PLEASE HELP URGENT

Which equation could be solved using this application of the quadratic formula?

[tex] \qquad \qquad \bf \huge\star \: \: \large{ \underline{Answer} } \huge \: \: \star[/tex]

[tex]\textsf{ \underline{\underline{Steps to solve the problem} }:}[/tex]

Take HJ = x, GH = y and GJ = z

put the value of x from equation 1 in equation 2

[tex]{ \qquad❖ \: \sf \:z = (y + 2) + y - 17} [/tex]

[tex]{ \qquad❖ \: \sf \:z = 2y - 15} [/tex]

now, put the value of x and z in equation 3

[tex]{ \qquad❖ \: \sf \:y + 2 + y + 2y - 15 = 73} [/tex]

[tex]{ \qquad❖ \: \sf \:4y - 13 = 73} [/tex]

[tex]{ \qquad❖ \: \sf \:4y = 86} [/tex]

[tex]{ \qquad❖ \: \sf \:y = 21.5 \: \: in} [/tex]

Now, we need to find HJ (x)

[tex]{ \qquad❖ \: \sf \:x = y + 2} [/tex]

[tex]{ \qquad❖ \: \sf \:x = 21.5 + 2} [/tex]

[tex]{ \qquad❖ \: \sf \:x = 23.5 \: \: in} [/tex]

[tex]{ \qquad \large \sf {Conclusion} :} [/tex]

Answer:

We lend 100% - 9.2% and this will equal 90.8%, then we will make this percentage of 42,600 which will give us 38,680 and we will add the 42,600 and this will give us a total of 81,460 copies.

Answer:

Two Equal Angles = 45°

Third Angle = 90°

Step-by-step explanation:

Given information

Two equal angles of an isosceles triangle are half the third angle

Set variables

Let x be the angle of the equal angles of the isosceles triangle

Let 2x be the angle of the third angle

Set equations

x + x + 2x = 180 (Triangle angle sum theorem)

Combine like terms

2x + 2x = 180

4x = 180

Divide 4 on both sides

4x / 4 = 180 / 4

[tex]\Large\boxed{x=45^\circ}[/tex]

Substitute the x value into the expression to find the third angle

Third angle = 2x

                   = 2 (45)

                   = [tex]\Large\boxed{90^\circ}[/tex]

Hope this helps!! :)

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The area of the region enclosed by the curves x + y = 1, x - 3 = y, y = √x and x = 0 is 16.815 square units.

In this question we must determine the area generated by four functions: three linear functions and a radical function. First, we plot the four functions to determine the required combinations of definite integrals need for calculation:

[tex]A = \int\limits^{0.382}_0 {[f(x) - g(x)]} \, dx + \int\limits^2_{0.382} {[h(x) - g(x)]} \, dx[/tex], where f(x) = √x, g(x) = x - 3 and h(x) = - x + 1.

[tex]A = \int\limits^{0.382}_{0} {[\sqrt{x} - x + 3]} \, dx + \int\limits^{2}_{0.382} {[- 2 \cdot x + 4]} \, dx[/tex]

[tex]A = \int\limits^{0.382}_{0} {\sqrt{x}} \, dx - \int\limits^{0.382}_{0} {x} \, dx + 3 \int\limits^{0.382}_{0}\, dx - 2 \int\limits^{2}_{0.382} {x} \, dx + 4 \int\limits^{2}_{0.382}\, dx[/tex]

[tex]A= 2 \cdot x^{\frac{3}{2} }|_{0}^{0.382} - \frac{x^{2}}{2}|_{0}^{0.382} + 3\cdot x |_{0.382}^{2} - x^{2} |_{0.382}^{2}+4\cdot x |_{0.382}^{2}[/tex]

[tex]A = 2 \cdot (0.382^{\frac{3}{2} }-0^{\frac{3}{2} })-\left(\frac{1}{2} \right)\cdot (0.382^{2}-0^{2}) + 3 \cdot (2 - 0.382) - (2^{2}-0.382^{2})+4\cdot (2^{2}-0.382^{2})[/tex]

A = 16.815

The area of the region enclosed by the curves x + y = 1, x - 3 = y, y = √x and x = 0 is 16.815 square units.

