Which is a valid prediction about the continuous
function f(x)?
Of(x) ≥ 0 over the interval [5, ∞).
Of(x) ≤0 over the interval [-1, %).
Of(x) > 0 over the interval (-∞, 1).
Of(x) < 0 over the interval (-∞. –1).

Answer:

92

Step-by-step explanation:

46 + 46 = 92

Answer: Option (4)

Step-by-step explanation:

We know that the radius of the circle is 3.

Using the distance formula, the distance from the given point to the center is

[tex]\sqrt{(\sqrt{6}-0)^{2}+(2-0)^{2}} = \sqrt{10} \neq 3[/tex]

Answer:d

Step-by-step explanation:

When the bird is 2 feet away from the first fence post, it reaches its minimum height of 4 feet  , Minimum at (2,4)

Therefore Option C and D is the correct answer.

The equation for a quadratic function is given by

y = a(x-h)² +k²

where (h,k) is the extreme value(vertex).

Here the equation given is

y = x²-4x +8

y = x²-4x+4+4

y = (x-2)² +4

Here the value of a = 1 , h = 2 and k = 4

The extreme value is at (2,4)

Therefore When the bird is 2 feet away from the first fence post, it reaches its minimum height of 4 feet  , Minimum at (2,4)

Therefore Option C and D is the correct answer.

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Answer:

38 is the correct solution

Answer:

The answer is C. -2x -4

Step-by-step explanation:

Your given information.

-3 - 3x - 1 + x

Combine and simplify common numbers/variables.

(-3x + x) + (-3 - 1)

-2x - 4

The solutions to the system of equation f(x) and g(x) are [tex]\frac{3+\sqrt{89}}{4} \ or \ \frac{3-\sqrt{89}}{4}[/tex] and

[tex]\frac{2\sqrt{15}}{3} \ or \ \frac{-2\sqrt{15}}{3}\\[/tex]

Quadratic equations are equations that has a leading degree of 2. Given the following functions:

f(x)=2x^(2)-3x-10

g(x)=-3x^(2)+20

To solve, we will use the quadratic formula as shown below;

[tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]

For the quadratic equation f(x)=2x^(2)-3x-10

a = 2

b = -3

c = -10

Substitute

[tex]x=\frac{-(-3)\pm \sqrt{(-3)^2-4(2)(-10)}}{2(2)}\\x=\frac{3\pm\sqrt{89}}{4}\\x= \frac{3+\sqrt{89}}{4} \ or \ \frac{3-\sqrt{89}}{4}[/tex]

For the quadratic equation -3x² + 20

a =-3

b = 0

c = 20

[tex]x=\frac{0\pm \sqrt{(0)^2-4(-3)(20)}}{2(-3)}\\x=\frac{\pm4\sqrt{15}}{-6}\\x=\frac{2\sqrt{15}}{3} \ or \ \frac{-2\sqrt{15}}{3}\\[/tex]

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The coordinates of the point that is 4/5 of the way from T to S is (2,10)

The given parameters are:

T = (6,-2)

S = (1,13)

Location = 4/5

The location 4/5 means

Ratio, m : n = 4 : 1

The coordinate is then calculated as:

[tex](x,y) = \frac{1}{m +n} * (mx_2 + nx_1,my_2 + ny_1)[/tex]

This gives

[tex](x,y) = \frac{1}{4 +1} * (4 * 1 + 1 * 6 , 4 * 13 + 1 * -2)[/tex]

Evaluate

[tex](x,y) = \frac{1}{5} * (10 , 50)[/tex]

This gives

(x,y) =  (2,10)

Hence, the coordinates of the point that is 4/5 of the way from T to S is (2,10)

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Answer:

Hey there : y= 3x - 3

Answer:

y = [tex]\frac{3}{2}[/tex] x + 3

Step-by-step explanation:

the equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

calculate m using the slope formula

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

with (x₁, y₁ ) = (- 2, 0) and (x₂, y₂ ) = (0, 3) ← 2 points on the line

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

the line crosses the y- axis at (0, 3 ) ⇒ c = 3

y = [tex]\frac{3}{2}[/tex] x + 3 ← equation of line

Answer:

4.1968

Explanation:

Given that:

angle: 25°

opposite: x

adjacent: 9

So, use the tan rule:

opposite/adjacent = tan(x)

x/9 = tan(25)

x = 9tan(25)

x = 4.1968

Option A

Answer:

I will assume the equation is supposed to be f(x) = x^2 – 5x + 12 since it is said to be a quadratic equation.

Step-by-step explanation:

See attached graph.

The value of f(–10) = 82     False

      f(-10) = (-10)^2 - 5*(-10) + 12

       f(-10) = (100) +50 + 12

      f(-10) = 162

The graph of the function is a parabola.  True

The graph of the function opens down.   False

The graph contains the point (20, –8).      False

The graph contains the point (0, 0).          False

Answer:

Vans - 7

Cars - 15

Step-by-step explanation:

$200 * 7 = $1400

$100 * 15 = $1500

$1400 + $1500 = $2900

1. 6 km ≈ 3.73 mi

2. 2.5 L ≈ .66 gal

3. 90 lb ≈ 40.82 kg

Answer:

A. (6x + 7)(6x - 7)

Step-by-step explanation:

use difference of squares, which says a^2 - b^2 =(a+b)(a-b)

take the square roots of the two numbers, 6x and 7, and use them as a and b

Answer:

pete

Step-by-step explanation:

i think its pete???

Using two triangle concepts, it is found that the solution for y is of y = 70º.

Hence:

Then:

x + y + 70 = 180

x + 190 - 3x + 70 = 180.

2x = 80.

x = 40.

Hence:

y = 190 - 3x = 190 - 3(40) = 70º.

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Answer:

25

Step-by-step explanation:

This is an equilateral triangle so all of its angle measures, and side lengths likewise, are equal.

To find x we can use the following equation:

3x - 5 = 2x + 20 export like terms to the same side of the equation

3x - 2x = 20 + 5 (notice how terms change from "+" to "-" and vice versa)

add/subtract like terms

x = 25

Answer:

4x² + x - 7

Step-by-step explanation:

Given f(x) = -3 and g(x) = 4x² + x - 4, solve for (f + g)(x).

Step 1: Replace the function designators with the actual functions in f(x) and g(x); that gives us 4x² + x - 4 - 3.

Step 2: Simplify; that gives us the answer 4x² + x - 7.

Answer:

625 × 48 = 30,000

[tex]\sqrt[4]{30000} = 10 \sqrt[4]{3} [/tex]

answer = third option

Answer:

(h, k )

Step-by-step explanation:

given the standard form of an absolute function

f(x) = a| x - h | + k , then

vertex = (h, k )

The standard form of an absolute value function is f(x) = a|x - h| + k, where (h,k) is the vertex of the absolute value function.

The correct option is (h,k).

The term "function" refers to a relationship between a set of inputs and outputs. Just one output is connected to each input in a function, which connects inputs in that manner. Every function has a domain, co-domain, and range.

In general, the standard form of an absolute value function is given as:

f(x) = a |x - h| + k

where:

a is a non-negative constant that determines the shape and direction of the V-shaped graph of the absolute value function

h is the horizontal shift of the graph of the function from the origin (if h > 0, the graph shifts to the right; if h < 0, the graph shifts to the left)

k is the vertical shift of the graph of the function from the origin (if k > 0, the graph shifts up; if k < 0, the graph shifts down)

The vertex of the absolute value function is the point where the graph changes direction, that is, the point where the graph changes from sloping downwards to sloping upwards or vice versa.

Since the vertex is the point of the absolute value function where the graph changes direction, it lies at the center of the V-shaped graph. Thus, the vertex is located at the point (h,k).

Therefore, in the standard form f(x) = a |x - h| + k, (h,k) represents the coordinates of the vertex of the absolute value function.

