Which of the following values are in the range of the function graphed below?
Check all that apply.
first answer gets brainliest need asap!!!

Answer:

48 Square Centimeters

Step-by-step explanation:

The red triangular exterior pieces are all equilateral triangles with some side length.

We don't need to know what that side length is. It doesn't matter.

The area of one red equilateral triangle is some area A.

There are 6 of these red triangles, so the red exterior triangular parts combine to a total area of 6*A.

The hexagon is composed of equilateral triangles as well.

Each of these blue equilateral triangles is congruent to any outer red triangle because the side length is the same.

Therefore, the 6 blue equilateral triangles composing this hexagon combine to get a total area of 6*A

The area of the star overall is 12*A because we have 6 blue triangles combining with the 6 red triangles.

There is a total of 12 triangles.

Answer:  [tex]h(x)=\sqrt{x+4}[/tex]

Step-by-step explanation:

The graph of h(x) is the graph of f(x) shifted 4 units left.

perimeter of octagon

Now area

Area of square

Area of shaded region

Answer:

C.  647 square units

Step-by-step explanation:

To find the shaded area, subtract the area of the unshaded square from the area of the octagon.

Area of the octagon

[tex]\textsf{Area of a regular polygon}=\dfrac{n\:l\:a}{2}[/tex]

where:

Given:

Substitute the given values into the formula and solve for A:

[tex]\implies \textsf{Area}=\sf \dfrac{8 \cdot 13 \cdot 15.69}{2}[/tex]

[tex]\implies \textsf{Area}=\sf \dfrac{1631.76}{2}[/tex]

[tex]\implies \textsf{Area}=\sf 815.88\:\:square \:units[/tex]

Area of the square

[tex]\implies \textsf{Area}=\sf 13^2=169 \:\:square \:units[/tex]

Area of the shaded region

= area of the octagon - area of the square

= 815.88 - 169

= 646.88

= 647 square units (nearest square unit)

The exponential model best represents the data is B. f(x) = 5(0.2)^x

From the table of values, we have the following points:

(x, y) = (0,5) and (1,1)

An exponential model is represented as:

y = ab^x

Substitute (x, y) = (0,5)

5 = ab^0

5 = a

This gives

a = 5

So, we have:

y = 5b^x

Substitute (x, y) = (1,1)

1 = 5b^1

This gives

5b = 1

Divide by 5

b = 0.2

Substitute b = 0.2 in y = 5b^x

y = 5(0.2)^x

Hence, the exponential model best represents the data is B. f(x) = 5(0.2)^x

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Answer:  2/6+5/6 + 1.167

Answer:

1.16

Step-by-step explanation:

Answer:

z⁴

Step-by-step explanation:

(z x z x z x z x z x z) / (z x z)

z⁶ / z² = z⁴

See attached images for work.

The linear equation is y = (1/7)*x + 11/7.

The general linear equation is:

y = a*x + b

Where a is the slope.

Here we know that a = (1/7), then the line is:

y = (1/7)*x + b

We also know that the line contains the point (-4, 1), this means that when x = -4, we must have y = 1.

Replacing that, we get:

1 = (1/7)*(-4) + b

1 + 4/7 = b

7/7 + 4/7 = b

11/7 = b

So the linear equation is:

y = (1/7)*x + 11/7.

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The equation of the linear function of the line is: y = 1/7x + 11/7.

A linear function equation is modelled as, y = mx + b, where m is the slope and b is the y-intercept.

Slope (m) = 1/7, and the line passes through (-4, 1), therefore, substitute (x, y) = (-4, 1) and m = 1/7 into y = mx + b:

1 = 1/7(-4) + b

1 = -4/7 + b

1 + 4/7 = b

11/7 = b

b = 11/7

Plug in the values of m and b into y = mx + b:

y = 1/7x + 11/7

The equation of the linear function is: y = 1/7x + 11/7

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x-intercept -4.8

y-intercept 12

slope 2.5

plot points

x y

-2, 7

-1, 9.5

0, 12

1, 14.5

2 ,17

Answer:

[tex]x = -\frac{15}{2}[/tex]

Step-by-step explanation:

Hello!

