Please help, i really need this.

15 points

the center is at the origin of a coordinate system and the foci are on the y-axis, then the foci are symmetric about the origin.

The hyperbola focus F1 is 46 feet above the vertex of the parabola and the hyperbola focus F2 is 6 ft above the parabola's vertex. Then the distance F1F2 is 46-6=40 ft.

In terms of hyperbola, F1F2=2c, c=20.

The vertex of the hyperba is 2 ft below focus F1, then in terms of hyperbola c-a=2 and a=c-2=18 ft.

Use formula c^2=a^2+b^2c

2

=a

2

+b

2

to find b:

\begin{gathered} (20)^2=(18)^2+b^2,\\ b^2=400-324=76 \end{gathered}

(20)

2

=(18)

2

+b

2

,

b

2

=400−324=76

.

The branches of hyperbola go in y-direction, so the equation of hyperbola is

\dfrac{y^2}{b^2}- \dfrac{x^2}{a^2}=1

b

2

y

2

−

a

2

x

2

=1 .

Substitute a and b:

\dfrac{y^2}{76}- \dfrac{x^2}{324}=1

76

y

2

−

324

x

2

=1 .

Answer: 3

happy learning

Answer:

B.

Step-by-step explanation:

From the point (-1,0) the next point on the graph is up 3, right 1, making the slope a positive 3.

Answer:

182/3,3 8/3, 152/3

Step-by-step explanation:

a+b+c=124

a trừ c=  10

4b=c

Answer:

a=29,b=79,c=19

Step-by-step explanation:

a=c+10

b=4c

=> a+b+c=c+10+4c+c=124

=> c=19

=> a= 29, b=79

Answer:

The right answer is:

(a) 0.1456

(b) 18.125, 69.1202, 8.3139

Step-by-step explanation:

Given:

N = 24

n = 5

r = 7

The improperly drilled gearboxes "X".

then,

⇒ [tex]P(X) = \frac{\binom{7}{x} \binom {17}{5-x}}{\binom{24}{5}}[/tex]

(a)

P (all gearboxes fit properly) = [tex]P(x=0)[/tex]

                                               = [tex]\frac{\binom{7}{0} \binom{17}{5}}{\binom{24}{5}}[/tex]

                                               = [tex]0.1456[/tex]

(b)

According to the question,

[tex]X = 91+5[/tex]

Mean will be:

⇒ [tex]\mu = E(x)[/tex]

       [tex]=E(91+5)[/tex]

       [tex]=9E(1)+5[/tex]

       [tex]=9.\frac{nr}{N}+5[/tex]

       [tex]=9.\frac{5.7}{24} +5[/tex]

       [tex]=18.125[/tex]

Variance will be:

⇒ [tex]\sigma^2=Var(X)[/tex]

         [tex]=V(9Y+5)[/tex]

         [tex]=81.V(Y)[/tex]

         [tex]=81.n.\frac{r}{N}.\frac{N-r}{N}.\frac{N-n}{N-1}[/tex]

         [tex]=81.5.\frac{7}{24}.\frac{24-7}{24}.\frac{24-5}{24-1}[/tex]

         [tex]=69.1202[/tex]

Standard deviation will be:

⇒ [tex]\sigma = \sqrt{69.1202}[/tex]

       [tex]=8.3139[/tex]      

A shape of a trapezoid has exactly one line of symmetry. The correct option is B.

An open, flat object with four straight sides and one pair of parallel sides is referred to as a trapezoid or trapezium.

A balanced and proportionate likeness between an object's two halves is referred to as symmetry in geometry. It implies that one half is the other's mirror image.

A trapezium's non-parallel sides are referred to as the legs, while its parallel sides are referred to as the bases. The legs of a trapezium can also be parallel. The parallel sides may be vertical, horizontal, or angled.

Therefore, the shape of a trapezoid has exactly one line of symmetry. The correct option is B.