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The equation that represents this number sentence is 1/5n - 3 = 28

Let the unknown number be "n". One-fifth of a number is expressed as 1/5n

Three less than one-fifth of a number is expressed as 1/5n - 3

Finally, if three less than one-fifth of a number is equal to 28, then the final expression will be 1/5n - 3 = 28

The equation that represents this number sentence is 1/5n - 3 = 28

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Step-by-step explanation:

thr standard form for a quadratic equation is:

ax²+bx+c

in this example, a=1, b=(-1), c=(-42)

Answer:

yes,you can buy the textbook

Step-by-step explanation:

determine discount

60 × 0.2(decimal form of 20%)

=12

apply discount

60-12

=48

50>48✓

Answer:

distance between -5 to 0 = 5 units

The management, organisation, and technology factors contributed to this problem are listed below.

The  problem in the airways was that corporation didn't know its customers, the airline hasn't been able to take use of its market opportunity in recent years.

Airways was unable to evaluate and keep track of its marketing efforts.

The technology factor that contributed were:

The Organisation factors that contributed were

The Management factors that contributed were

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Answer:

============================

[tex]\dfrac{\sqrt{14} }{\sqrt{7}-2} -\dfrac{\sqrt{14} }{\sqrt{7}+2}[/tex]

Bring both fractions into common denominator:

[tex]\dfrac{\sqrt{14} (\sqrt{7}+2)}{(\sqrt{7}-2)(\sqrt{7}+2)} - \dfrac{\sqrt{14} (\sqrt{7}-2)}{(\sqrt{7}-2)(\sqrt{7}+2)}[/tex]

Simplify:

[tex]\dfrac{\sqrt{14} ((\sqrt{7}+2) - (\sqrt{7}-2))}{(\sqrt{7}-2)(\sqrt{7}+2)} =[/tex]

[tex]\dfrac{\sqrt{14} (\sqrt{7}+2 - \sqrt{7}+2)}{(\sqrt{7}-2)(\sqrt{7}+2)} =[/tex]

[tex]\dfrac{4\sqrt{14} }{(\sqrt{7}-2)(\sqrt{7}+2)} =[/tex]

[tex]\dfrac{4\sqrt{14} }{(\sqrt{7})^2-2^2} =[/tex]

[tex]\dfrac{4\sqrt{14} }{7-4} =[/tex]

[tex]\dfrac{4}{3} \sqrt{14} }[/tex]

Compare the result with given expression to get:

[tex]\large\sf{\sqrt{\frac{1+sin\theta}{1-sin\theta}}+\sqrt{\frac{1-sin\theta}{1+sin\theta}}}[/tex]

[tex]\\[/tex]

[tex]\large \sf{=\frac{\left(\sqrt{1+sin\theta}\right)^{2}+\left(\sqrt{1-sin\theta}\right)^{2}}{\sqrt{(1-sin\theta)(1+sin\theta)}}}[/tex]

[tex]\\[/tex]

[tex]\large\sf{=\frac{1-sin\theta+1+sin\theta}{\sqrt{1^{2}-sin^{2}\theta}}}[/tex]

[tex]\\[/tex]

[tex]\large \sf{=\frac{2}{\sqrt{cos^{2}\theta}}}[/tex]

[tex]\\[/tex]

[tex]\large\sf{= \frac{2}{cos\theta}}[/tex]

[tex]\\[/tex]

[tex]\large\sf{= 2sec\theta}[/tex]

[tex]\\[/tex]

[tex]\large\sf{=RHS}[/tex]

Therefore,

[tex]\large\sf\red{\sqrt{\frac{1+sin\theta}{1-sin\theta}}+\sqrt{\frac{1-sin\theta}{1+sin\theta}}=2sec\theta}[/tex]

Answer:

B. 70

Explanation:

A total of 182 pairs of jeans and shorts were sold and the number of shorts sold was 28 fewer than twice the number of jeans. This can be expressed with j being the number of jeans as:

j2 - 28 = number of shorts

As the number of shorts is 28 lower than twice the number of jeans, this means it can not be C or D, as these numbers are larger than 182 when doubled. The process of elimination can be done with the remaining using our equation.

51(2) - 28 = 102 - 28 = 74
70(2) - 28 = 140 - 28 = 112

Now that we have the number of shorts sold by using the equation, we can add them with what number was used as their number of jeans.

74 + 51 = 125
112 + 70 = 182

So the number of jeans sold is B. 70.