how to graph

y=-1/2x + 3
y=5

The matching of the inequality with the statement will be 1-B, 2-D, 3-A, and 4-C.

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 matching of inequality with the statement is given as,

The matching of the inequality with the statement will be 1-B, 2-D, 3-A, and 4-C.

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A TV has a width of 45 inches and a height of 30inches the length of the diagonal of the TV is 54.08inches.

What is Rectangle?

A rectangle is a quadrilateral with four right angles in the Euclidean plane. It can alternatively be described as a parallelogram with a right angle or an equiangular quadrilateral, where equiangular denotes that all of its angles are equal. A square is a rectangle with four equally long sides.

Make a diagonal and a rectangle. Two parallel triangles appear. Pay attention to one of the triangles.

We have a right triangle with a = 45 and b = 30

we have to find the c = ?
Let's find b using the Pythagorean Theorem;

a^2 + b^2 = c^2

45^2 + 30^2 = c^2

2025 + 900 = c^2

2925 = c^2

c = 54.08 inches

Therefore,the diagonal of the TV is 54.08inches.

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The solution to the system of equations in this problem is given as follows:

(2,8).

The system of equations for this problem is defined as follows:

To eliminate the variable y, we multiply the first equation by 3 and the second by 5, hence:

Adding the two equations, the solution for x is obtained as follows:

47x = 94

x = 94/47

x = 2.

Hence the solution for y is obtained as follows:

4(2) + 5y = 48

5y = 40

y = 40/5

y = 8.

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The line inside the circle depicted in the figure is radius.

A circle is defined as the round shape lacking corners and line. However, it does comprise of all the points in a plane. The center of the circle is the middle point which is equidistant from the outer surface of the circle. The outer surface or boundary of circle is called circumference.

A line joining the centre of the circle with circumference is called radius and the same is exhibited in the adjoining figure. A line that passes through the centre of the circle and touches the circumference at two ends is called diameter. Diameter is double in size to the radius.

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The complete question is attached in the figure.

An expression to represent the number of sheets Alexia used is y = 3x + 40.

In math, expressions are mathematical statements with a minimum of two terms containing numbers or variables, or both, connected by an operator in between.

The mathematical operators can be any of addition, subtraction, multiplication, or division. For example, p + q is an expression where p and q are terms having an addition operator in between.

There are two types of expressions, numerical expressions - which contain only numbers, and algebraic expressions- which contain both numbers and variables.

let

x = the number of sheets per ream

y = the total number of sheets

Alexia used three reams of paper, so the total number of sheets per ream will be 3x since each ream has the same number of sheets.

After that, she used another 40 sheets. Therefore the expression will become as

y = 3x + 40 represents the total number of sheets Alexia used.

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The surface area of the rectangular prism is 112 square metes.

The sum of all area of each surface that make up an object is referred to as its surface area.

So that in the given question,

Area of a rectangle = length x width

i. area of its base = length x width

                            = 8 x 2

                            = 16 sq. m

ii. area of its front surface = length x width

                                      = 4 x 8

                                      = 32 sq. m

iii. area of it one of its sides = length x width

                                    = 4 x 2

                                    = 8 sq. m

Therefore,

surface area of the rectangular prism = (2 x 16) + (2 x 32) + (2 x 8)

                                       = 32 + 64 + 16

                                       = 112

The surface area of the rectangular prism is 112 square meters.

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The resultant of the given expression (x² - 5x) (2x²+x-3) is 2x^4 - 7x^3 -8 x^2 + 15x .

What is Algebraic expression ?

Algebraic expression can be defined as the combination of variables and constants.

Given expression,

(x² - 5x) (2x²+x-3)

= x^2 ( 2x^2+x-3) - 5x(2x^2 + x - 3)

= 2x^4 + x^3 - 3x^2 - 5x * 2x^2 + x*-5x - 3 * -5x

= 2x^4 + x^3 -3x^2  - 10 x^3 - 5x^2 + 15x

= 2x^4 + x^3 - 10x^3 -3x^2 - 5x^2 + 15x

= 2x^4 - 7x^3 -8 x^2 + 15x .