The statement reports an inconsistency with at least one function and needs to be modified in order to apply definite integrals in a consistent manner. New form is shown below:

Sketch the region enclosed by the given curves and find its area: (i) x + y = 1, (ii) x - 3 = y, (iii) y = √x, (iv) x = 0.

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Answer:

1/4 ,- 4, 1/2 2 1/2 4 1/4

Step-by-step explanation:

1/4, 4 1/2, 2,1/2,4, 1/4,

The graph of the option in the question has a domain of -∞ < x < 3.5.

The domain of a graph are the possible x-values that can be obtained from the graph.

A graph that has a domain given by the inequality, -∞ < x < ∞ does not have a vertical asymptote.

An asymptote is a straight line to which a graph approaches, as either the x or y-value approaches infinity.

The given graph has a vertical asymptote at y ≈ 3.5

The domain of the given graph is therefore, -∞ < x < 3.5

Similarly, the graph has a horizontal asymptote at x ≈ 3

The range of the given graph is therefore, -∞ < y < 3.

A graph that has a domain of -∞ < x < ∞, extends to infinity to the left and the right of the graph.

A function that has a graph with a domain of -∞ < x < ∞ is one of direct proportionality.

An example is, y = x

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1)  The empirical formula for this compound is; K₂Cr₂O₇

2) It is an Ionic compound because it has two potassium ions (K+) and a negatively charged dichromate ion (Cr2O7-).

1) Elements of the unknown compound are;

Potassium = 40 g

Chromium = 52 g

Oxygen = 56 g

Let us find the number of moles of each element;

Number of moles of Potassium = 40/39 = 1.026 mol K

Number of moles of Chromium = 52/52 = 1 mol Cr

Number of moles of oxygen = 56/16 = 3.5 mol O

Multiply through the number of moles by 2 and round up to the nearest whole number to get the empirical formula at; K₂Cr₂O₇

2) The compound K₂Cr₂O₇ is called Potassium dichromate. It is classified as an ionic compound because it has two potassium ions (K+) and a negatively charged dichromate ion (Cr2O7-).

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Answer: [tex]\pm1,\pm3,\pm5, \pm15[/tex]

Step-by-step explanation:

We can list the possible rational roots of this polynomial using the Rational Root Theorem. This theorem states that all the possible rational roots of an equation follow the structure [tex]\frac{p}{q}[/tex], where p is any of the factors of the constant term and q is any of the factors of the leading coefficient.

In this example, -15 is the constant term and 1 is the leading coefficient ([tex]x^4[/tex] has a coefficient of 1).

The factors of -15 are [tex]\pm1,\pm3,\pm5, \pm15[/tex], while the factors of 1 are [tex]\pm1[/tex]. p is can be any one of the factors of -15, while q can be any of the factors of 1.

[tex]\frac{\pm1,\pm3,\pm5, \pm15}{\pm1}[/tex]

The possible roots can be any of the numbers on the top divided by any of the numbers on the bottom. Since dividing by 1 or -1 won't change any of the numbers on the top, the rational roots of this function are [tex]\pm1,\pm3,\pm5, \pm15[/tex].

[tex]X[/tex] and [tex]Y[/tex] are independent and identically distributed with PMF

[tex]\mathrm{Pr}(X = x) = \begin{cases}1/6 & \text{if } x \in\{1,2,3,4,5,6\} \\ 0 & \text{otherwise}\end{cases}[/tex]

If [tex]Z=X-Y[/tex], then [tex]Z[/tex] has range/support

{-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}

where we can get

and so on, so that the PMF of [tex]Z[/tex] is

[tex]\mathrm{Pr}(Z=z) = \begin{cases}1/36 & \text{if } z\in\{-5,5\} \\ 2/36 & \text{if } z\in\{-4,4\} \\ 3/36 & \text{if }z\in\{-3,3\} \\ 4/36 & \text{if } z\in\{-2,2\} \\ 5/36 & \text{if } z\in\{-1,1\} \\ 6/36 & \text{if } z =0 \\ 0 & \text{otherwise}\end{cases}[/tex]

Answer:

1 +-3i

Step-by-step explanation:

Answer:

answer is 1) x=±3i

Step-by-step explanation:

The answer is 4) 140.