The formula for the AOS is given to you, and you simply have to plug in the information given by the equation.

Equation of a Parabola (Standard): [tex]y = ax^2 + bx + c[/tex]

Comparing it to our equation:

Plug it in to the AOS formula:

You only have to input 15 in to the correct box since the negative sign is already placed for you.

Answer:

452/2 = 226

Step-by-step explanation:

45 divided by 2 is 22 and the remainder would be 12, which could be divided by 2 again to get 6. Add this to the total and you get 226.

Hope this helps! :D

The domain of the given graph is [−3, ∞) and the range is (−∞, 4].

We need to find the domain and range of the given graph.

The range of values that we are permitted to enter into our function is known as the domain of a function. The x values for a function like f make up this set (x). A function's range is the collection of values that it can take.

We can observe that the graph extends horizontally from −3 to the right without a bound, so the domain is [−3, ∞). The vertical extent of the graph is all range values 4 and below, so the range is (−∞, 4].

Therefore, the domain of the given graph is [−3, ∞) and the range is (−∞, 4].

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The simplification of the algebraic expression to its simplest form is -3x³ + 2x² + 13x - 10

The simplest form of an algebraic expression is a stage at which the expression can no longer be further simplified.

From the given expression:

= -(x − 2)(3x² + 4x − 5)

= (-x +2)(3x² + 4x − 5)

Open brackets

= -3x³ - 4x² + 5x +6x² + 8x - 10

= -3x³ + 2x² + 13x - 10

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Answer:

-150

or 150 degrees below 0

Step-by-step explanation:

The temp is -110. Dropping another 40 degrees means subtracting 40.

-110 - 40 = -150

Step-by-step explanation:

if cospheta =3/5 and pheta is located in the fourth quadrant, then sin2pheta=_____

-8/25

24/5

-24/25

8/25

The correct answer for this question is 8/25

Answer and Step-by-step explanation:

There is a cube with side length 3. The volume of this cube is 27.

Answer:

h = A / ( 1/2 a - 1/2 b)

Step-by-step explanation:

A = ½ ah - ½ bh

Factor out an h

A = h( 1/2 a - 1/2 b)

Divide each side by ( 1/2 a - 1/2 b)

A / ( 1/2 a - 1/2 b) = h ( 1/2 a - 1/2 b)/ ( 1/2 a - 1/2 b)

A / ( 1/2 a - 1/2 b) = h

h = A / ( 1/2 a - 1/2 b)

Answer:

7+45/n or 5

Step-by-step explanation:

45 divided by n or technically 5 is 9 plus 7 is 16 :)

Answer:7+45/n or 5

Step-by-step explanation:

Answer:

(-3, 0, 4, 6) (option D)

Step-by-step explanation:

The "range" of a function is all possible y-values. In this image, there are x's entered on the left side--which I know to be x values because a function, f(x) is essentially the outcome of what happens when x is entered.

So, the y-values are featured on the right. All of these values are the possible values for y, or, what we call the range.

Meaning that the range of this function is (-3, 0, 4, 6)

Answer:

The answer is A

Step-by-step explanation:

They both have a maximum of 5!

Answer:

okkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk

will you be my friend

Step-by-step explanation:

.......... okkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk

Answer:

no solutions

Step-by-step explanation:

we use the quadratic formula.

for an equation ax²+bx+c=0, the solutions for x are:

x = (-b±sqrt(b²-4ac))/2a

in our case, a=2, b=-6 and c=5, so:

x= (6±sqrt(36-40))/4

we see that we get a negative number inside the sqrt, so we have no real solutions

Answer:

No solution

Step-by-step explanation:

1) Use the quadratic formula

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

Once in standard form, identify a, b, and c from the original equation and plug them into the quadratic formula.