Hence, The resultant of the given expression (x² - 5x) (2x²+x-3) is 2x^4 - 7x^3 -8 x^2 + 15x .

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The angle PAB is given as 40 degrees so we can use the sine rule to calculate the lengths of the chords is CQD = 17.7 degrees.

The sine rule states that the ratio of the length of a side of a triangle to the sine of its opposite angle is equal to the ratio of any other side to the sine of its opposite angle.

Let us denote the length of chord AB by x and the length of chord CD by y.

Therefore x/sin 40 = y/sin (180-40)

=> x/0.766 = y/0.939

=> x = 0.766*0.939 = 0.723

y = 0.939*0.766 = 0.723

Since AB=CD, x=y=0.723.

We can now use the cosine rule to calculate the angle CQD.

The cosine rule states that the square of the length of a side of a triangle is equal to the sum of the squares of the other two sides minus twice the product of the other two sides times the cosine of the included angle.

Therefore CQ^2 = 8^2 + 8^2 - 2(8^2)cosCQD

=> CQ^2 = 64 - 64cosCQD

=> 64 - 64cosCQD = 0.723^2

=> 64 - 64cosCQD = 0.521

=> 64cosCQD = 63.479

=> cosCQD = 63.479/64

=> cosCQD = 0.988

Therefore CQD = cos^-1(0.988)

=> CQD = 17.7 degrees.

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Answer:

y=-3.5(x−5)^2+17

Step-by-step explanation:

Let's assume that since this equation has a vertex that it is a parabola with the form y = ax^2 + bx + c.

Let's rewrite it in vertex form as:

y=a(x−h)^2+k

where h and k are the horizontal (h) and vertical (k) coordinates of the vertex.

The vertex of (5,17) gives us both h (5) and k (17),  Substituting these values gives us:

  y=a(x−5)^2+17

To find a, use the one given point the parabola intersects (11,-109) and solve for a:

y=a(x−5)^2+17

-109=a((11)−5)^2+17

-109=a(6)^2+17

-109-17=a(36)

36a = -126

a = -3.5

This leads us to:

y=-3.5(x−5)^2+17

See the attached graph.

"The domain of 9(x) = f(x-2) - 8 is -3 ≤ x ≤ 9 and the range is -6 ≤ y ≤ 7" is correct.

What is the Domain and Range of the function?

The domain of a function is the set of all input values (x-values) for which the function is defined. It is the set of all values of x for which the function is defined.

The range of a function is the set of all output values (y-values) that the function can produce. It is the set of all possible values that the function can take on.

The domain of a function is the set of all input values (x-values) for which the function is defined. The range of a function is the set of all output values (y-values) that the function can produce.

When we shift a function horizontally, the domain remains unchanged, but the range will change. In the case of a shift of -2, the domain of f(x) is still -3≤x≤9

For 9(x) = f(x-2) - 8, it means that we are taking the output of f(x-2) and subtracting 8 from it. So, the range of 9(x) will be (2-8) ≤ y ≤ (15-8) = -6 ≤ y ≤ 7.

In summary,

The domain of 9(x) = f(x-2) - 8 is -3 ≤ x ≤ 9

The range of 9(x) = f(x-2) - 8 is -6 ≤ y ≤ 7

Hence, statement "The domain of 9(x) = f(x-2) - 8 is -3 ≤ x ≤ 9 and the range is -6 ≤ y ≤ 7" is correct.

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The variable 'time to destination' quantitative.

Quantitative variables are those whose values are obtained through measurement or counting. A qualitative variable, also referred to as a category variable, is a non-numerical variable. It describes information that can be categorized. Examples like Eye colors, States, Dog breeds etc.

Any variable that isn't numerical is referred to as a qualitative variable or a category variable. It describes information that falls into a category. Generally speaking, a variable is considered to be quantitative if it can be used mathematically (such as addition). Otherwise, it's top-notch. You cannot, for instance, add blue and green.