If we closely examine the pattern of the series, we see that after a number is subtracted by a value, it is multiplied by the same value, and then it moves on to the next natural number.

The next step, according to the pattern, would be to multiply 4.

Compute the divergence of [tex]\vec F[/tex].

[tex]\mathrm{div}(\vec F) = \dfrac{\partial(2x^3+y^3)}{\partial x} + \dfrac{\partial (y^3+z^3)}{\partial y} + \dfrac{\partial(3y^2z)}{\partial z} = 6x^2 + 3y^2 + 3y^2 = 6(x^2+y^2)[/tex]

By the divergence theorem, the integral of [tex]\vec F[/tex] across [tex]S[/tex] is equivalent to the integral of [tex]\mathrm{div}(\vec F)[/tex] over the interior of [tex]S[/tex], so that

[tex]\displaystyle \iint_S \vec F\cdot d\vec S = \iiint_{\mathrm{int}(S)} \mathrm{div}(\vec F)\,dV[/tex]

The paraboloid meets the [tex]x,y[/tex]-plane in a circle with radius 3, so we have

[tex]\mathrm{int}(S) = \left\{(x,y,z) \mid x^2+y^2\le3 \text{ and } 0 \le z \le 9-x^2-y^2\right\}[/tex]

and

[tex]\displaystyle \iiint_{\mathrm{int}(S)} \mathrm{div}(\vec F) \,dV = \int_{-3}^3 \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} \int_0^{9-x^2-y^2} 6(x^2+y^2)\,dz\,dy\,dx[/tex]

Convert to cylindrical coordinates, with

[tex]\begin{cases}x = r\cos(\theta) \\ y = r\sin(\theta) \\ z = \zeta \\ dV = dx\,dy\,dz = r\,dr\,d\theta\,d\zeta\end{cases}[/tex]

so that [tex]x^2+y^2=r^2[/tex], and the domain of integration is the set

[tex]\left\{(r,\theta,\zeta) \mid 0 \le r \le 3\text{ and } 0 \le \theta\le2\pi \text{ and } 0 \le \zeta \le 9-r^2\right\}[/tex]

Now compute the integral.

[tex]\displaystyle \int_0^3 \int_0^{2\pi} \int_0^{9-r^2} 6r^2\cdot r\,d\zeta\,d\theta\,dr = 12\pi \int_0^3 \int_0^{9-r^2} r^3\, d\zeta \, dr \\\\ ~~~~~~~~~~~~ = 12\pi \int_0^3 r^3 (9 - r^2) \, dr \\\\ ~~~~~~~~~~~~ = 12\pi \int_0^3 (9r^3 - r^5) \, dr \\\\ ~~~~~~~~~~~~ = 12\pi \left(\frac94 r^4 - \frac16 r^6\right)\bigg|_0^3 = 12\pi \left(\frac94\cdot3^4-\frac16\cdot3^6\right) = \boxed{729\pi}[/tex]

Answer:

40 apples

Step-by-step explanation:

x = (4·2)/(1/5) = 8/(1/5) = (8/1):(1/5) = (8/1)·(5/1) = 40/1 = 40 apples

It takes 10 apples to make a gram of fat. Multiply the 10 apples by 4 grams of fat to find that it takes 40 apples to make 4 grams of fat.

Answer: 22

Step-by-step explanation:

Number of floor tiles = (5.4m × 4.2m) ÷ 1m²

Number of floor tiles = 22.68

Minimum number of floor tiles = 22

Answer: [tex]x=2, y=1[/tex]

Step-by-step explanation:

Subtracting the second equation from the first yields [tex]2x=4 \longrightarrow x=2[/tex].

This means that [tex]y=1[/tex].

So, the solution is [tex]x=2, y=1[/tex]

Answer:

-7

Step-by-step explanation:

8-15 = -7 but since it's an absolute value, it is 7 and then you add the negative sign in front since it is not in the absolute value.

Answer:

<-3,3>

Step-by-step explanation:

The required number is 7. The expression formed with the given information is x - 21 = 2(-x).

An expression is the combination of variables, constants, and coefficients.

If two expressions are related by an equal in between them, then that is said to be an equation.

