Determine the point of intersection of the lines x+y=4 and 2x+3y=12 algebraically

[tex]x=57^{\circ}[/tex] (alternate segment theorem)

[tex]y=67^{\circ}[/tex] (alternate segment theorem)

The scale factor of the image under a dilation centered at the origin is 2.

Transformation is the movement of a point from its initial location to a new location. Types of transformation are reflection, translation, rotation and dilation.

Translation is the movement of a point either up, down, left or right on the coordinate plane.

Let us assume that quadrilateral ABCD have vertices at A(2, 1) while its image has vertices at A'(4, 2), hence:

Scale factor = 4/2 = 2/1 = 2

The scale factor of the image under a dilation centered at the origin is 2.

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The statement that is true about the estimate is that the number will be less than 1 because when adding the exponents, 11 + (negative 14) = negative 3, and a number in scientific notation with a negative exponent is less than 1 (option D).

According to this question, Bethany wants to estimate the value of (4.296 times 10 Superscript 11) (1.8614 times 10 Superscript negative 14).

This expression can be mathematically written as:

(4.296 × 10¹¹) (1.8614 × 10-¹⁴)

The result of this expression is 7.996 × 10-³, which is value less than 1.

Therefore, the number will be less than 1 because when adding the exponents, 11 + (negative 14) = negative 3, and a number in scientific notation with a negative exponent is less than 1.

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Answer:

gradient = [tex]\frac{1}{3}[/tex]

Step-by-step explanation:

calculate the gradient m using the gradient formula

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

with (x₁, y₁ ) = (- 6, 0) and (x₂, y₂ ) = (0, 2) ← 2 points on the line

m = [tex]\frac{2-0}{0-(-6)}[/tex] = [tex]\frac{2}{0+6}[/tex] = [tex]\frac{2}{6}[/tex] = [tex]\frac{1}{3}[/tex]

Answer:

x=7.48 cm, y=6.94

Step-by-step explanation:

Start by finding the value of x.

Using the Pythagorean Theorem, [tex]13^{2}+x^{2} =15^{2}[/tex]

[tex]169+x^{2} =225\\x^2=56\\x=\sqrt{56}\\x=2\sqrt{14}[/tex]

Using this, we can again use the Pythagorean Theorem to find the value of y.

[tex]2.8^{2} +y^{2} =(2\sqrt{14} )^{2} \\7.84+y^{2} =56\\y^{2} =48.16\\y=\sqrt{48.16}[/tex]

(I rounded to two decimal places, not sure what is required here)

x=7.48 cm

y=6.94 cm

Answer:

4x+22

Step-by-step explanation:

distribute 4(x+4)

it would become 4x+16+6

16+6 = 22

4x+22

Therefore, the measures of the angles in triangle ABC is as follows:

The measure of the angles in the right triangle can be found as follows:

Using trigonometric ratios,

sin ∅ = opposite / hypotenuse

sin ∅ = 7 / 25

∅ = sin ⁻¹ 0.28

∅ = 16.2602047083

∅ = 16.3°

Hence,

m∠B = 16.3°

m∠C = 90°

m∠A = 180 - 90 - 16.3 = 73.7°

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Answer:

B

Step-by-step explanation:

cable subscribers

This shows that 154mm is equivalent to 15centimeters 4millimeters

Units are measure of a quantity. From the given question, we are to convert 15cm4mm to millimeters

Since 10mm =1cm, hence;

15cm4mm = (15 * 10) + 4

15cm4mm = 150+ 4

15cm4mm = 154mm

This shows that 154mm is equivalent to1 5centimeters 4millimeters

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The rate at which the distance from the plane to the station increases is 537 miles/hour. Computed using Pythagoras Theorem and Differentiation.

We assume the station to be at point A.

The plane when directly over the station, takes point C.

As it is flying horizontally continuously at an altitude of 1 mile, AC = b = 1 mile.

The plane flies continuously, and when it is at a distance of 3 miles from the station, it is at point B.

The distance between the plane and the station of 3 miles can be shown as AB = c = 3 miles.

The movement of the plane from C to B can be shown as BC = a.

Now, from the given case, we get a right triangle ABC.

By Pythagoras' Theorem,

a² + b² = c² ... (i).

or, a² + 1² = 3²,

or, a² = 9 - 1 = 8,

or, a = 2√2 miles.

We are told that the plane is moving at a speed of 570 miles/hour.

This can be shown as, da/dt = 570 miles/hour.

As the plane is moving horizontally, the vertical distance between the plane and the station doesn't change, that is, db/dt = 0.

In the question, we are asked to find the rate at which the distance from the plane to the station is increasing, that is, we are asked to find dc/dt.

Differentiating (i), with respect to time t, we get:

2a(da/dt) + 2b(db/dt) = 2c(dc/dt).

Substituting values of a = 2√2 miles, b = 1 mile, c = 3 miles, da/dt = 570 miles/hour, and db/dt = 0, we get:

2(2√2)(570) + 2(1)(0) = 2(3)(dc/dt),

or, 2280√2 + 0 = 6(dc/dt),

or, dc/dt = 380√2 miles/hour = 537.40 miles/hour ≈ 537 miles/hour.

Thus, the rate at which the distance from the plane to the station increases is 537 miles/hour. Computed using Pythagoras Theorem and Differentiation.

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The equation of the parabola which has a vertex at (-2,0) is; y = x² +4x +4.

From the task content, it follows that the vertex of the of the parabola described by the graph in discuss is given by the point; (-2,0).

Hence, the vertex form of the equation is;

y = (x-(-2))² +0

Hence, the equation is; y = x² +4x +4.

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Answer: B

Step-by-step explanation:

Edge 2023

In the given diagram, the value of the side labelled x is 10.8 cm

From the question, we are to determine the value of side x

Using the law of sines

In any given triangle XYZ,

[tex]\frac{x}{sinX}=\frac{y}{sinY}=\frac{z}{sinZ}[/tex]

Thus,

In the given triangle,

[tex]\frac{x}{sin81^\circ} =\frac{7}{sin40^\circ}[/tex]

∴ [tex]x=\frac{7 \times sin81^\circ}{sin40^\circ}[/tex]

x = 10.75599

x ≈ 10.8 cm

Hence, the value of the side labelled x is 10.8 cm

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The coordinates of point P along AB so that the ratio of AP to PB is 3 to 1 is (4,13).

A coordinate system in geometry is a system that employs one or more integers, or coordinates, to define the position of points or other geometric components on a manifold such as Euclidean space.

Given that the ratio of the two lines is 3:1, therefore, we can write,

m:n = 3:1

Also, the coordinate of the endpoints of the line are,

A = (x₁, y₁) = (1, 7)

B = (x₂, y₂) = (5, 15)

Now, Using the Section formula the coordinate of the point P can be written as,

[tex]P = (\dfrac{mx_2+nx_1}{m+n}\ , \dfrac{my_2+yx_1}{m+n})\\\\P = (\dfrac{3(5)+1(1)}{4}\ , \dfrac{3(15)+1(7)}{4})\\\\P = (4, 13)[/tex]

Hence, the coordinates of point P along AB so that the ratio of AP to PB is 3 to 1 is (4,13).

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The value of k is  [tex]-h^2 -2xh -5x + 6[/tex]

Given;

x^(2)-5x+6=(x+h)^(2)+k

Expand the expression on the right side, we have

[tex]x^2 - 5x + 6 = (x+ h) (x+ h) + k[/tex]

[tex]x^2 -5x + 6 = x^2 +xh + xh + h^2 +k[/tex]

Collect like terms

[tex]x^2 -5x + 6 = x^2 + 2xh + h^2 + k[/tex]

[tex]x^2 -5x + 6 -x^2 - 2xh -h^2 = k[/tex]

We then have

[tex]-5x + 6 -2xh -h^2 = k[/tex]

[tex]k = -h^2 -2xh -5x + 6[/tex]