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Using the constant of proportionality the value for B in the table is obtained as option C: B = 40.

What is a COP?

The ratio between two quantities that are directly proportional is the constant of proportionality. When two quantities grow and shrink at the same rate, they are directly proportional.

The table contains three columns A, C, D.

The first row contains all the part values - 14, 28, 63

The second row contains all the whole values - 20, B, 90

It is necessary to calculate the constant of proportionality.

In this case, for A, we have 14 part and 20 whole, the constant will be -

COP = Whole / Part

COP = 20 / 14

COP = 1.42857

In this case, for C, we have 63 part and 90 whole, the constant will be -

COP = Whole / Part

COP = 90 / 63

COP = 1.42857

Since, the constant of proportionality is same for A and C, it will be same for B also.

In this case, for B, we have 28 part and B whole, the constant will be -

COP = Whole / Part

1.42857 = B / 28

B = 1.42857 × 28

B = 40

Therefore, the value for B is obtained as 40.

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A three-column table is given. Part A C D Part 14 28 63 Whole 20 B 90 What is the value of B in the table? 64 50 40 12

The sum of a+b+c+d+e+f=57.

Apparently, you want to simplify: [tex]\frac{1}{\sqrt{2} +\sqrt{3} +\sqrt{7} }[/tex]

so the denominator is rational. It looks like the form you want is:

[tex]\frac{A\sqrt{2}+B\sqrt{3} +C\sqrt{7} +D\sqrt{E} }{F}[/tex]

And you want to know the sum A+B+C+D+E+F.

We can start by multiplying the numerator and denominator by a conjugate of the denominator. Then we can multiply the numerator and denominator by a conjugate of the resulting denominator.

[tex]\frac{1}{\sqrt{2} +\sqrt{3} +\sqrt{7}} .\frac{\sqrt{2} +\sqrt{3} -\sqrt{7}}{\sqrt{2} +\sqrt{3} -\sqrt{7}} =\frac{\sqrt{2} +\sqrt{3} -\sqrt{7}}{2\sqrt{6-2} } \\\\\\=\frac{\sqrt{2} +\sqrt{3} -\sqrt{7}}{2\sqrt{6} -2} .\frac{\sqrt{6}+1 }{\sqrt{6} +1} =\frac{(1+\sqrt{6} )(\sqrt{2} +\sqrt{3} -\sqrt{7})}{10} \\\\\\=\frac{\sqrt{2} +\sqrt{3} -\sqrt{7}+2\sqrt{3}+3\sqrt{2} -\sqrt{42} }{10} =\frac{4\sqrt{2} +3\sqrt{3} -\sqrt{7} -\sqrt{42} }{10}[/tex]

Comparing this to the desired form we have:

A = 4, B = 3, C = -1, D = -1, E = 42, F = 10

Then the sum is : A +B +C +D +E +F = 4 + 3 -1 -1 +42 +10 = 59 -2 = 57

The sum of interest is 57.

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[tex]\frac{3x^{2} }{x^{2} +x-6}[/tex] is the product of the rational expressions.

What are rational expressions?

Two polynomials' ratio is shown through rational expressions. It denotes a polynomial in the denominator as well as the numerator. It is an algebraic expression that contains an unknown variable and is similar to a fraction in that it is a ratio.

We are given [tex]\frac{3x^{2} }{x^{2} +x-6}[/tex]

This means that the expression has to be multiplied by the number 1 for obtaining the product of the rational expression.

We know that when anything is multiplied by the number 1, we get the number itself.

Hence, [tex]\frac{3x^{2} }{x^{2} +x-6}[/tex] is the product of the rational expressions.

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The height of the building is 30 feet.

Annette is standing with a tree directly between her and a building. She stands so that, in her eyes, the top of the tree appears to "touch" the top of the building.  

The distance between her and the tree is 30 ft, the distance between the tree and the building is 15 ft, and the tree is 20 ft tall.

Therefore, the height of the building can be calculated as follows:

The situation forms two right angle triangle.

