symmetric about the origin.

An even function : is a function that is symmetric with respect to the y-axis.

Since , f(x) is an odd function, it has rotational symmetry about the origin.

its meaning that its graph remains unchanged after rotation of 180 degrees about the origin.

Therefore, It has rotational symmetry about the origin.

Final answer:

Carol practices for 8.5 hours per week, so in a 5-week period, she would practice for a total of 42.5 hours.

Explanation:

The student asked how many hours Carol practices her culinary skills in a 5-week period, if she practices for 17 hours in a 2-week period. To find the answer, we calculate how many hours Carol practices per week by dividing the total hours she practices in two weeks by two. Then we multiply the weekly hours by the number of weeks in question, which is five.

The calculation is as follows: Carol practices for 17 hours / 2 weeks = 8.5 hours per week. Then, 8.5 hours/week x 5 weeks = 42.5 hours in total for a 5-week period.

3-1 = 2

 2 is greater than 1/4

 so > is the answer

The area of the composite figure is 6π + 16 cm²

Composite figures are composed of different dimensional figures. The area of a composite figure is the sum of the whole 2 dimensional figures that forms the composite figure.

Therefore, the figure above has 3 semi circle and 1 square.

Therefore, the area can be calculated as follows;

area = 1 / 2 πr² + 1 / 2 πr² + 1 / 2 πr² + L²

area = 3 / 2 (πr²) + L²

where

r = 2 cm

L = 4 cm

Therefore,

area of the composite figure = 3 / 2(π × 4) + 4²

area of the composite figure = 3 / 2(4π) + 16

area of the composite figure =  6π + 16 cm²

learn more on composite figures here: https://brainly.com/question/1639299

Yes, it is possible to have more than one bisector in a line segment.

Bisector is a line that divides a line or an angle in to two equivalent parts. There are two types of Bisectors based on what geometrical shape it bisects.

 In general 'to bisect' something means to cut it into two equal parts. The bisector  is the one that doing the cutting process.

With a line bisector, we cut a line segment into two equal parts with another line - the bisector. Just imagine the line PQ is being cut into two equal lengths (PF and FQ) by the bisector line AB.

Whenever AB intersects at a right angle, it is called the "perpendicular bisector" of PQ. If it crosses at any other angle it is simply called a bisector. Drag the points A or B and see both types.

For obvious reasons, the point F is called the midpoint of the line PQ,