Thus, we have k as [tex]-h^2 -2xh -5x + 6[/tex]

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Answer:

D

21√2/2

C

C

C

Step-by-step explanation:

For all problems, the ratios of the lengths of the sides are:

30-60-90 triangle 1 : √3 : 2

45-45-90 triangle 1 : √2 : √2

Problem 3:

30-60-90 triangle at left

Hypotenuse = 11

Short leg = 11/2

Long leg = 11/2 × √3

45-45-90 triangle at right

x = 11/2 × √3 × √2 = 11√6/2

Answer: D

Problem 4:

30-60-90 triangle at top

Hypotenuse = 7√3

Short leg = 7√3/2

Long leg = 7√3/2 × √3 = 7 × 3/2 = 21/2

45-45-90 triangle at right

x = 21/2 × √2 = 21√2/2

Answer: 21√2/2

Problem 7:

30-60-90 triangle

Long leg = 4√3

Short leg = y = 4√2/2 = 2√2

Answer: C.

Problem 6:

30-60-90 triangle at left

Hypotenuse = 12√6

Short leg = 12√6/2 = 6√6

Long leg = 6√6 × √3 = 6√2√3√3 = 18√2

45-45-90 triangle at right

x = 18√2/√2 = 18

Answer: C.

Problem 8:

45-45-90 triangle

Hypotenuse = 2

Leg = y = 2/√2 = 2√2/(√2√2) = √2

Answer: C.

The inequality equation of the lines are given as 4x – y ≤ 4 and 2x – 2y ≤ 3. Then the correct option is A.

Inequality is defined as an equation that does not contain an equal sign. Inequality is a term that describes a statement's relative size and can be used to compare these two claims.

The graph is given below.

From the graph, equation of the lines are given as

4x - y = 4 and 2x - 2y = 3

But for the inequality is shown in graph. Then the inequality equation of the lines are given as

4x – y ≤ 4 and 2x – 2y ≤ 3

Then the correct option is A.

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The graphs show a parabola with a vertex that has Roots are -2 and 4.

We have given that,

vertex of (1,-9) and solutions of (-2,0) and (4,0).

y=k*(x+2)(x-4)

Vertex = (1,-9) is a point of the parabola

-9=k*(1+2)(1-4) implies k=1

Equation of the parabola is y=(x+2)(x-4).

The graphs show a parabola with a vertex that has Roots are -2 and 4.

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Answer:

The year in which population will be 54000, is 2020.

Step-by-step explanation:

In the question, it is given that the city's population is growing linearly. In 2008 , population was 28200 and in 2012 , it is 36800 .

It is required to identify the year in which population will be 54000 . equation.

Step 1 of 2

The population in 2008 was 28200 and in 2012 it was, 36800 .

The equation relating population and year is,

[tex]$$\begin{aligned}&y-28200=\left(\frac{36800-28200}{2012-2008}\right)(x-2008) \\&y-28200=\left(\frac{8600}{4}\right)(x-2008) \\&y-28200=2150(x-2008)\end{aligned}$$[/tex]

Step 2 of 2

The year in which population will be 54000 , can be determined by substituting 54000 for y,

[tex]54000-28200=2150(x-2008)$\\ $25800=2150(x-2008)$\\ $x-2008=12$\\ $x=2020$[/tex]

Answer:

Step-by-step explanation:

Comment

Find the number of square yards

Area = L * w

L = 2

w = 1

Area = 2 * 1

Area = 2

Now for the counter top

1 sq yard costs 204.63

2 sq yards cost 2 * 204.63

2 sq yards cost 409.26

Answer: 409.26

The pattern in the values as the exponents increase is multiply the previous value by 2.

Exponents and powers are ways used to represent very large numbers or very small numbers in a simplified manner.

