help me plsss i really need did that but still dont get it

The grades of Path A and Path B are calculated by dividing the vertical change by the horizontal distance. Path A has a grade of approximately 35%, while Path B has a grade of approximately 71%.

The grade of a path or road is calculated by taking the vertical rise and dividing it by the horizontal distance, typically expressed as a percentage. In Path A, the vertical change is 11 meters and the horizontal distance is 31 meters. Therefore, the grade of Path A is calculated as follows:

So, Path A has a grade of approximately 35% when rounded to the nearest percent.

For Path B, the vertical change is 3 meters and the horizontal distance is 4.25 meters. Therefore, the grade of Path B is calculated as follows:

So, Path B has a grade of approximately 71% when rounded to the nearest percent.

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Answer:  The value of 'h' is 3.

Step-by-step explanation:  Given that the vertex form of a function is given by

[tex]g(x)=a(x-h)^2+k~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]

We are to find the value of 'h' when the following function is converted to the vertex form.

[tex]g(x)=x^2-6x+14~~~~~~~~~~~~~~~~~~~~~~(ii)[/tex]

From equation (ii), we have

[tex]g(x)=x^2-6x+14\\\\\Rightarrow g(x)=x^2-2\times x\times 3+3^2-3^2+14\\\\\Rightarrow g(x)=(x-3)^2-9+14\\\\\Rightarrow G(x)=(x-3)^2+5.[/tex]

Comparing it with the vertex form (i), we get

[tex]h=3.[/tex]

Thus, the value of 'h' is 3.

The value of h in the vertex form is -3.

in the vertex form:

g(x)=a(x−h)^2 + k

h is the x-value of the vertex.

Remember that for the general quadratic equation:

y = a*x^2 + b*x + c

The vertex is at:

h = -b/2a

So in our equation:

g(x) = x^2 - 6x + 14

We will have:

h = -(-6)/2*1 = 3

h = 3

That is the value of h.

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Answer:

-64

Step-by-step explanation:

Plato/Edmentum users -64 is correct.

we are given

[tex]f(x)=4(3\sqrt{81} )^x[/tex]

Domain:

we know that domain is all possible values of x for which any function is defined

Here , since, x is only exponent

so, we can take any values of x

it will be defined for all values of x

so, domain is all real numbers

{x | x is a real number}

Range:

we know that

range is all possible values of f(x) or y

Since, there is no negative sign here

so, f(x) will always be positive and greater than 0

so, range is y>0

Answer: {x| x is a real number}; {y| y>0}

Step-by-step explanation:

we know that

The volume of a rectangular prism as

[tex]V=A*h[/tex]

where

V is volume

A is area

H is height

now, we are given

[tex]V=2x^3+9x^2-8x-36[/tex]

[tex]h=x+2[/tex]

now, we can find A

[tex]V=A*h[/tex]

[tex]A=\frac{V}{h}[/tex]

now, we can plug it

[tex]A=\frac{2x^3+9x^2-8x-36}{x+2}[/tex]

now, we can synthetic division method

so, we can write it as

[tex]A=\frac{2x^3+9x^2-8x-36}{x+2}=(2x^2+5x-18)[/tex]

so, the area of base is

[tex]=(2x^2+5x-18)[/tex].............Answer

Pythagorean theorem is used only for the right angled triangles.

For ΔABC,

Cosine rule is

[tex] a^2=b^2 + c^2-2bc$ cosA $  [/tex]

[tex] b^2=a^2 + c^2-2ac$ cosB $  [/tex]

[tex] c^2=a^2 + b^2-2ab$ cosC $  [/tex]

Here we have used one of these formulae

Hence definition of cosine or Cosine rule is the right answer

Option 2) is the right answer

The missing reason in Step 8 is substitution of a² = b² – 2bc·cosA + c²

The law of cosines relates the lengths of the sides of a triangle to the cosine of one of its angles.

