Find the value of x and the value

A: x=20,y=45
B:x=45,y=20
C:x=60,y=120
D:x=90,y=60

Answer:  The correct option is (C) 62500.

Step-by-step explanation:  We are given to find the 7-th term of the following geometric sequence :

4,   −20,   100,    .    .     .

We know that

the n-th term of a geometric sequence with first term a and common ratio r is given by

[tex]a_n=ar^{n-1}.[/tex]

For the given geometric sequence, we have

first term, a = 4

and common ratio r is given by

[tex]r=\dfrac{-20}{4}=\dfrac{100}{-20}=~~.~~.~~.~~=-5.[/tex]

Therefore, the 7-th term of the given geometric sequence is

[tex]a_7=ar^{7-1}=4\times(-5)^6=4\times 15625=62500.[/tex]

Thus, the required 7-th term is 62500.

Option (C) is CORRECT.

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

The value of x is 9.6.

To find the value of x to the nearest tenth, we can use the angle bisector theorem, which states that in a triangle, an angle bisector divides the opposite side into two segments that are proportional to the other two sides of the triangle.

Given that:

[tex]\[\frac{6.3}{5.4} = \frac{11.2}{x}\][/tex]

We can solve for x by cross-multiplying:

[tex]\[6.3 \cdot x = 5.4 \cdot 11.2\][/tex]

[tex]\[6.3x = 60.48\][/tex]

Now, divide both sides by 6.3 to isolate x:

[tex]\[x = \frac{60.48}{6.3}\][/tex]

[tex]\[x \approx 9.6\][/tex]

So, the value of x to the nearest tenth is 9.6. Therefore, the correct option is x = 9.6.

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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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:

40B+30M≤300

200B+80M≤1200

Step-by-step explanation:

Answer:

The slope is [tex]-\frac{350}{month}\[/tex]

Step-by-step explanation:

This saving account is declining a fixed amount every month, this means that the function that represents this decline is linear.

The other thing we know is that originally (september 1), the account has a balance of $4900, and this is the intercept.

Therefore, for a linear function depending of the variable "x", we can write that

[tex]f(x)= (-\frac{350}{month}x+4900)\[/tex]

where the units were factored. Then the slope is

[tex]-\frac{350}{month}\[/tex]

Use the theorem above to find the measure of angle formed by the intersection of the tangent that intersects chord AC.

By the theorem, the measure of angle is half of the intercepted arc which is 126.

Then, (1/2) · 126 = 63;

The correct answer is C. 63;

Answer:63

Step-by-step explanation:

Answer:

-64

Step-by-step explanation:

Plato/Edmentum users -64 is correct.

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:

Option D is correct.

The fourth graph shows a system of equation with two solutions.

Step-by-step explanation:

From the given figure,

we can see that we have a parabola and a straight line.

Since, the line is touching the parabola at two points, one point and no point.

In the first graph, the line touches the parabola at one point.

In the second graph, the line does not touches the parabola at no point.

also,In the third graph,the line touches the parabola at one point and

In fourth graph, the lines touches the parabola at two points.

For the system of equations :

Therefore, the graph which is most likely shows a system of equation with two solutions is: In fourth graph

Answer:

Two solution to a system of equation means when we look at the graph of the function the graph of the two equations must intersect  at exactly two points.

The graph satisfying such condition is attached to the answer.

in this graph the curve that is downward parabola and a line intersect at exactly two points and hence result in two solutions.

Final answer:

To find the product of (2x + 9) and (x + 1), we use the distributive property and combine like terms, resulting in 2x² + 11x + 9.

Explanation:

The student is asking for the correct product when multiplying the binomials (2x + 9) and (x + 1). This is a math problem involving algebraic multiplication.

To find the product, we apply the distributive property (also known as FOIL method in binomials):

Combine the terms to get the final product: 2x² + 2x + 9x + 9, which simplifies to 2x² + 11x + 9.

The correct product of (2x + 9)(x + 1) is 2x² + 11x + 9.

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.

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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The average time it takes to fill one bag of leaves is approximately 0.17 hours.

To find the average time to fill one bag of leaves, we use the formula

Average Time=Total Time÷Total Bags. In this scenario, the total time spent is the average workday hours, which is 9 hours, and the total number of bags filled is 36.

So,

Average Time=9hours÷36bags=0.25 hours/bag.

​Therefore, the average time it takes to fill one bag of leaves is 0.25 hours or 15 minutes. This means, on average, it takes 15 minutes to fill each bag of leaves.

In summary, the staff at Larry's Lawns spends 9 hours a day on lawn-related tasks, with mowing and raking taking 6 hours. Since they fill an average of 36 bags of leaves in that time, each bag takes approximately 0.25 hours or 15 minutes to fill, assuming a constant rate of filling.

To find the average time it takes to fill one bag of leaves at Larry's Lawns, divide the total time spent filling bags by the number of bags filled. Time per bag = 1/6 hours per bag.

To find the average time it takes to fill one bag of leaves, we need to divide the total amount of time spent filling bags by the number of bags filled. In this case, the staff at Larry's Lawns spend 6 hours mowing and raking and fill 36 bags of leaves. So to find the average time per bag, we divide 6 hours by 36 bags:

Time per bag = Total time / Number of bags = 6 hours / 36 bags

Since we want the average time per bag, we can simplify the fraction:

Time per bag = 1/6 hours per bag.

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: