Cameron bought four begonias in 6-in pots and a $19 fern at a fundraiser. he spent a total of $63. how much is each begonia?

41 is the best estimate for the number of people in line at 9am

Two or more expressions with an Equal sign is called as Equation.

y equal to minus five point one three one x square plus Thirty one point eight two one x minus three point three three three.

y=-5.131x²+ 31.821x-3.333 approximates the number of people standing in line to catch commuter train x hours after 5am.

There is 4 hours in between 9 am and 5 am. so x=4

Substitute x=4 into the equation

y=-5.131(4)²+ 31.821(4)-3.333

y=-5.131(16)+127.284-3.333

y=-82.096+127.284-3.333

y=41.855

Hence 41 is the best estimate for the number of people in line at 9am

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Answer:

B. seven $1 bills and eight $5 bills.

Step-by-step explanation:

n = number of $1 bills

m = number of $5 bills

Given:

n + m = 15

n($1) + m($5) = $47

Just solve 2 equations in 2 unknowns n and m:

(i)  n + m = 15

(ii) n + 5m = 47

Subtract (i) from (ii) and get

4m = 32

m = 32/4 = 8

Then from (i) get

n = 15 - m = 15 - 8 = 7

There are 7 $1 bills and 8 $5 bills.

replace the numbers inside () for x

 f(4) = -4(4)-3 = -16 -3 = -19

G(-4) = 2(-4)^2 -3(-4) = 44

Answer:  The correct description is (C). the product of a number and 45.

Step-by-step explanation:  We are given to select the correct description that matches the algebraic expression 45n.

Let, 'n' be a number.

Then, the sum of 'n' and 45 is written as (45 + n).

So, option (A) is incorrect.

The quotient of 'n' and 45 is given by [tex]\dfrac{n}{45}~\textup{or}~\dfrac{45}{n}.[/tex]

So, option (B) is incorrect.

The product of 'n' and 45 is

45 × n = 45n, which means 45 times of a number 'n' or n times of 45.

So, option (C) is correct.

The difference of 'n' and 45 is (n - 45) or (45 - n), according as 'n' is greater or 45 is greater.

So, option (D) is incorrect.

Thus, the correct option is (C).

Answer:

Median = $1.84

The difference of the first and third quartiles = $0.34

Step-by-step explanation:

Median

How to calculate the median of a small data or ungrouped data;

1. Arrange the data in ascending or descending order.

2. Choose the middle value. The middle value can be a single value or two value. If the middle value is a single value, then, that value is the median but if the middle values are two, the mean of the two middle values is evaluated to get the median.

Arranging the gasoline prices given in ascending order;

$1.61,  $1.75, $1.79,  $1.84,  $1.96, $2.09, $2.11

The middle value is a single number $1.84 in bold. Therefore, the median is $1.84.  

                         Median = $1.84

First (Lower) Quartile

To calculate for the First (Lower) quartile,

1. Arrange the data give in ascending order.

2. The First (Lower) quartile, Q1 is then calculated using;  

                 [tex]Q1 =\frac{1}{4} (n + 1)^{th}[/tex] term

where n is the total number of terms of the data. In the given question, the total number of terms, n = 7

                 [tex]Q1 =\frac{1}{4} (7 + 1)^{th}[/tex] term

                 [tex]Q1 =\frac{8}{4}^{th}[/tex] term

                 Q1 = 2nd term

3. Select the 2nd term of the given data already arranged in ascending order.

                 $1.61,  $1.75, $1.79,  $1.84,  $1.96, $2.09, $2.11

                 The First Quartile, Q1 = $1.75

Third (Upper) Quartile

To calculate for the third (upper) quartile,

1. Arrange the data give in ascending order.

2. The third (upper) quartile (Q3) is then calculated using;  

                 [tex]Q3 =\frac{3}{4} (n + 1)^{th}[/tex] term

where n is the total number of terms. In the given question, the total number of terms, n = 7

                 [tex]Q3 =\frac{3}{4} (7 + 1)^{th}[/tex] term

                 [tex]Q3 =\frac{24}{4}^{th}[/tex] term

                 [tex]Q3 = 6^{th} term[/tex]

3. Select the [tex]6^{th} term[/tex] of the given data already arranged in ascending order.

                 $1.61,  $1.75, $1.79,  $1.84,  $1.96, $2.09, $2.11

The third (upper) quartile Q3 = $2.09

The difference between the first and third quartiles is called interquartile range.

Now, calculating the difference of the first and third quartiles

                  Interquartile Range = Q3 - Q1

                              $2.09 - $1.75 = $0.34

The difference of the first and third quartiles, Interquartile Range = $0.34

Answer:

Therefore, The correct option is D. 0.069

Step-by-step explanation:

Amount of the money paid in the state property taxes every year = $2849.84

The property owned has an assessed value.

Now, the assessed value of the property is $41302

We need to find the state's property tax rate.

