Once an antibiotic is given, number of bacteria decreases at a rate of 15%/day. There were about 15,000 bacteria prior to the treatment. The graph models the number of bacteria after x days of treatment. What are the key features of the exponential function modeled in terms of this situation and how can they be interpreted? Select each correct answer. 30 The line y = 0 is an asymptote of the graph. The x-intercept is 30. The line x = 0 is an asymptote of the graph. The asymptote indicates that there will never be more than 15,000 bacteria. The y-intercept is 15,000. The y-intercept represents the number of bacteria when treatment began. The asymptote indicates that as time increases, the number of bacteria approaches 0.​

The slope of the line is -32, and the y-intercept is 480 if the equation of the line is y = -32x + 480.

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:

A linear equation has the form y = mx + b in the slope-intercept format. X and Y are the variables in the equation. The values m and b represent the line's slope (m) and the value of y when x is 0.

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

The equation of the line uses two points:

(0, 480) and (15, 0)

y = (-480/15)[x - 15]

y = -32(x - 15)

y = -32x + 480

y = mx + c

m = -32

c = 480

Thus, the slope of the line is -32, and the y-intercept is 480 if the equation of the line is y = -32x + 480.

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Answer:

3(b-2)=24

Step-by-step explanation:

3b - 6 = 24

3b = 30

b= 10

Answer: Value of x is B equals 6

Step-by-step explanation:

Seven huge boxes and two tiny boxes total were bought by the contractor.

A linear equation is defined as an algebraic equation of maximum degree 1 is a linear equation.

According to the question, a builder needs to purchase nails to construct a house.

Both small and large boxes of nails are provided. There are 50 nails in each small box and 350 in each large box.

The contractor purchased 5 more large than tiny boxes, hence it is assumed that he purchased a certain number of little boxes.

He purchased (x+5) huge cartons.

So, 50(x) + 350(x+5) = 2590.

50x + 350x + 1750 = 2590.

400x = 2590 - 1750.

400x = 840.

x = 2.1

The bulk of the boxes he bought is 7.

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In contrast to the signs of the real x-intercepts on the graph, we must remember the signs of the values in a quadratic's factored form. because the cartesian coordinate points are shifted.

Any function of the form [tex]\rm f(x) =ax^2+bx+c[/tex].where x is variable and a, b, and c are any real numbers where a ≠ 0 is called a quadratic function.

When factoring by grouping is used, the replacement terms will both share the same middle-term sign. The factor signs will have one of each sign with the "stronger" number (the larger absolute value) matching the middle term if the last term is negative.

We can locate the function's x-intercepts or zeros with the aid of factored form.

Thus, in contrast to the signs of the real x-intercepts on the graph, we must remember the signs of the values in a quadratic's factored form. because the cartesian coordinate points are shifted.

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In this question, line1 and line2 are parallel to each other by finding the slope.

What is the slope of line?

Given two line-side coordinates, use the slope formula to determine the slope of the line. The slope is defined as the ratio of the change in the y values to the change in the x values using the formula m=(y2-y1)/(x2-x1). x1 and y1 are represented by the first point's coordinates. The second points are located at x2, y2, and their locations. Whichever point you choose to classify as the first and which as the second doesn't matter.

We know that formula two find the slope of the line joining two points which is

Given four lines are

Line 1: y=1/2x -6

(Line 2: x-2y=-1

(Line 3: y=4x+6

(Line 4: y+3=1/4 (x-6)

We have to find out the slope of the corresponding lines such that

the slope of line 1 = 1/2

the slope of line 2 = 1/2

the slope of line 3 = 4

the slope of line 4 = 14

So the line has the same slope they are parallel to each other.

Hence in this question, line1 and line2 are parallel to each other.

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For having a discriminant equal to or larger than zero we must have:

k ≤ 2 and k ≥ 10

For a quadratic equation of the form:

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

The discriminant is:

D = b^2 - 4ac

And the equation has real roots only if the discriminant is equal to or larger than zero.

Here our equation is:

x^2 + (k - 2)*x + (2k - 4) = 0

The discriminant is:

D = (k - 2)^2 -4*1*(2k - 4)

D = k^2 -  4k + 4 - 8k + 16

D = k^2 -12k + 20

The solutions of :

k^2 -12k + 20 = 0

are:

k = (+12 ± √( (-12)^2 - 4*1*20))/(2)

k = (+12 ± 8)/2

The two values of k are:

k = 20/2 = 10

k = 4/2 = 2

The possible values of k are:

k ≤ 2 and k ≥ 10

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Answer:

7.81

Step-by-step explanation:

x1= 1

x2= 6

y1= -4

y2= 2

31.The first one just has one minor error which is not putting a square root.

after putting the square root on  [tex]\sqrt{61}[/tex] you will get an answer which is 7.81 units

the second one has an error in putting the correct values

32. the equation for distance is [tex]\sqrt{(x_2-x_1 )+(y_2-y_1)} }[/tex] however the values are inputted wrong as x2=6 and x1=1 but here x1 is taken as 2 which is the value of y2 and instead of inputting y2 as 2 it is written as 1 which is the value of x1

Comment if you still don't understand

Answer:

31.  Omission of the square root sign in step 1.

32.  Subtracting the y-value from the x-value in each parentheses in step 1.

Step-by-step explanation:

[tex]\boxed{\begin{minipage}{7.4 cm}\underline{Distance between two points}\\\\$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$\\\\\\where $(x_1,y_1)$ and $(x_2,y_2)$ are the two points.\\\end{minipage}}[/tex]

Given points:

The error was the omission of the square root sign in the first step of the calculation:

[tex]\textsf{Error}: \quad AB=(6-2)^2+(1-(-4))^2[/tex]

[tex]\textsf{Correction}: \quad AB=\sqrt{(6-2)^2+(1-(-4))^2}[/tex]

Correct calculation:

[tex]\begin{aligned}AB&=\sqrt{(6-1)^2+(2-(-4))^2}\\&=\sqrt{5^2+6^2}\\&=\sqrt{25+36}\\&=\sqrt{61}\\& \approx 7.8\end{aligned}[/tex]

The error was subtracting the y-value from the x-value in each parentheses in the first step of the calculation, rather than subtracting the x-values and the y-values separately:

[tex]\begin{aligned}\textsf{Error}: \quad AB&=\sqrt{(x_A-y_A)^2+(x_B-y_B)^2}\\ &=\sqrt{(6-2)^2+(1-(-4))^2}\end{aligned}[/tex]

[tex]\begin{aligned}\textsf{Correction}: \quad AB&=\sqrt{(x_A-x_B)^2+(y_A-y_B)^2}\\&=\sqrt{(6-1)^2+(2-(-4))^2\end{aligned}[/tex]

Correct calculation:

[tex]\begin{aligned}AB&=\sqrt{(6-1)^2+(2-(-4))^2}\\&=\sqrt{5^2+6^2}\\&=\sqrt{25+36}\\&=\sqrt{61}\\& \approx 7.8\end{aligned}[/tex]

The congruence theorem that can be used to prove that the triangles are congruent is F. Not enough information is given to prove that the triangles are congruent

If all three corresponding sides are equal and all three corresponding angles are equal in measure, two triangles are said to be congruent.

According to the Angle-Side-Angle Theorem (ASA), if two angles and their included side are congruent to two angles and their included side, then these two triangles are congruent.

The hypotenuse is always the opposite side of the right angle. The legs of the triangle are the other two sides of a right triangle. Because a right triangle has two legs and a hypotenuse, it is also known as a hypotenuse-leg (or HL) triangle.

In this situation, enough information aren't given about the triangles. Therefore, the correct option is F.

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Answer:a

Step-by-step explanation:

Answer:

The correct answer is D.

Step-by-step explanation:

1. You add all 4 of the given closing prices with answer choice D ($56.24)  Getting $271.35

2. Then divide the total of $271.35 by 5. which gives you $54.27.

meaning that your answer is D.

Sorry if this didn't help very much... Would appreciate a 5 star if it did :)

Check the picture below.

[tex]\stackrel{\measuredangle MNQ}{4x+16}\implies 4(14)+16\implies \stackrel{\measuredangle MNQ}{\text{\LARGE 72}}[/tex]

Step-by-step explanation:

You need to replace x in the equation with x in the table.

The domain of the relationship graphed, Domain: {x|x € N} is from -4 and ended at 4.