Given:

Powers of 2      Values

[tex]2^{-5}[/tex]                         1/32

[tex]2^{-4}[/tex]                           1/16

[tex]2^{-3}[/tex]                           1/8

[tex]2^{-2}[/tex]                           1/4

[tex]2^{-1}[/tex]                           1/2

As, from the given data we can see that negative powers of 2 reducing by 1.

as, 2^-5 * 2= 2^-4

Hence, multiply the previous value by 2 to obtain the next value.

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Based on quantitative reasoning, Alden should take the classes from Dr. Baeno in his next semester.

Quantitative Reasoning Behind Choosing Dr. Baeno

Alden talked to 17 friends that previously took course from Dr. Alvarez.

Eight out of 17 friends recommended Dr. Alvarez.

∴ The fraction of his friends in favor of Dr. Alvarez = 8/17

Also, Alden talked to 18 friends who took course from Dr. Baeno.

Eleven out of 18 friends recommended Dr. Baeno.

∴ The fraction of his friends in favor of Dr. Baeno = 11/18

Making Comparison

Now, since we know that,

11/18 > 8/17

The capacity for using knowledge and mathematics to address problems in the real world is known as quantitative reasoning.

Hence, applying quantitative reasoning, we can infer that it is more favorable for Alden to take classes from Dr. Baeno.

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The width of the rectangle is 10 inches

The given parameters are:

Length = 3x + 6

Width = 3x + 4

Area = 120

The area of a rectangle is

Area = Length * Width

So, we have:

(3x + 6) *(3x + 4) = 120

Expand

9x^2 + 12x + 18x + 24 = 120

Evaluate the like terms

9x^2 + 30x - 96 = 0

Using a graphing calculator, we have:

x = 2

Substitute x = 2 in Width = 3x + 4

Width = 3 * 2 + 4

Evaluate

Width = 10

Hence, the width of the rectangle is 10 inches

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x+3 is not a factor of f(x)=x^4+6x^3+5x^2-15x-2 because there is a remainder of 7 when we use the long division method to divide x+3 with the given function f(x).

The factors of a quadratic equation are the simplified form of the quadratic equation, that if multiplied together, results in the original quadratic equation.

Given that:

[tex]\mathbf{=\dfrac{x^4+6x^3+5x^2-15x-2}{x+3} }[/tex]

[tex]\mathbf{=x^3+3x^2-4x-3+\dfrac{7}{x+3} }[/tex]

x+3 is not a factor of f(x)=x^4+6x^3+5x^2-15x-2 because there is a remainder of 7 when we use the long division method to divide x+3 with the given function f(x).

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Answer:

9 sq cm

Step-by-step explanation:

If you join the two triangles it will become a square of side 3cm.

3cm * 3cm = 9cm squared.

Hope you have a nice day :)

Any questions please comment.

Answer:

table A

Step-by-step explanation:

A direct variation can be expressed as: [tex]\frac{y}{x}=k[/tex]. The y/x is equal to a constant value k. Multiply both sides by x you get: [tex]y=kx[/tex]. So as x increases by 1, y increases by some constant value. So to find which one will create a direct variation, see which of the tables has a constant y/x

Table A:
   7/2 = 3.5

   14/4 =3.5

   17.5/5=3.5

   21/6 = 3.5

   this has a constant y/x so it is a direct variation

Table B:

   7/2 = 3.5

   14/4 = 3.5

   21 / 5 = 4.2

   this is not a direct variation, it doesn't have a constant y/x

table C:

   7 / 2 = 3.5

   14 / 4 = 3.5

   15/5 = 3

   This is not a direct variation, it doesn't have a constant y/x

table d:
   7/2 = 3.5

   14/4 =3.5

   17/5=3.4

   This is not a direction variation, it doesn't have a constant y/x

Answer:

c=104.1 ft

Step-by-step explanation:

Law of Cosines

c^2 = 66^2 + 120^2 - 2 (66)(120)cos 60

c = 104.1 ft

Answer:

7.76

Step-by-step explanation:

0.22×20=4.4

3.36(base amount ,average of the others)+4.4=7.76

Answer:

3rd option : 5

Step-by-step explanation:

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its a slope question

f(x) is y

y2-y1/x2-x1

-10+25/-1+4

15/3

5

Since we know that the equation of a line in slope-intercept form is: , where, m= Slope and b= y-intercept.