Therefore, using similar triangle ratio,

20 / 30 = x / 45

where

Hence,

45 × 20 = 30x

900 = 30x

x = 900 / 30

x = 30

Therefore,

height of the building  = 30 ft

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By cancelative property, the roots of quadratic equation x² + 4 · x - 20 are: Exact: - 2 + 2√6, - 2 - 2√6, Approximate: 2.899, -6.899.

In this problem we must factorize a quadratic equation and use an algebraic property known as cancelative property, defined as:

If a · b = 0, then a = 0 or b = 0.

In addition, we can apply the following factorization rule:

x² - (r₁ + r₂) · x + r₁ · r₂ = (x - r₁) · (x - r₂)

First, factorize the quadratic equation:

x² - (- 2 + 2√6 - 2 - 2√6) · x + (- 2 + 2√6) · (- 2 - 2√6)

(x + 2 - 2√6) · (x + 2 + 2√6)

Second, apply cancelative property:

x + 2 - 2√6 = 0

x = - 2 + 2√6

x + 2 + 2√6 = 0

x = - 2 - 2√6

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Answer:

$1,684.25

Step-by-step explanation:

(.85 x 1250) = 1062.50  If we take off 15% we leave on 85% which as a decimal is .85

Now add the tax

1062.50 + .06(1062.50)

1620.50 + 63.75

1684.25

Answer:

F(x)=-x^2

Step-by-step explanation:

The one example of function with range of only negative number is:

F(x)=-x^2

Why this? We know for every x, x^2 is positive. So -x^2 will be negative.

The result of the division of the number by 100 gives 3.05109 in decimal.

A decimal is really a number that is divided into two parts: another whole and a fraction.

For the given number.

= 305.109 ÷ 100

Then point will shift to 2 decimals towards left.

= 3.05109

Thus, the result of the division of the number by 100 gives 3.05109 in decimal.

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A graph of the solutions of this inequality 3x + 2 ≥ 1/2(x + 5) is shown on the number line attached below.

In this exercise, we would solve the given inequality by making x the subject of formula as follows;

3x + 2 ≥ 1/2(x + 5)

By multiplying both sides of the inequality by 2, we have the following:

(3x + 2) × 2 ≥ 1/2(x + 5) × 2

6x + 4 ≥ x + 5

By rearranging and collecting like terms, we have the following:

6x - x ≥ 5 - 4

5x ≥ 1

x ≥ 1/5

Next, we would use an online graphing calculator to plot the inequality as shown in the graph attached below. Additionally, the circle on the number line (graph) is closed because of the greater than or equal to (≤) symbol.

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The roots of quadratic equations:

In this question we need to determine sixteen cases of quadratic equations, whose roots should be determined. This can be done by following procedure:

Now we proceed to show the solution to each case:

Case 1

b² - 4 = 0

b² = 4

b = ± 4

Case 2

4 · a² - 1 = 0

4 · a² = 1

a² = 1 / 4

a = ± 1 / 2

Case 3

2 · x² = 21 + 11 · x

2 · x² - 11 · x - 21 = 0

2 · [x² - (11 / 2) · x - 21 / 2] = 0

2 · [x² - (11 / 2) · x + 121 / 16] = 2 · (121 / 16) + 2 · (21 / 2)

2 · (x - 11 / 4)² = 289 / 8

(x - 11 / 4)² = 289 / 16

x - 11 / 4 = ± 17 / 4

x = 11 / 4 ± 17 / 4

Case 4

2 · n² = 32

n² = 16

n = ± 4

Case 5

5 · x² = 3 · x + 92

5 · x² - 3 · x - 92 = 0

5 · [x² - (3 / 2) · x - 92 / 5] = 0

5 · [x² - (3 / 2) · x] = 5 · (92 / 5)

5 · [x² - (3 / 2) · x + 9 / 16] = 5 · (92 / 5 + 9 / 16)

5 · (x - 3 / 4)² = 1517 / 16

(x - 3 / 4)² = 1517 / 80

x - 3 / 4 = ± √(1517 / 80)

x = 3 / 4 ± √(1517 / 80)