Given is a proof using triangles property and rule,

In △ ABC, BD ⊥ AC

Prove: the formula for the law of cosines, a² = b² + c² – 2bccos

         Statements                                                    Reasons

1.  In △ ABC, BD ⊥ AC                                               1. given

2.  In △ ADB, c²  = x²  + h²                                 2. Pythagorean thm.

3. In △ BDC, a² = (b – x)² + h²                           3. Pythagorean thm.  

4.  a²  = b²  – 2bx + x²  + h²                               4. prop. of multiplication

5. a² = b² – 2bx + c²                                           5. substitution

6. In △ ADB, cos(A) = x/6                                   6. def. cosine

7. ccos(A) = x                                                       7. mult. prop. of equality

8.  a² = b² – 2bccos(A) + c²                                8. ?

9. a² = b² + c² – 2bccos(A)                               9. commutative property

We are asked to find the reason used in step 8,

From step 5) we have, a² = b² – 2bx + c².......(i)

From step 7) we have, ccos(A) = x

So, by substituting ccosA for x in eq(i), we get the equation for step 8)

a² = b² – 2bc·cosA + c²

Hence, the missing reason in Step 8 is substitution of a² = b² – 2bc·cosA + c²

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The polynomial [tex]\( 4x^4 - 4x^3 + 10x^6 \)[/tex] is classified as a 6th degree trinomial. The correct option is d) 6th degree trinomial.

To classify the polynomial [tex]\( 4x^4 - 4x^3 + 10x^6 \)[/tex]:

Identify the degree of each term:

  - [tex]\( 4x^4 \)[/tex] has a degree of 4.

  - [tex]\( 4x^3 \)[/tex] has a degree of 3.

  - [tex]\( 10x^6 \)[/tex] has a degree of 6.

Determine the degree of the polynomial:

The degree of the polynomial is the highest degree of its terms, which is 6.

Count the number of terms: The polynomial has three terms.

Thus, the polynomial has 6th degree and number of terms trinomial.

The complete question is:

Choose the correct classification of [tex]4x^4-4x^3+10x^6.[/tex]

a) 4th degree trinomial

b) 12th degree trinomial

c) 3rd degree trinomial

d) 6th degree trinomial

Answer:

40B+30M≤300

200B+80M≤1200

Step-by-step explanation:

Final answer:

The expected value of the game where you pay $3.00 to roll a die and win $5.00 for rolling a 1 or a 6 is -$0.33 per roll, indicating a loss over time.

Explanation:

The problem is asking us to calculate the expected value of the game in which you pay $3.00 to roll a fair die with the potential of winning $5.00 for rolling a 1 or a 6. To find the expected value, we need to multiply the outcomes by their respective probabilities and then sum these products.

The probability of rolling a 1 or a 6 is 1/6 for each number, so the combined probability for these winning rolls is 1/6 + 1/6 = 1/3. The probability of rolling a 2, 3, 4, or 5 is therefore 2/3 since these are the non-winning rolls.

Let's calculate the expected value (EV):

Winning: (1/3) * $5.00 = $1.67

Losing: (2/3) * -$3.00 = -$2.00

Now we add the two values:

EV = $1.67 - $2.00 = -$0.33

Therefore, the expected value of playing the game is a loss of $0.33 per roll. This means that in the long run, you can expect to lose an average of $0.33 for each time you play this game.

since u already have a answer and brainly isn't letting me ask a question i have to ask it here...... how did you figure out what the max height was for the rocket.

The rocket will reach it's maximum height in 4.98 seconds.

The rocket would return to earth in 9.96 seconds from the time it was launched.

Equations of motion are the equations which relate quantities like velocity, displacement, time and acceleration.

There are three equations of motion, which are:

First equation of motion, v = u + at

Second equation of motion, s = ut + [tex]\frac{1}{2}[/tex] at²

Third equation of motion, v² = u² + 2as

where, u is the initial velocity, v is the final velocity, a is the acceleration, t is the time and s is the displacement.