Upon substituting m = 5 in slope-intercept form of equation we will get,

Now let us find y-intercept by substituting coordinates of one point in our given line.

So the equation of given line will be,

Now let us find value of n by substituting x = n and y = 20 in our equation.

Therefore, the value of n is 5.

The confidence interval of employes individuals from home

Construction of hypothesis:

0.10 - 0.047 < p < 0.10 + 0.047

A simple random sample size n = 250 . Of the 250 employed individuals ​surveyed,42 responded that they did work home a minimum of once per week.

Construct 99% confidence interval for population

For proportion : 42 / 250 = 1/10 = 0.16

Mean = 2.5 * sqrt [ 0.1 * 0.9 / 250]

= 2.5 * 0.01

= 0.47

Construction of hypothesis:

0.10 - 0.047 < p < 0.10 + 0.047

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Answer:

60 in²

Step-by-step explanation:

Area of a Regular Polygon

  [tex]\textsf{A}=\dfrac{n\:s\:a}{2}[/tex]

where:

From inspection of the diagram:

Substitute the values into the formula and solve for A:

[tex]\implies \sf A = \dfrac{5 \times 6 \times 4}{2}[/tex]

[tex]\implies \sf A = \dfrac{120}{2}[/tex]

[tex]\implies \sf A = 60\:in^2[/tex]

Therefore, the area of the given regular pentagon IJKLM is 60 in².

[tex]\huge\text{Hey there!}[/tex]

[tex]\huge\textbf{First thing, let's get the formula for}\\\huge\textbf{the regular pentagon/polygon.}[/tex]

[tex]\mathbf{\dfrac{1}{2}\times \boxed{a}pothem\times\boxed{p}erimeter = {\boxed{a}rea}}[/tex]

[tex]\huge\textbf{Or we could simply say....}[/tex]

[tex]\mathbf{\dfrac{\boxed{n}umber\ of\ sides\ you\ have \times \boxed{s}ide \ length\ on\ one\ side\times \boxed{a}pothem}{2} = \boxed{a}rea}[/tex]

[tex]\huge\textbf{Let's plug the pieces to the equation,}\\\huge\textbf{since we have that information out the}\\\huge\textbf{way.}[/tex]

[tex]\bullet\mathbf{\ Number\ of\ sides \rightarrow \boxed{\bf 5}}[/tex]

[tex]\bullet\mathbf{\ Side\ length\ of\ one\ side \rightarrow \boxed{\bf 6\ inches}}[/tex]

[tex]\bullet\mathbf{\ Apothem \rightarrow \boxed{\bf 4\ inches}}[/tex]

[tex]\huge\textbf{Making the equation for you:}[/tex]

[tex]\mathbf{\dfrac{5\times 6 \times 4}{2} = \boxed{a}rea}[/tex]

[tex]\huge\textbf{Solving for the result:}[/tex]

[tex]\mathbf{\dfrac{5\times 6 \times 4}{2} = \boxed{a}rea}[/tex]

[tex]\mathbf{\dfrac{30\times 4}{2} = \boxed{a}rea}[/tex]

[tex]\mathbf{\dfrac{120}{2} =\boxed{a}rea}[/tex]

[tex]\mathbf{\dfrac{120\div2}{2\div2} = \boxed{a}rea}[/tex]

[tex]\mathbf{\dfrac{60}{1} = \boxed{a}rea}[/tex]

[tex]\mathbf{60\div1 = \boxed{a}rea}[/tex]

[tex]\mathbf{60 = \boxed{a}rea}[/tex]

[tex]\huge\textbf{Met the result....}[/tex]

[tex]\huge\text{Thus, your answer should be: \boxed{\mathsf{60\ in^2}}}\huge\checkmark[/tex]

[tex]\huge\text{Good luck on your assingment \& enjoy your day!}[/tex]