FLE.
= a
=t.
18. The table shows the amount of fabric
Camille used based on the number of
headbands that she made. Find the slope, and
explain what it means in the context of the
situation.
Headbands
Fabric (yd)
3
3
5
5
1
7775
5015
9
q

Answer:

3(b-2)=24

Step-by-step explanation:

3b - 6 = 24

3b = 30

b= 10

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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Three mathematical terms with no definitions: (1) The terms (ii) and (iii) are correct., (2) The term (ii) is correct., (3) The term (i) and (ii) is correct.

What is parallel and perpendicular lines?

The definition of a parallel line is a pair of lines that never cross and always have the same distance between them. Lines that cross at right angles are referred to as perpendicular lines (90 degrees).

(1) The mathematical term parallel lines explicitly uses the undefined term(s).

The terms (ii) and (iii) are correct.

(2) The mathematical term perpendicular lines explicitly uses the undefined term(s)

The term (ii) is correct.

(3) The mathematical term line segment explicitly uses the undefined term(s).

The term (i) and (ii) is correct.

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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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Hope this helps you with your question.

Answer:

Step-by-step explanation:

I did it on a graphing app

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]

Answer: D. Y-2 = -7/4(X+9)

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]

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.

The rigid transformation of the pre-image ΔABC, that produces the image ΔDEF is option E as follows;

E. [tex]R_{(90^{\circ},\,(0,\, 0))}[/tex]

A rigid transformation is one in which the distances between each pair of points on the pre-image is preserved on the image, such that the pre-image and the image have the same size.

The coordinates of the vertices of ΔABC are;

A(-3, -3), B(-1, -4), C(-3, -4)

The coordinates of the vertices of ΔDEF are; D(-3, 3), E(-4, 1), F(-4, 3)

The formula for the rotation of a point (x, y) 90° clockwise about the origin is presented as follows;

Pre-image point [tex]{}[/tex] (Rotation 90° clockwise)  Image point after rotation

(x, y)            [tex]{}[/tex]                        ⇒               [tex]{}[/tex]           (-y, x)

The points in the image ΔDEF of pre-image triangle ΔABC which corresponds to points on ΔABC are;

Vertex A in triangle ΔABC corresponds to vertex D on triangle ΔDEF

Vertex B in triangle ΔABC corresponds to vertex E on triangle ΔDEF

Vertex C in triangle ΔABC corresponds to vertex F on triangle ΔDEF

The transformation of (x, y) ⇒ (-y, x), (which is the transformation of a 90° clockwise rotation about the origin), on the vertices of ΔABC are;

A(-3, -3)    [tex]\underset \longrightarrow {R_{(90^{\circ}, (0, 0)}}[/tex]  D(-3, 3)

B(-1, -4)    [tex]\underset \longrightarrow {R_{(90^{\circ}, (0, 0)}}[/tex]  D(-4, 1)

C(-3, -4)    [tex]\underset \longrightarrow {R_{(90^{\circ}, (0, 0)}}[/tex]  F(-4, 3)

Therefore, the triangle ΔDEF can be obtained from the triangle ΔABC by a rotation of 90° clockwise about the origin.

The correct option is; E [tex]R_{(90^{\circ}, (0, \,0))}[/tex]

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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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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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The measure of angle 4 is 35 degrees and the measure of angle 5 is 81 degrees

The sum of angles on a straight line is equal to 180 degrees

Then the equation will be

∠3 + 81 +∠4 = 180  degrees

Substitute the value of ∠3 in the equation

64 + 81 + ∠4 = 180  

145 + ∠4 = 180

∠4 = 180 - 145

∠4 = 35 degrees

The sum of interior  angles of the triangle is 180 degrees

∠4 + ∠5 + 64 = 180

Substitute the values in the equation

35 + ∠5 + 64 = 180

∠5 + 99 = 180

∠5 = 180 - 99

∠5 = 81 degrees

Hence, the measure of angle 4 is 35 degrees and the measure of angle 5 is 81 degrees

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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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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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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 required standard equation is given as y(t) = 5.39 sin(4πt + 0.379π).

Given that,

To rewrite y(t)=2×sin 4pi t+5×cos 4pi t in the form y(t)=Asin (wt+Ф).

These are the equation that contains trigonometric operators such as sin, cos.. etc.. In algebraic operations.

Here
We know that.
Asin(α + β) = Asinαcosβ + Acosαsinβ,
Comparing it with the given equation,
Asinα = 5  
Acosα = 2
Taking the ratio of the equation,
sinα/cosα = 5/2
tanα = 5/2
α = 0.379π
Now,
Asin0.379π = 5
A = 5 / sin0.379
A = 5.39

Now,
The standard form equation is given as, y(t) = 5.39 sin(4πt + 0.379π) where ω = 4π.

Thus, the required equation is given as y(t) = 5.39 sin(4πt + 0.379π).

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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:

450 centimeters

Step-by-step explanation:

We know that,

1 meter = 100 centimeters

so 4.5 meters = 4.5*100 centimeters

                      =450 centimeters

Step-by-step explanation:

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

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]

From the procedure, the simplest form of the radical is  5√7.

A radical can also be called a surd. The surd is an irrational number. We say that it is irrational because it can not be expressed as a terminating decimal.

Now we have to look at the radical √175. We have to think of two numbers that we can multiply to obtain 175 and one of them would be a perfect square.

If we choose the numbers 7 and 25, we can write;

√175 = √7 * √25

Thus we have in the  simplest radical form 5√7.

The expression √5/12 is already in the simplest form.

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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:

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 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:

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 :)

Answer: Value of x is B equals 6

Step-by-step explanation:

If Nita scored 62% of the marks then the maximum marks of the examination was 600  .

In the question ,

it is given that,

the number of marks scored by Nita in the examination = 372 marks

percent of marks scored by Nita in the examination = 62% ,

let the maximum marks of the examination = "x" ,

So , according to the questions

372 = 62% of x

372 = 0.62 * x       ....because 62% = 0.62

x = 372/0.62

x = 600

Therefore , If Nita scored 62% of the marks then the maximum marks of the examination was 600  .

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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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