[tex]\text{So, State's property tax rate = }\frac{\text{state's property tax}}{\text{assessed value}}\\\\\text{State's property tax rate = }\frac{2849.84}{41302}\approx 0.069[/tex]

Therefore, The correct option is D. 0.069

we know that

The CPCTC states that corresponding parts of congruent triangles are congruent

so

△FEB≅△CNB

that means

[tex]FE=CN\\FB=CB\\EB=NB[/tex]

we have

[tex]FE=6.1\ units\\FB=4.8\ units\\EB=3.8\ units[/tex]

therefore

[tex]BN=3.8\ units[/tex]

the answer is

the length of BN is [tex]3.8\ units[/tex]

the right answer is actually 30,60,90, I got the answer right on the test

The measures of the angles are 30°, 60°, and 90° for the given right triangle. The correct answer would be option (A).

A right triangle is defined as a triangle in which one angle is a right angle or two sides are perpendicular.

The right triangle is given in the figure, as shown.

According to the figure,  

perpendicular = 7√3

hypotenuse = 14√3

As per the question, we have

Since sin θ = perpendicular/hypotenuse

Substitute the values in the above formula, and we get

sin θ = 7√3/14√3

sin θ = 1/2

θ = 30°

This means the measures of the angles are 30°, 60°, and 90°.

Hence, the correct answer would be an option (A).

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The question seems to be incomplete, the missing part has been attached below.

The local extreme values of [tex]\( f(x, y) = x^2 - y^2 - 2 \)[/tex] subject to the constraint [tex]\( x^2 + y^2 = 16 \)[/tex] are -18  at points (0, 4) and (0, -4).

To find the local extreme values of [tex]\( f(x, y) = x^2 - y^2 - 2 \)[/tex] subject to the constraint [tex]\( x^2 + y^2 = 16 \)[/tex] using Lagrange multiplier techniques, we need to set up the Lagrangian function:

[tex]\[ L(x, y, \lambda) = f(x, y) - \lambda(g(x, y) - c) \][/tex]

where [tex]\( g(x, y) \)[/tex] is the constraint, c is a constant, and [tex]\( \lambda \)[/tex] is the Lagrange multiplier.

In this case, [tex]\( g(x, y) = x^2 + y^2 \)[/tex] is the constraint, and c = 16 .

So, the Lagrangian function is:

[tex]\[ L(x, y, \lambda) = x^2 - y^2 - 2 - \lambda(x^2 + y^2 - 16) \][/tex]

Now, find the partial derivatives of L with respect to ( x, y, ) and[tex]\( \lambda \)[/tex] and set them equal to zero to find critical points:

[tex]\[ \frac{\partial L}{\partial x} = 2x - 2\lambda x = 0 \]\[ \frac{\partial L}{\partial y} = -2y - 2\lambda y = 0 \]\[ \frac{\partial L}{\partial \lambda} = x^2 + y^2 - 16 = 0 \][/tex]

Solving the system of equations, we get two critical points:

1. When [tex]\( x = 0, y = 4, \lambda = -\frac{1}{4} \)[/tex]

2. When [tex]\( x = 0, y = -4, \lambda = -\frac{1}{4} \)[/tex]

Now, evaluate f(x, y) at these critical points:

1. f(0, 4) = 0 - 16 - 2 = -18

2. f(0, -4) = 0 - 16 - 2 = -18

Therefore, the local extreme values of [tex]\( f(x, y) = x^2 - y^2 - 2 \)[/tex] subject to the constraint [tex]\( x^2 + y^2 = 16 \)[/tex] are -18  at points (0, 4) and (0, -4).

Answer: (2,1) (4,2) (-2,-1)

Step-by-step explanation:

thats apex, my dude

Answer:

Option A, C, F are the correct options.

Step-by-step explanation:

The given equation of line is [tex]y=\frac{1}{2}x[/tex]

Now we will plug in the values of points and validate the equation.

Option A. (2, 1)

[tex]y=\frac{1}{2}x[/tex]

[tex]1=\frac{2}{2}=1[/tex]

So this points lies on the given equation of line.

Option B. (3, 15)

[tex]15\neq \frac{3}{2}[/tex]

So the point is not on the line.

Option C. (4, 2)

[tex]2=\frac{4}{2}=2[/tex]

Point lies on the line.

Option D. (3, 6)

[tex]6\neq \frac{3}{2}[/tex]

Therefore point is not on the line.

Option E. (-2, 1)

[tex]1\neq \frac{-2}{2}[/tex]

Point does not lie on the line.

Option F. (-2, -1)

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

So this point lies on the line.

There are three points (2, 1), (4, 2) and (-2, -1) are on the line.

Answer:

y ≥ 0

Step-by-step explanation:

The real absolute value function is defined on the set of all real numbers, assigning each real number its respective absolute value. Formally, the absolute value of any real number x, is defined by:

[tex]|x|=x,\hspace{3}if\hspace{3}x\geq0\\|x|=-x,\hspace{3}if\hspace{3}x<0[/tex]

The domain of the function is all real numbers, and by definition, the absolute value of x will always be greater than or equal to zero and never negative. Hence:

[tex]Domain:\\x\in R\\Range:\\y\in R : y\geq0[/tex]

I attached you a picture of the graph.