How to identify domain and range from co-ordinates?

If x,y be a co-ordinate pair then x is the domain and y is the range .

The domain of a function is the set of all possible inputs for the function.

The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain.

Here

x starts from -4 and ended at 4.

domain is the set of natural numbers

Hence,  the domain of the relationship graphed, Domain: {x|x € N} is from -4 and ended at 4.

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Answer:

D. 80 degree

Step-by-step explanation:

According to the question:

[tex]\triangle ABC \cong \triangle DEF[/tex]

[tex] \rm m \angle A = (x + 30) \degree \\ \rm m \angle C = 45 \degree \\ \rm m \angle D = 80 \degree[/tex]

From above congruence we can conclude:

[tex] \rm m \angle A = m \angle D[/tex]

[tex] \therefore [/tex]

[tex] \rm m \angle A = 80 \degree[/tex]

The distance of each station from the ship is = 250 miles.

acc to our question-

hence,The distance of each station from the ship is = 250 miles.

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Answer:

y - 6 = 8(x + 4)

Step-by-step explanation:

The point-slope form is [tex]y-y_{1}=m(x-x_{1})[/tex], where y1 and x1 are any point on the line and m is the slope.

The formula for slope is [tex]\frac{y_{2}-y_{1} }{x_{2}-x_{1} }[/tex]

By plugging in the points we have, we get:  [tex]\frac{22-6}{-2-(-4)}=8=m[/tex]

We can use any of the two pairs of points to represent y1 and x1 so if we choose (-4,6), we get:

[tex]y-6=8(x-(-4))\\y-6=8(x+4)[/tex]

The equivalent resistance currents passing through each resistance  are: 1.8 A , 1.29A , 1A.

The total current passing through the circuit is  19.83 A.

The resistance 5Ω , 7Ω , 9Ω  which are are connected in parallel, then,

R1 =5Ω,

R 2=7Ω,

R3 =9Ω; supply voltage V=9 volt.

Effective resistance of parallel combination can be calculated as :

1/RP = 1/R1 + 1/R2 + 1/R3

1/RP= 1/5 + 1/7 +1/9

Rp = 0.45397Ω

Then the Current through  resistors

are R1,  I1 = V/R1 = 9/5 = 1.8A

R2,  I2 = V/R2 = 9/7 = 1.29A

R3,  I3 = V/R3 = 9/9 = 1A

Then Total current I = V/Rp = 9/ 0.45397 = 19.83 A

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The quadratic form of (x² - 1)² + (x² - 1) - 12 = 0 is (x² + 3)(x² - 4) = 0 and the roots are x = ± i√3 Or x = ± 2.

A quadratic equation is an algebraic expression in the form of variables and constants.

A quadratic equation has two roots as its degree is two.

Given, (x² - 1)² + (x² - 1) - 12 = 0.

∴ x⁴ - 2x² + 1 + x² - 1 - 12 = 0.

x⁴ - x² - 12 = 0.

x⁴ + 3x² - 4x² - 12 = 0.

x²(x² + 3) - 4(x² + 3) = 0.

(x² + 3)(x² - 4) = 0.

Now (x² + 3) = 0 Or (x² - 4) = 0.

x² = - 3 Or x² = 4.

x = ± √-3 Or x = ± 2.

x = ± i√3 Or x = ± 2.

We can also solve this equation by assuming x² = t and substitute we will get the roots in terms of t and then we'll replace x² again in t.

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The completed-square (vertex) form 2x² + 12x + 1 is 2(x+3)^2−17. Completing the Square is a technique to find maximum or minimum values of quadratic functions.

Completing the square is a highly helpful technique or way to change a quadratic equation from also known as the "standard form," to its "vertex form,".

For some combinations of h and k, completing the square is a method for transforming a quadratic polynomial. In other words, the quadratic expression is completed by inserting a perfect square trinomial.

Finding the maximum or lowest values of quadratic functions can be done using the square method, sometimes known as "completing the square."

Use the form ax^2 + bx + c, to find the values of a, b, and c.

Consider the vertex form of a parabola.

a(x+d)^2+e

Find the value of d using the formula

d = b/2a.

d = 3

Find the value of e using the formula

e = c− b^2 / 4a.

e = −17

Substitute the values of a, d, and e into the vertex form 2(x+3)^2−17.