Case 6

4 · x² = 7 · x + 92

4 · x² - 7 · x - 92 = 0

4 · [x² - (7 / 4) · x - 23] = 0

4 · [x² - (7 / 4) · x + 49 / 64] = 92 + 49 / 64

4 · (x - 7 / 8)² = 5937 / 64

x - 7 / 8 = ± √(5937 / 256)

x = 7 / 8 ± √(5937 / 256)

Case 7

5 · k² = - 7 · k + 138

5 · k² + 7 · k - 138 = 0

5 · [k² + (7 / 5) · k - 138 / 5] = 0

5 · [k² + (7 / 5) · k] = 5 · (138 / 5)

5 · [k² + (7 / 5) · k + 49 / 100] = 5 · (138 / 5 + 49 / 100)

5 · (k + 7 / 10)² = 2809 / 20

k + 7 / 10 = ± √(2809 / 100)

k = - 7 / 10 ± √(2809 / 100)

Case 8

m² - 88 = 3 · m

m² - 3 · m - 88 = 0

m² - 3 · m - 352 / 4 = 0

m² - 3 · m = 352 / 4

m² - 3 · m + 9 / 4 = 361 / 4

(m - 3 / 2)² = 361 / 4

m - 3 / 2 = ± √(361 / 4)

m = 3 / 2 ± √(361 / 4)

Case 9

4 · n² - 7 = 12 · n

4 · n² - 12 · n = 7

4 · (n² - 3 · n + 9 / 4) = 7 + 9

4 · (n - 3 / 2)² = 16

n - 3 / 2 = ± 4

n = 3 / 2 ± 4

Case 10

6 · n² - 2 · n = 140

6 · [n² - (1 / 3) · n] = 140

6 · [n² - (1 / 3) · n + 1 / 36] = 140 + 1 / 6

6 · (n - 1 / 6)² = 841 / 6

n - 1 / 6 = ± 841 / 36

n = 1 / 6 ± 841 / 36

Case 11

4 · p² = 64

p² = 16

p = ± 4

Case 12

6 · v² - 10 · v = 84

6 · [v² - (5 / 3) · v] = 84

6 · [v² - (5 / 3) · v + 25 / 36] = 84

6 · (v - 5 / 6)² = 84

(v - 5 / 6)² = 14

v - 5 / 6 = ± 14

v = 5 / 6 ± 14

Case 13

3 · r² - 77 = - 10 · r

3 · r² + 10 · r = 77

3 · [r² + (10 / 3) · r] = 77

3 · [r² + (10 / 3) · r + 25 / 9] = 77 + 25 / 9

3 · (r + 5 / 3)² = 718 / 9

(r + 5 / 3)² = 718 / 27

r + 5 / 3 = ± √(718 / 27)

r = - 5 / 3 ± √(718 / 27)

Case 14

5 · x² - 11 · x = - 6

5 · [x² - (11 / 5) · x + 121 / 100] = - 6 + 5 · (121 / 100)

5 · (x - 11 / 10)² = 1 / 20

(x - 11 / 10)² = 1 / 100

x - 11 / 10 = ± √(1 / 10)

x = 11 / 10 ± √(1 / 10)

Case 15

2 · n² = 11 · n + 6

2 · n² - 11 · n = 6

2 · [n² - (11 / 2) · n] = 6

2 · [n² - (11 / 2) · n + 121 / 16] = 6 + 2 · (121 / 16)

2 · (n - 11 / 4)² = 169 / 8

(n - 11 / 4)² = 169 / 16

n - 11 / 4 = ± 13 / 4

n = 11 / 4 ± 13 / 4

Case 16

4 · x² = 100

x² = 25

x = ± 5

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Answer:

5,7

Step-by-step explanation:

There is not enough paint for all the canvases. 26.7 cm² of more white paint is required.