To know more about Trapezoids follow

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Answer:

a) 75

b) 4.33

c) 0.75

d) [tex]3.2 \times 10^{-13}[/tex] probability that no one in a simple random sample of 100 young adults owns a landline

e) [tex]6.2 \times 10^{-61}[/tex] probability that everyone in a simple random sample of 100 young adults owns a landline.

f) Binomial, with [tex]n = 100, p = 0.75[/tex]

g) [tex]4.5 \times 10^{-8}[/tex] probability that exactly half the young adults in a simple random sample of 100 do not own a landline.

Step-by-step explanation:

For each young adult, there are only two possible outcomes. Either they do not own a landline, or they do. The probability of an young adult not having a landline is independent of any other adult, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

The expected value of the binomial distribution is:

[tex]E(X) = np[/tex]

The standard deviation of the binomial distribution is:

[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]

75% of all young adults between the ages of 18-35 do not have a landline in their homes and only use a cell phone at home.

This means that [tex]p = 0.75[/tex]

(a) On average, how many young adults do not own a landline in a random sample of 100?

Sample of 100, so [tex]n = 100[/tex]

[tex]E(X) = np = 100(0.75) = 75[/tex]

(b) What is the standard deviation of probability of young adults who do not own a landline in a simple random sample of 100?

[tex]\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{100(0.75)(0.25)} = 4.33[/tex]

(c) What is the proportion of young adults who do not own a landline?

The estimation, of 75% = 0.75.

(d) What is the probability that no one in a simple random sample of 100 young adults owns a landline?

This is P(X = 100), that is, all do not own. So

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 100) = C_{100,100}.(0.75)^{100}.(0.25)^{0} = 3.2 \times 10^{-13}[/tex]

[tex]3.2 \times 10^{-13}[/tex] probability that no one in a simple random sample of 100 young adults owns a landline.

(e) What is the probability that everyone in a simple random sample of 100 young adults owns a landline?

This is P(X = 0), that is, all own. So

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{100,0}.(0.75)^{0}.(0.25)^{100} = 6.2 \times 10^{-61}[/tex]

[tex]6.2 \times 10^{-61}[/tex] probability that everyone in a simple random sample of 100 young adults owns a landline.

(f) What is the distribution of the number of young adults in a sample of 100 who do not own a landline?

Binomial, with [tex]n = 100, p = 0.75[/tex]

(g) What is the probability that exactly half the young adults in a simple random sample of 100 do not own a landline?

This is P(X = 50). So

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 50) = C_{100,50}.(0.75)^{50}.(0.25)^{50} = 4.5 \times 10^{-8}[/tex]

[tex]4.5 \times 10^{-8}[/tex] probability that exactly half the young adults in a simple random sample of 100 do not own a landline.

The work required for the given task of pumping all of the oil out over the top of the tank is 7,500 ft·lb

The known parameters;

The length of the rectangular tank, l = 4 feet

The width of the tank, w = 2 feet

The depth of the tank, h = 5 feet

The weight density of the oil with which the tank is filled, ρ × g = 50 lb/ft³

The unknown parameter

The work required to pump all of the oil out over the top of the tank

Method;

Calculate the force required to lift each slice (layer) of the oil to the top multiplied by the distance, y, the slice moves and summing the result as an integration as follows;

The volume of each slice, [tex]\mathbf{V_i}[/tex] = l × w × dy

The force required to move each slice, [tex]\mathbf{F_i}[/tex] = ρ × g × l × w × dy

The work done, [tex]\mathbf{W_i}[/tex], in moving the slice a distance, y, is given as follows;

[tex]\mathbf{W_i}[/tex] = ρ × g × l × w × y × dy

Therefore, the total work done, W, in pumping all the water located from y = 0, to y = 5, to the top of the tank, is given as follows;

[tex]\mathbf{W = \int\limits^5_0 {(\rho \times g \times l \times w \times y) } \, dy}[/tex]

Therefore;

W = (ρ × g × l × w × y²)/2

Plugging in the values, gives;

W = (50 lb/ft³ × 4 ft. × 3 ft. × (5 ft.)²)/2 = 7,500 ft·lb

The work required to pump all of the oil out over the top of the tank, W = 7,500 ft·lb.

Learn more about the use of integration to calculate the amount of work required for a given task here;

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Answer:

640

Step-by-step explanation:

muntiply all the number length width height

Answer:

[tex]5xy[/tex]

Step-by-step explanation:

[tex]\mathrm{Factor\:}:5x^2y[/tex]

[tex]5\cdot \:x\cdot \:x\cdot \:y[/tex]

[tex]\mathrm{Factor\:}:10xy^2[/tex]

[tex]2\cdot \:5\cdot \:x\cdot \:y\cdot \:y[/tex]

Common factor:-

[tex]5\cdot \:x\cdot \:y[/tex]

OAmalOHopeO

Answer:

team a and team c scored 74 points which is 48 points more than team b, scoring 26 points.

Step-by-step explanation:

Answer:

Its the first one | -9 | < | 9 |

Step-by-step explanation:

its false because both sides are identical

Hope this helps!

Answer:

Step-by-step explanation:

Part (a)

The standard error (SE) formula is

[tex]\text{SE} = \frac{\sigma}{\sqrt{n}}\\\\[/tex]

where n is the sample size. We're given SE = 2 and sigma = 34, so,

[tex]\text{SE} = \frac{\sigma}{\sqrt{n}}\\\\2 = \frac{34}{\sqrt{n}}\\\\2\sqrt{n} = 34\\\\\sqrt{n} = \frac{34}{2}\\\\\sqrt{n} = 17\\\\n = 17^2\\\\n = 289\\\\[/tex]

So we need a sample size of n = 289 to have an SE value of 2.

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

Part (b)

We'll use SE = 1 this time

[tex]\text{SE} = \frac{\sigma}{\sqrt{n}}\\\\1 = \frac{34}{\sqrt{n}}\\\\1*\sqrt{n} = 34\\\\\sqrt{n} = 34\\\\n = 34^2\\\\n = 1156\\\\[/tex]

Because we require greater precision (i.e. a smaller SE value), the sample size must be larger to account for this. In other words, as SE goes down, then n must go up, and vice versa.

Answer:

[tex]x=2[/tex]

Step-by-step explanation:

[tex]\frac{3x-2}{7}-\frac{5x-8}{4}=\frac{1}{14}[/tex]

In order to factor an integer, we need to divide it by the ascending sequence of primes 2, 3, 5.

The number of times that each prime divides the original integer becomes its exponent in the final result.

In here,  Prime number 2 to the power of 2 equals 4.

[tex]\frac{3x-2}{7}-\frac{5x-8}{2^{2} }=\frac{1}{14}[/tex]

First, We need to add fractions-

Rule:-

[tex]\frac{A}{B} +\frac{C}{D} =\frac{\frac{LCD}{B}+\frac{LCD}{D}C }{LCD}[/tex]

LCD = [tex]7 \cdot 2^{2}[/tex]

[tex]\frac{4(3x-2)+7(-(5x-8))}{7*2^{2} } =\frac{1}{14}[/tex]

[tex]x=2[/tex]

OAmalOHopeO

Answer:

[tex]C. \ \ \ 86[/tex]°

Step-by-step explanation:

1. Approach

In order to solve this problem, one must first find a relationship between arc (a) and arc (c). This can be done using the congruent arcs cut congruent segments theorem. After doing so, one can then use the secants interior angle to find the precise measurement of arc (a).

2. Arc (a) and arc (c)

A secant is a line or line segment that intersects a circle in two places. The congruent segments cut congruent arcs theorem states that when two secants are congruent, meaning the part of the secant that is within the circle is congruent to another part of a secant that is within that same circle, the arcs surrounding the congruent secants are congruent. Applying this theorem to the given situation, one can state the following:

[tex]a = c[/tex]

3. Finding the degree measure of arc (a),

The secants interior angle theorem states that when two secants intersect inside of a circle, the measure of any of the angles formed is equal to half of the sum of the arcs surrounding the angles. One can apply this here by stating the following:

[tex]86=\frac{a+c}{2}[/tex]

Substitute,

[tex]86=\frac{a+c}{2}[/tex]

[tex]86=\frac{a+a}{2}[/tex]

Simplify,

[tex]86=\frac{a+a}{2}[/tex]

[tex]86=\frac{2a}{2}[/tex]

[tex]86=a[/tex]

Answer:

[tex]0.04 = 4 \div 100 [/tex]

Answer:

Check the image

Answer:

B. 0

Step-by-step explanation:

Rate of change from x = -3 to x = 5

Rate of change = [tex] \frac{f(b) - f(a)}{b - a} [/tex]

where, from the graph, we have:

a = -3, f(a) = -1,

b = 5, f(b) = -1,

Plug in the values

Rate of change = [tex] \frac{-1 -(-1)}{5 - (-3)} [/tex]

Rate of change = [tex] \frac{0}{8} [/tex]

Rate of change = 0

Answer:

2^(1/6) (cos(-pi/12)+i sin(-pi/12))

2^(1/6) (cos(3pi/12)+i sin(3pi/12))

2^(1/6) (cos(7pi/12)+i sin(7pi/12))

2^(1/6) (cos(11pi/12)+i sin(11pi/12))

2^(1/6) (cos(5pi/4)+i sin(5pi/4))

2^(1/6) (cos(19pi/12)+i sin(19pi/12))

Step-by-step explanation:

Let's convert to polar form.

-2i=2(cos(A)+i sin(A) )

There is no real part so cos(A) has to be zero and since we want -2 and we already have 2 then we need sin(A)=-1 so let's choose A=-pi/2.

So z=2(cos(-pi/2)+i sin(-pi/2)).

There are actually infinitely many ways we can write this polar form which we will need.

z=2(cos(-pi/2+2pi k)+i sin(-pi/2+2pi k))

where k is an integer

Now let's find the 6 6th roots or z.

2^(1/6) (cos(-pi/12+2pi k/6)+i sin(-pi/12+2pi k/6))

Reducing

2^(1/6) (cos(-pi/12+pi k/3)+i sin(-pi/12+pi k/3))

Plug in k=0,1,2,3,4,5 to find the 6 6th roots.

k=0:

2^(1/6) (cos(-pi/12+pi (0)/3)+i sin(-pi/12+pi (0)/3))

=2^(1/6) (cos(-pi/12)+i sin(-pi/12))

k=1:

2^(1/6) (cos(-pi/12+pi/3)+i sin(-pi/12+pi/3))

2^(1/6) (cos(3pi/12)+i sin(3pi/12))

k=2:

2^(1/6) (cos(-pi/12+2pi/3)+i sin(-pi/12+2pi/3))

2^(1/6) (cos(7pi/12)+i sin(7pi/12))

k=3:

2^(1/6) (cos(-pi/12+3pi/3)+i sin(-pi/12+3pi/3))

2^(1/6) (cos(11pi/12)+i sin(11pi/12))

k=4:

2^(1/6) (cos(-pi/12+4pi/3)+i sin(-pi/12+4pi/3))

2^(1/6) (cos(15pi/12)+i sin(15pi/12))

2^(1/6) (cos(5pi/4)+i sin(5pi/4))

k=5:

2^(1/6) (cos(-pi/12+5pi/3)+i sin(-pi/12+5pi/3))

2^(1/6) (cos(19pi/12)+i sin(19pi/12))

Answer:

h = 6[tex]\sqrt{3}[/tex]

Step-by-step explanation:

The given is the special right triangle with angle measures : 90-60-30

and the side lengths for the given angles are represented by :