2(x+3)^2−17

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11 dummy variables will be needed.

A dummy variable is a numerical variable used in regression analysis to represent subgroups of the sample in your study. In research design, a dummy variable is often used to distinguish different treatment groups.

we know that,

dummy variable formula = K- 1,

here, K = 12 (months in a year)

this implies, K - 1 = 12 - 1 = 11

Therefore, no. of dummy variables needed = 11.

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The percent sales tax on perchaging new pants is 7%. Percent is very useful for comparing two different things.

What is percent?

It is a figure or ratio stated as a fraction of 100 in mathematics. It is frequently represented by the percent sign, "%".

The formula to calculate the percent sales tax is as follows:

[tex]\textnormal{Percent sales tax}=\frac{\textnormal{Tax amount}}{\textnormal{Price without tax}} \times 100[/tex]

Total price of new pants with tax = $21.40

Price of new pants without tax = $20

The amount of tax on perchanging pants = $21.40 - $20 = $1.40

Now, calculate the percent tax.

[tex]\textnormal{Percent sales tax}=\frac{\textnormal{Tax amount}}{\textnormal{Price without tax}} \times 100\\\textnormal{Percent sales tax}=\frac{\$1.40}{\textnormal\$20}\times 100\\=7\%[/tex]

Hence, the percent sales tax on perchagin new pants is 7%.

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Answer to part (A) is y = 42x+9

Answer to part (B) is 98

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

Explanation:

Part (A)

Let's plug x = 0 into the 1st derivative of f(x)

[tex]f'(\text{x}) = \cos(\pi \text{x}) + \text{x}^5 + 6\\\\f'(0) = \cos(\pi *0) + (0)^5 + 6\\\\f'(0) = 1 + 0 + 6\\\\f'(0) = 7\\\\[/tex]

We'll use that later in the steps below, which show computing the derivative value of h(x) at x = 0.

[tex]h(\text{x}) = ( f(\text{x}) )^2\\\\h'(\text{x}) = 2( f(\text{x}) )*f'(\text{x}) \ \ \text{ ... chain rule}\\\\h'(0) = 2( f(0) )*f'(0)\\\\h'(0) = 2( 3 )*7\\\\h'(0) = 42\\\\[/tex]

This is the slope of the tangent line to h(x) at x = 0.

Now plug x = 0 into the h(x) function itself, without any derivatives applied.

[tex]h(\text{x}) = ( f(\text{x}) )^2\\\\h(0) = ( f(0) )^2\\\\h(0) = ( 3 )^2\\\\h(0) = 9\\\\[/tex]

This is the y intercept of the line, i.e. the b value.

We found that

We go from y = mx+b to y = 42x+9 as the equation of the tangent line.

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

Part (B)

In the previous part, we already calculated the first derivative. Differentiate that with respect to x to get the second derivative.

[tex]h'(\text{x}) = 2( f(\text{x}) )*f'(\text{x})\\\\h''(\text{x}) = \frac{d}{dx}\left[h'(x)\right]\\\\h''(\text{x}) = \frac{d}{dx}\left[2( f(\text{x}) )*f'(\text{x})\right]\\\\h''(\text{x}) = \frac{d}{dx}\left[2( f(\text{x}) )\right]*f'(\text{x})+2*f(\text{x})*\frac{d}{dx}\left[f'(\text{x})\right] \ \ \text{ ... product rule}\\\\h''(\text{x}) = 2f'(\text{x})*f'(\text{x})+2*f(\text{x})*f''(\text{x})\\\\h''(\text{x}) = 2\left(f'(\text{x})\right)^2+2*f(\text{x})*f''(\text{x})\\\\[/tex]

The second derivative is useful to determine where the function is concave up or concave down. And also to determine points of inflection.

The h''(x) function involves f''(x), so we'll need to find the second derivative of the f(x) function.

[tex]f'(\text{x}) = \cos(\pi \text{x}) + \text{x}^5 + 6\\\\f''(\text{x}) = \frac{d}{dx}\left[\cos(\pi \text{x}) + \text{x}^5 + 6\right]\\\\f''(\text{x}) = -\pi\sin(\pi \text{x}) + 5\text{x}^4\\\\[/tex]

Then plug in x = 0

[tex]f''(\text{x}) = -\pi\sin(\pi \text{x}) + 5\text{x}^4\\\\f''(0) = -\pi\sin(\pi *0) + 5(0)^4\\\\f''(0) = 0\\\\[/tex]

We have enough info to find h''(0) finally.