From the given data,

Consider the required number = x

The number is decreased by 21 i.e., x - 21

Twice the opposite of the number = 2(-x)

So, on equating,

x - 21 = 2(-x)

On simplifying,

⇒ x - 21 = -2x

⇒ x + 2x = 21

⇒ 3x = 21

∴ x = 7

Thus, the required number is 7.

Check:

7 - 21 = -14 and 2(-7) = -14.

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Answer:

[tex]\boxed {AB = 52 m}[/tex]

Step-by-step explanation:

This forms a right triangle.

Therefore, by using the Pythagorean Theorem, we can find AB.

AB = √AC² + BC²

I hope it helped you solve the problem.

Good luck on your studies!

Answer: Distance AB = 52 m

Step-by-step explanation:

Given information

Point A = 20 m north of point C

Point B = 48 m east of point C

Please refer to the attachment below for a graphical understanding

Concept

According to the graph drawn, Point A, Point B, and Point C form a right angle, and the distance between Point A and B would form a right triangle.

Therefore, we can use the Pythagorean theorem to find the distance between points A and B.

Given formula

a² + b² = c²

Substitute values into the formula

a² + b² = c²

(20)² + (48)² = c²

Simplify exponents

400 + 2304 = c²

Simplify by addition

2704 = c²

c = √2704

c = 52  or  c = -52 (reject, since no distance can be negative)

Therefore, the distance AB is [tex]\Large\boxed{52~m}[/tex]

Hope this helps!! :)

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Answer:

$ 296 732 .10

Step-by-step explanation:

FV = PV ( 1 + i)^n      FV = future value = 531402     PV = ?

                                 i = decimal interest per period = .06

                                 n = periods = 10

531 402 = PV ( 1.06)^10         Solve for PV = 296 732 .10

The probability of 11 voters out of 15 voting candidate A is 0.9770

According to the statement

we have given that the percentage of voters is 81% and poll conducts are 15 and we have to find the probability that 11 of the 15 voters will prefer Candidate

So, For this purpose,  

the question in based on binomial probability which means the probability of exactly x successes on n repeated trials in an experiment which has two possible outcomes p and q and p+q=1

let probability of choosing candidate A be p

and probability of choosing candidate B be q

In question p=0.81 and q=0.16

n be total number of voters=15

x be number of successes

then P(x=x)= [tex]\frac{n}{x} p^{x} q^{n-x}[/tex]

P(x=11)= [tex]\frac{11}{15} 0.81^{11} 0.16^{4}[/tex]

= 0.9770

So, the probability of 11 voters out of 15 voting candidate A is 0.9770

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The expression that will help him to determine the total cost is as follows:

y = 60 + 4x

He signed up to provide snacks for his daughter and her school group's camping trip.

He found a package of granola bars in bulk for $60 that he thought would be enough for everyone for a few days.

He also wanted to buy some boxes of fruit snacks, which were $4 each.

The number to buy and the total cost can be expressed as follows:

let

y = total cost

x = number of boxes of fruit juice.

Therefore, the  expression that will help him to determine the total cost is as follows:

y = 60 + 4x

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Answer:

This is an example of correlation because there is an obvious relationship between the two scenarios

Step-by-step explanation:

This is not cause-and-effect therefore cannot be causation or "cause"

The amount of the loss exists 1350.

Amount of Loss means an amount equivalent to the outstanding balance of the principal amount, less any amounts recognized by perfecting rights under a security agreement, together with such interest as the executive director shall permit, to a maximum of such interest as may be permitted by rule.

Given: Mr. Black purchased a television set for $450.00. He subsequently sold the television set at a defeat of 30%.

From the given information, we get

[tex]$4500\cdot \frac{30}{100}[/tex]

simplifying, we get

[tex]$4500\cdot \frac{30}{100}[/tex]

= 1350

Therefore, the amount of the loss exists 1350.

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A)  Rotate by 90° counterclockwise.

B) Reflection transformation about the line y = 5.

C) Parallel Lines are C₂D₂ and A₂B₂; EF and E₁F₁.

D) Perpendicular lines are;  D₁F₁ and  E₁F₁; DF and EF; DC and AB.

E) Line segments with slope of 0 are; AB, C₂D₂, A₂B₂, EF and E₁F₁

F) Line segments on the boat that have an undefined slope are; Lines A₁B₁, DF and DC.