2a-a[tex]\sqrt{3}[/tex]-a

the side length that sees 60 degrees is represented by a[tex]\sqrt{3}[/tex] (h in this case)

the area of a triangle is calculated by multiplying height and base and that is divided by 2

a[tex]\sqrt{3}[/tex]*a/2 = 18[tex]\sqrt{3}[/tex] multiply both sides by 2

a^2[tex]\sqrt{3}[/tex] = 36[tex]\sqrt{3}[/tex] divide both sides by [tex]\sqrt{3}[/tex]

a^2 = 36 find the roots for both sides

a = 6

since h sees angle measure 60 and is represented by a[tex]\sqrt{3}[/tex]

h = 6[tex]\sqrt{3}[/tex]

Answer:

y = -1

Step-by-step explanation:

Answer:

The student is completely incorrect because there is no solution to this inequality.

Answer:

D on edge

Step-by-step explanation:

9514 1404 393

Answer:

  C. EGD

Step-by-step explanation:

A major arc is typically named using the end points and a point on the arc. Here, the end points are E and D, and points on the major arc include C, G, and F. The major arc ED could be named any of

Of course, the reverse of any of these names could also be used: DCE, DGE, DFE.

The ratio/proportion is young / old = young / old.

We know that one ratio is 3 / 5, so we need to complete the other.

3 / 5 = young / 40

5 goes into 40, 8 times, therefore we need to multiply the numerator by 8 also.

3 x 8 = 24

The younger student is 24 years old.

Hope this helps!

Answer:

24

Step-by-step explanation:

younger : older

3             :5

The older is 40

40/5 = 8

Multiply each by 8

younger : older

3 *8       :5 *8

24         : 40

The younger is 24

Answer:

[tex]x = y^3+5y[/tex]

Step-by-step explanation:

Complete question

[tex]\frac{x - 5y}{y^3} - 1=0\\[/tex]

Required

Solve for x

We have:

[tex]\frac{x - 5y}{y^3} - 1=0[/tex]

Collect like terms

[tex]\frac{x - 5y}{y^3} = 1[/tex]

Multiply through by [tex]y^3[/tex]

[tex]x - 5y = y^3[/tex]

Make x the subject

[tex]x = y^3+5y[/tex]

Answer:

Base = 21 while Height = 16

Answer:

Step-by-step explanation:

Answer:

Hello,

[tex]\boxed{y'=2-\dfrac{200}{x^2} }\\[/tex]

Step-by-step explanation:

[tex](f(x)+g(x))'=f'(x)+g'(x)\\\\(2x)'=2*(x)'=2*1=2\\\\(\dfrac{200}{x} )'=200*(x^{-1})'=200*(-1)*x^{-1-1})=-\dfrac{200}{x^2} \\\\\\\boxed{y'=2-\dfrac{200}{x^2} }\\[/tex]

Answer:

[tex] \frac{dy}{dx } = 2 - \frac{200}{ {x}^{2} } [/tex]

Step-by-step explanation:

[tex]the \: equation \: can \: be \: rewriten \: as \\ y = 2x + 200 {x}^{ - 1} \\ \\ now \: differentiate \: the \: equation\ \\ \frac{dy}{dx} = 2 - 200 {x}^{ - 2} \\ \frac{dy}{dx} = 2 - \frac{200}{ {x}^{2} } [/tex]

Answer:

1 inch= 20 miles. 5*20=100 miles. The answer is 100 miles.

Step-by-step explanation:

On these types of questions just do that every time, then you don't need to ask, for example:

1 foot = 50 miles

If it measures 3 feet.

3*50=150 miles.

If you have any questions regarding my answer, tell me in the comments, and I will answer them.

Answer:

The AP is 1, 11/2, 10, 29/2, 19, ....

Step-by-step explanation:

Let the first term be a and d be the common difference of the arithmetic progression.

ATQ, a+2d+a+6d=38, 2a+8d=38 and a+8d=37. Solving this, we will get a=1 and d=9/2. The AP is 1, 11/2, 10, 29/2, 19, ....