[tex]h''(\text{x}) = 2\left(f'(\text{x})\right)^2+2*f(\text{x})*f''(\text{x})\\\\h''(0) = 2\left(f'(0)\right)^2+2*f(0)*f''(0)\\\\h''(0) = 2\left(7\right)^2+2*3*0\\\\h''(0) = 98\\\\[/tex]

Side notes:

Answer:

Step-by-step explanation:

Let the dimensions are width - w, length - l and height - h.

Formula for volume:

Variable expressions for length, height and width:

Equation for the volume of this box:

Substitute the variables:

or

Answer:

Step-by-step explanation:

a) Formula we use,

→ V = l × w × h

b) The variable expressions are,

→ l = (h + 5)

→ w = (h + 2)

→ h = h

c) Forming required equation,

→ V = l × w × h

→ V = (h + 5)(h + 2)(h)

→ 216 = (h + 5)(h + 2)(h)

These are the required answers.

Answer:

[tex]y=4x-6[/tex]

Step-by-step explanation:

The slope is [tex]\frac{-2-2}{1-2}=4[/tex].

[tex]y-2=4(x-2) \\ \\ y-2=4x-8 \\ \\ y=4x-6[/tex]

The given figure is pentagon, and the perimeter of pentagon is 47 cm.

What is pentagon and its perimeter?

The perimeter of a pentagon is the total length of the pentagon’s boundaries. Therefore, we can find the perimeter by adding the lengths of the five sides of the pentagon.

The length of side with radius = 2rsin(180/n)

Here, n = 5

length of side with r = 8 = 2*8sin(180/5)

=16 sin 32

=16 (0.58779)

=9.4

Perimeter is given as 5(length) = 47 cm.

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Answer:

8

Step-by-step explanation:

Corresponding sides of similar triangles are proportional, so:

[tex]\frac{A'B'}{12}=\frac{4}{6} \implies A'B'=8[/tex]

The equation for the total cost of the dog tags will be 348 + 2t.

The equation to show the revenue would be Revenue = 5t

The number of dog tags that Logan needs to sell to break even is 116 dog tags.

Assuming that the number of dog tags sold by Logan to raise money is represented by t, the total cost would be:

= Flat fee charge + Cost per tag x Number of tags

= 348 + 2t

The total revenue would be:

= Number of dog tags x Selling price per dog tag

= 5t

The breakeven point would be the point where total profit would be zero and so can be shown as:

0 = Sales - total cost

0 = 5t - (348 + 2t)

0 = 5t - 348 - 2t

3t = 348

t = 348 / 3

t = 116 tags

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The triangles ΔABC and ΔCDE are congruent to each other using the SSS rule.

We are given two triangles. The vertices of the first triangle are A, B, and C. The vertices of the second triangle are C, D, and E. We need to prove that the triangles are congruent to each other.

The length of side AB is equal to that of side CD. The length of side BC is equal to that of side DE. We know that C is the midpoint of line segment AE. This means that the length of AC is equal to the length of CE.

All three corresponding sides of the two triangles are equal. Hence, the triangles are congruent to each other using the SSS rule for congruency.

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The half-life of this substance that weighs 32 mg and one year later, the sample weighs 25.5 mg will be 3.053 year.

The duration it takes for a given quantity to fall to half of its initial value is known as the half-life. The phrase can be used to refer to other types of decay, whether or not they are exponential, but it is most frequently used in connection with atoms going through radioactive decay. The amount of time needed for the reactant concentration to drop to half its initial value is known as the half-life of a reaction.

Here,

The initial weight of radioactive substance=32 mg

The final weight of radioactive substance=25.5 mg

Duration=1 year

The formula,

N(t)=N(0)*1/2^(t/t₁/₂)

25.5=32*1/2^(1/t₁/₂)

half-life, t₁/₂ = 3.053 year

The half-life of this substance, which weighs 32 mg, will be 3.053 years when the sample weighs 25.5 mg.

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