(a) Given that a model rocket is launched from the ground with an initial velocity of 160 feet/sec.

So, u = 160 feet/sec. = 160 × 0.305 meter/second = 48.8 meter/sec.

At maximum height velocity will be equal to zero.

So, v = 0

Here value of the acceleration will be g, acceleration of gravity = 9.8 m/s². And the sign of g will be negative because motion is in upward direction against gravity.

So, a = -g = -9.8 m/s².

From first equation of motion,

v = u + at

0 = 48.8 + (-9.8)t

9.8 t = 48.8

t = 48.8 / 9.8

t = 4.98 seconds.

(b) If the model rocket’s parachute failed to deploy and the rocket fell back to the ground, the time taken to reach the ground from it's maximum height will be the same as the time taken to reach the maximum height from the ground.

So total time taken = 4.98 + 4.98 = 9.96 seconds

Hence, the rocket will reach it's maximum height in 4.98 seconds.

The rocket would return to the Earth from the time it was launched in 9.96 seconds.

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Equivalent expressions are expressions with the same value.

The values of the variables are:

[tex]\mathbf{a = 1}[/tex]      [tex]\mathbf{b = 9}[/tex]      [tex]\mathbf{c = -2}[/tex]       [tex]\mathbf{d = 4}[/tex]

The expression is given as:

[tex]\mathbf{\frac{3x^2 + 2x - 21}{-2x^2 -2x + 12} \cdot \frac{2x^2 + 25x + 63}{6x^2 + 7x - 49}}[/tex]

Expand

[tex]\mathbf{\frac{3x^2 + 2x - 21}{-2x^2 -2x + 12} \cdot \frac{2x^2 + 25x + 63}{6x^2 + 7x - 49 } = \frac{3x^2 + 9x - 7x - 21}{-2x^2 +4x -6x + 12} \cdot \frac{2x^2 + 14x + 9x + 63}{6x^2 + 21x - 14x - 49 } }[/tex]

Factorize

[tex]\mathbf{\frac{3x^2 + 2x - 21}{-2x^2 -2x + 12} \cdot \frac{2x^2 + 25x + 63}{6x^2 + 7x - 49 } = \frac{3x(x + 3) - 7(x + 3)}{-2x(x -2) -6(x - 2)} \cdot \frac{2x(x + 7) + 9(x + 7)}{3x(2x + 7) - 7(2x - 7) } }[/tex]

Factor out the terms

[tex]\mathbf{\frac{3x^2 + 2x - 21}{-2x^2 -2x + 12} \cdot \frac{2x^2 + 25x + 63}{6x^2 + 7x - 49 } = \frac{(3x - 7) (x + 3)}{(-2x -6)(x - 2)} \cdot \frac{(2x + 9)(x + 7)}{(3x - 7) (2x - 7) } }[/tex]

Cancel out 3x - 7

[tex]\mathbf{\frac{3x^2 + 2x - 21}{-2x^2 -2x + 12} \cdot \frac{2x^2 + 25x + 63}{6x^2 + 7x - 49 } = \frac{(x + 3)}{(-2x -6)(x - 2)} \cdot \frac{(2x + 9)(x + 7)}{ (2x - 7) } }[/tex]

Factor out -2

[tex]\mathbf{\frac{3x^2 + 2x - 21}{-2x^2 -2x + 12} \cdot \frac{2x^2 + 25x + 63}{6x^2 + 7x - 49 } = \frac{(x + 3)}{-2(x +3)(x - 2)} \cdot \frac{(2x + 9)(x + 7)}{ (2x - 7) } }[/tex]

Cancel out x + 3

[tex]\mathbf{\frac{3x^2 + 2x - 21}{-2x^2 -2x + 12} \cdot \frac{2x^2 + 25x + 63}{6x^2 + 7x - 49 } = \frac{1}{-2(x - 2)} \cdot \frac{(2x + 9)(x + 7)}{ (2x - 7) }}[/tex]