G) Slope of Line ED = 1

A) The blue segment CD is seen on the graph as a perpendicular line with 2 units while the line segment C₂D₂ is seen as a horizontal line. Thus, to match CD with C₂D₂, we will rotate by 90° counterclockwise.

B) The transformations that would be used on the blue triangle to get it to match with the red triangle is a reflection transformation about the line y = 5.

C) The line segments that are parallel to each other are; C₂D₂ and A₂B₂; EF and E₁F₁.

D) The line segments that are perpendicular are;  D₁F₁ and  E₁F₁; DF and EF; DC and AB.

E) Horizontal lines that are parallel to the x-axis have zero slope. Thus, AB, C₂D₂, A₂B₂, EF and E₁F₁ all have zero slopes.

F) Undefined slope is the slope of a vertical line. Thus, Lines A₁B₁, DF and DC have undefined slopes.

G) Slope of ED = (5 - 4)/(-10 - (-11))

Slope of ED = 1/1

Slope = 1

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Considering the definition of probability, 12 of the 48 students surveyed were in favor of the new dress code.

Probability is the greater or lesser chance that a given event will occur.

In other words, the probability establishes a relationship between the number of favorable events and the total number of possible events.

Then, the probability of any event A is defined as the ratio between the number of favorable cases (number of cases in which event A may or may not occur) and the total number of possible cases. This is called Laplace's Law.

P(A)=number of favorable events÷ number of total events

In this case, you know:

Replacing in the definition of probability:

P(A)=225 students÷ 900 students

Solving:

P(A)= 0.25

Expressed as a percentage:

P(A)= 25%

In this case, you know:

Replacing in the definition of probability:

0.25=students in favor of the new dress code÷ 48 students

Solving:

students in favor of the new dress code= 0.25×48 students

students in favor of the new dress code= 12 students

Finally, 12 of the 48 students surveyed were in favor of the new dress code.

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[tex](x^2 - y^2) \, dx + 2xy \, dy = 0[/tex]

Multiply both sides by [tex]\frac1{x^2}[/tex].

[tex]\left(1 - \dfrac{y^2}{x^2}\right) \, dx + \dfrac{2y}x \, dy = 0[/tex]

Substitute [tex]y=vx[/tex], so [tex]v=\frac yx[/tex] and [tex]dy=x\,dv+v\,dx[/tex].

[tex](1-v^2) \, dx + 2v (x\,dv + v\,dx) = 0[/tex]

[tex](1 + v^2) \, dx + 2xv \, dv = 0[/tex]

Separate the variables.

[tex]2xv\,dv = -(1 + v^2) \, dx[/tex]

[tex]\dfrac{v}{1+v^2}\,dv = -\dfrac{dx}{2x}[/tex]

Integrate both sides

[tex]\displaystyle \int \frac{v}{1+v^2}\,dv = -\frac12 \int \frac{dx}x[/tex]

On the left side, substitute [tex]w=1+v^2[/tex] and [tex]dw=2v\,dv[/tex].

[tex]\displaystyle \frac12 \int \frac{dw}w = -\frac12 \int\frac{dx}x[/tex]

[tex]\displaystyle \ln|w| = -\ln|x| + C[/tex]

Solve for [tex]w[/tex], then [tex]v[/tex], then [tex]y[/tex].

[tex]e^{\ln|w|} = e^{-\ln|x| + C}[/tex]

[tex]w = e^C e^{\ln|x^{-1}|}[/tex]

[tex]w = Cx^{-1}[/tex]

[tex]1 + v^2 = Cx^{-1}[/tex]

[tex]1 + \dfrac{y^2}{x^2} = Cx^{-1}[/tex]

[tex]\implies \boxed{x^2 + y^2 = Cx}[/tex]

Your mistake is in the first image, between third and second lines from the bottom. (It may not be the only one, it's the first one that matters.)

You incorrectly combine the fractions on the left side.

[tex]\dfrac1{-2v} -\dfrac v{-2} = \dfrac1{-2v} - \dfrac{v^2}{-2v} = \dfrac{1-v^2}{-2v} = \dfrac{v^2-1}{2v}[/tex]

Answer: 0

Step-by-step explanation:

Substituting in x=0, we get

[tex]\frac{0(0-2)}{2-2e^{2}(0)}=0[/tex]