Area of each canvas = Length × width

Area of each canvas = 5.3 × 8.5

Performing multiplication

Area of each canvas = 45.05 cm²

Total area of canvas in the class = 45.05 × 30

Performing multiplication

Total area of canvas in the class = 1351.5 cm²

Available amount of white paint = 6 × 220.8

Performing multiplication

Available amount of white paint = 1324.8 cm²

The available amount is less than canvas area in the class.

Amount of paint required = 1351.5 - 1324.8

Performing subtraction

Amount of paint required = 26.7 cm²

Thus, 26.7 cm² of white paint is required.

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Part 1

The area of one triangle is [tex](1/2)(7)(10)=35[/tex]. Multiplying this by 4 yields [tex]\boxed{140}[/tex].

Part 2

[tex](10+7)^2=\boxed{289}[/tex]

Part 3

[tex]289-140=\boxed{149}[/tex]

Part 4

[tex]10^2+7^2=\boxed{149}[/tex]

Part 5

The statement is missing.

1) The total combined area of the four triangles is: 140; 2) The area of the outer square is c² = c * c; 3) The area of the small inner square is: c² - 70; 4) The sum of the squares of the two legs of a right triangle is: 149.

Part 1:

The area of each triangle is (a * b) / 2, so the total combined area of the four triangles is 4 * (a * b) / 2 = 2 * 7 * 10 / 2 = 70.

Part 2:

The side length of the outer square is c, the hypotenuse, so the area of the outer square is c² = c * c.

Part 3:

The area of the small inner square is the area of the outer square minus the total combined area of the four triangles, so it is c² - 2 * 7 * 10 / 2 = c² - 70.

Part 4:

The formula for the sum of the squares of the two legs of a right triangle is a² + b² = c², so a² + b² = 7² + 10² = 49 + 100 = 149.

Part 5:

c > a and c > b, so c > 7 and c > 10. Therefore, c must be the largest of the three values: a, b, and c.

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To graph the algebraic rule y = 3x - 11, we need to consider input values ​​from -5 to 5 and also choose appropriate scaling for the x and y axes.

A common way to choose a scale is to use 1 or 2 increments on the x and y axes. This makes it easy to draw points and read charts easily.

The x-axis goes from -5 to 5, so you can scale from -5 -4 -3 -2 -1 0 1 2 3 4 5.

For the y-axis, I need to find the minimum and maximum values ​​of y given the input x values.

Substituting the x value into the formula y = 3x - 11 gives the y value.

y = 3(-5) - 11 = -16

y = 3(5) - 11 = 14

So for the y axis you can take -20 -15 -10 -5 0 5 10 15 20

So a good scale for the x-axis is -5 to 5 in increments of 1, and -20 to 20 in increments of 5 for the y-axis. Graphing the equation y = 3x - 11 using these scales makes the graph easier to read and understand.

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A mathematical expression is a phrase that has a minimum of two numbers or variables and at least one mathematical operation.

A term may be a number, a variable, the product of two or more variables, a number and a variable, or any combination of these. A single term or a collection of phrases can be used to create an algebraic expression. For instance, 4x and y are the two terms in the formula 4x + y.

Expressions that use numbers to perform operations. An example of a numerical expression is 2(3 + 8). At least one variable and one operation must be present in an algebraic expression (addition, subtraction, multiplication, division). One such algebraic expression is 2(x + 8y).

The following elements can be used to determine the algebraic expressions: a single variable or a group of related variables.

              = 6/6 /546/60

               =1 / 9.1

              =0.109

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The probability that he will choose a blue wristband is 6/25.

Probability is a method to know how likely something is going to happen. Whenever we're uncertain about the outcome of an event, we can talk about the probabilities of certain outcomes. The analysis of events overseen by probability is called statistics.

Total number of wristbands = 25

Number of blue colour wristbands = 6

As we know, the formula of experimental probability is

Probability of an Event P(E) = Total Number of times an event occurs/Total number of trials.

Number of blue colour wristbands = 6

Total number of wristbands = 25

∴ Probability of an Event P(Blue) = 6/25

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That X, w, and Z w are related we demonstrate that these are identical.

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