Rewrite as:

[tex]\mathbf{\frac{3x^2 + 2x - 21}{-2x^2 -2x + 12} \cdot \frac{2x^2 + 25x + 63}{6x^2 + 7x - 49 } = \frac{(2x + 9)(x + 7)}{ -2(x - 2)(2x - 7) } }[/tex]

Expand

[tex]\mathbf{\frac{3x^2 + 2x - 21}{-2x^2 -2x + 12} \cdot \frac{2x^2 + 25x + 63}{6x^2 + 7x - 49 } = \frac{2x^2 + 25x + 63}{ -4x^2 + 22x - 28}}[/tex]

Factorize again

[tex]\mathbf{\frac{3x^2 + 2x - 21}{-2x^2 -2x + 12} \cdot \frac{2x^2 + 25x + 63}{6x^2 + 7x - 49 } = \frac{(2x+ 7)(x + 9)}{(2x + 7)(-2x + 4)}}[/tex]

Cancel out common factors

[tex]\mathbf{\frac{3x^2 + 2x - 21}{-2x^2 -2x + 12} \cdot \frac{2x^2 + 25x + 63}{6x^2 + 7x - 49 } = \frac{x + 9}{-2x + 4}}[/tex]

From the question, we have:

[tex]\mathbf{\frac{ax + b}{cx + d}}[/tex]

So, we have:

[tex]\mathbf{\frac{ax + b}{cx + d} = \frac{x + 9}{-2x + 4}}[/tex]

By comparison, we have:

[tex]\mathbf{a = 1}[/tex]

[tex]\mathbf{b = 9}[/tex]

[tex]\mathbf{c = -2}[/tex]

[tex]\mathbf{d = 4}[/tex]

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Answer:

[tex](x+1)\cdot(x^2-9)[/tex]

[tex](x+1)\cdot(x-3)\cdot(x+3)[/tex]

Step-by-step explanation:

We can simplify the expression as follows:

[tex]x^3-9\cdot(x)+x^2-9[/tex]

[tex]x^3+x^2-9\cdot(x)-9[/tex]

we have a write the expression with a common factor of (x+1)

[tex]x^2\cdot(x+1)-9\cdot(x+1)[/tex]

[tex](x+1)\cdot(x^2-9)[/tex]

We can simplify (x²-9) as:

[tex](x-3)\cdot(x+3)=x^3-3\cdot(x)+3\cdot(x)-9[/tex]

Therefore the final form of the expression is:

[tex](x+1)\cdot(x-3)\cdot(x+3)[/tex]

The fourth option is the best option.

Answer:

d. total amount of commission earned

Step-by-step explanation:

A salesperson earns a weekly base salary plus a commission of 20% of all sales over the first $500.

This situation can be represented by the expression [tex]750+0.2(x-500)[/tex]

0.2(x-500) means total amount of commission earned.

This is because firstly its given that 20% is commission so we have 0.2.

The x-500 represents the sales over $500, where x is total sales. Like the person sold $800 worth of items so he will get 20% commission on $300.

Statement B is the one that is NOT always true.

Explanation of which statement is not always true among given mathematical statements.

To determine which statement is not always true among the given options, let's analyze each one:

Based on the analysis,

Statement B is the one that is NOT always true.

Final answer:

The function modeling the height of the cone in terms of its radius, given the volume of 180 in³, is h(r) = 540 / (πr²).

Explanation:

The volume V of a cone with radius r and height h is given by the formula V = (1/3)πr²h, where π is Pi (approximately 3.14159). Given that the volume of the cone is 180 in³, we can solve for the height h in terms of the radius r.

Starting with the formula:

V = (1/3)πr²h

We plug in the volume:

180 = (1/3)πr²h

To solve for h, we rearrange the equation:

h = 3*180 / (πr²)

h = 540 / (πr²)

So the function modeling the height h of the cone in terms of its radius r is:

h(r) = 540 / (πr²)