[tex]2x^2-6x+5=0[/tex]

[tex]a=2\\b=-6\\c=5[/tex]

[tex]x=\frac{-(-6)\pm\sqrt{(-6)^2-4*2*5} }{2*2}[/tex]

2) Simplify

Evaluate the exponent:

[tex]x=\frac{6\pm\sqrt{(-6)^2-4*2*5} }{2*2}[/tex]

[tex]x=\frac{6\pm\sqrt{36-4*2*5} }{2*2}[/tex]

Multiply the numbers:

[tex]x=\frac{6\pm\sqrt{36-4*2*5} }{2*2}[/tex]

[tex]x=\frac{6\pm\sqrt{36-40} }{2*2}[/tex]

Subtract the numbers:

[tex]x=\frac{6\pm\sqrt{36-40} }{2*2}[/tex]

[tex]x=\frac{6\pm\sqrt{-4} }{2*2}[/tex]

Multiply the numbers

[tex]x=\frac{6\pm\sqrt{-4} }{2*2}[/tex]

[tex]x=\frac{6\pm\sqrt{-4} }{4}[/tex]

3) No real solutions because the discriminant is negative

The square root of a negative number is not a real number

[tex]d=-4[/tex]

Result

No solution

The rule that describes the composition of transformations that maps figure PQRS to figure P"Q"R"S"is r y-axis • RO, 90° (x, y)

See attachment for complete question

From the image, we have:

P = (1,1)

Q = (1,5)

R = (3,5)

S = (3,1)

Start by reflecting PQRS across the y-axis. i.e. r y-axis

This is done by

(x, y) -> (-x,y)

So, we have:

P' = (-1,1)

Q' = (-1,5)

R' = (-3,5)

S' = (-3,1)

Next, we rotate by 90 degrees counterclockwise i.e. RO, 90° (x, y)

This is done by

(x, y) -> (-y, x)

So, we have:

P'' = (-1,-1)

Q'' = (-5,-1)

R'' = (-5,-3)

S'' = (-1,-3)

Hence, the rule that describes the composition of transformations that maps figure PQRS to figure P"Q"R"S"is r y-axis • RO, 90° (x, y)

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Answer:

C.

Step-by-step explanation:

Adding 2 fractions results in a fraction.

Answer:

B: The sum is a non-terminating and a non-reapeating decimal

Step-by-step explanation:

-11/4 +5/g

There is 15 steps you have to do after (look in the picture)

but this is what you get:

[tex]-\frac{11g-20}{4g}[/tex]

Answer:

(x,y)—>(6,2)

Step-by-step explanation:

Add both equations to eliminate y

9x=54

x=6

plug x=6 into either equation

2(6)+6y=24

6y=24-12

y=12/6=2

Graph (A) represents the equation of the line 5y = x + 5 shown in the graph option (A) is correct.

A straight line is a combination of endless points joined on both sides of the point.

The slope 'm' of any straight line is given by:

[tex]\rm m =\dfrac{y_2-y_1}{x_2-x_1}[/tex]

We have the equation of the line:

5y = x + 5

After plotting the line on the coordinate plane.

The above line passes through:

(-5, 0), (0,1), and (5,2)

And the line rises from (-8, -0.5), (-5, 0), (0,1), and (5,2)

Thus, graph (A) represents the equation of the line 5y = x + 5 shown in the graph option (A) is correct.

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Answer:  A. A line is graphed in an x y plane, where the x and the y axes range from negative 8 to 8 in increments of 2. The line rises through (negative 8, negative 0.5), (negative 5, 0), (0, 1), and (5, 2).

Step-by-step explanation:

Answer:

don't know but please keep trying up

Figure it out..............................................

The distance between the point (3, 5) and a line with the equation 5x – 3y + 10 = 0 will be 10 / √(34) units.

Let (x₁, y₁) be the point and ax + by + c = 0 be the line.

Then the distance between a point and a line will be

[tex]\rm d = \dfrac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}[/tex]

The point (3, 5) and a line with the equation 5x – 3y + 10 = 0.

Then the distance between them will be

[tex]\rm d = \dfrac{|5 *3 - 3 *5 + 10|}{\sqrt{5^2 + (-3)^2}}\\d = \dfrac{10 }{\sqrt{34}} \ units[/tex]

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