Find a function whose graph is a parabola with vertex (1, −2) and that passes through the point (5, 14)

Answer:

86 boxes

Step-by-step explanation:

trust me its right

Answer:

1.) A = (1/2bh) 4 + lxw

= 1/2 x 7 x 13 x 4 + 7 x 7

182 + 49 = 231cm2

l think u can use the formula to find the area of the other pyramids

Answer:

x = -30

Step-by-step explanation:

Use the direct variation equation, y = kx

Plug in 2 as y and -12 as x, and solve for k:

y = kx

2 = k(-12)

-1/6 = k

So, the equation is y = -1/6x

Plug in 5 as y and solve for x:

y = -1/6x

5 = -1/6x

-30 = x

So, when y = 5, x = -30

Answer:

the answer is ( -8,-8 )

Step-by-step explanation:

the first point is in -8 and the second point is also in -8 , therefore the answer is ( -8,-8 )

ಠ_ಠ (눈‸눈) (⌐■-■)

(ب_ب) ¯\_ಠ_ಠ_/¯

Answer:

(A) [tex]5\frac{1}{4}*4\frac{1}{5}[/tex]

Step-by-step explanation:

The area of a parallelogram is the same as the area of a rectangle which is A=bh where b is the base and h is the height. Therefore, Erica can use the expression [tex]5\frac{1}{4}*4\frac{1}{5}[/tex] to find the area of the parallelogram.

The exponential growth function to model represents as P = 236,000 (1+ 0.04)⁹ and 335,902 people in 2009.

The exponential growth function is given by:-

y = A(1 +r)ⁿ

where r is the rate of growth and A is the initial amount and n is the time period.

Let t represent the number of years since 2000 and P represent the population.

The population of a city is increasing at a rate of 4% each year. In 2000, there were 236,000 people in the city.

Put A = 236,000 , r = 0.04 and t = 9, we get

P = 236,000 (1+ 0.04)⁹

P = 335,902

Hence, the exponential growth function to model represents P = 236,000 (1+ 0.04)⁹ and 335,902 people in 2009.

Learn more about exponential growth and decay here:

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Answer:

There are 46 rows with 63 seats in each row

Step-by-step explanation:

I started looking for a whole number dividing seats and rows to make up the two pieces we need to multiply. I started backward from 70 (lucky guess) and then worked my way down to 63 and 46.

Now I was also looking for something more elegant in an algabraic formula and I stated with x being the number of rows and the seats being (x=17)

so I started with X(X+17)=2898 but that fif sot pan out other than to take me to x squared +17 = 2898 - subtract 17 from each side xsquared equals 2916

square root of 2916 is 54 which started my searching for a random number.

I got lucky

Answer:

The maximum volume of the box is:

[tex]V =\frac{5}{3}\sqrt{\frac{5}{3}}[/tex]

Step-by-step explanation:

Given

[tex]Surface\ Area = 10m^2[/tex]

Required

The maximum volume of the box

Let

[tex]a \to base\ dimension[/tex]

[tex]b \to height[/tex]

The surface area of the box is:

[tex]Surface\ Area = 2(a*a + a*b + a*b)[/tex]

[tex]Surface\ Area = 2(a^2 + ab + ab)[/tex]

[tex]Surface\ Area = 2(a^2 + 2ab)[/tex]

So, we have:

[tex]2(a^2 + 2ab) = 10[/tex]

Divide both sides by 2

[tex]a^2 + 2ab = 5[/tex]

Make b the subject

[tex]2ab = 5 -a^2[/tex]

[tex]b = \frac{5 -a^2}{2a}[/tex]

The volume of the box is:

[tex]V = a*a*b[/tex]

[tex]V = a^2b[/tex]

Substitute: [tex]b = \frac{5 -a^2}{2a}[/tex]

[tex]V = a^2*\frac{5 - a^2}{2a}[/tex]

[tex]V = a*\frac{5 - a^2}{2}[/tex]

[tex]V = \frac{5a - a^3}{2}[/tex]

Spit

[tex]V = \frac{5a}{2} - \frac{a^3}{2}[/tex]

Differentiate V with respect to a

[tex]V' = \frac{5}{2} -3 * \frac{a^2}{2}[/tex]

[tex]V' = \frac{5}{2} -\frac{3a^2}{2}[/tex]

Set [tex]V' =0[/tex] to calculate a

[tex]0 = \frac{5}{2} -\frac{3a^2}{2}[/tex]

Collect like terms

[tex]\frac{3a^2}{2} = \frac{5}{2}[/tex]

Multiply both sides by 2

[tex]3a^2= 5[/tex]

Solve for a

[tex]a^2= \frac{5}{3}[/tex]

[tex]a= \sqrt{\frac{5}{3}}[/tex]

Recall that:

[tex]b = \frac{5 -a^2}{2a}[/tex]

[tex]b = \frac{5 -(\sqrt{\frac{5}{3}})^2}{2*\sqrt{\frac{5}{3}}}[/tex]

[tex]b = \frac{5 -\frac{5}{3}}{2*\sqrt{\frac{5}{3}}}[/tex]

[tex]b = \frac{\frac{15 - 5}{3}}{2*\sqrt{\frac{5}{3}}}[/tex]

[tex]b = \frac{\frac{10}{3}}{2*\sqrt{\frac{5}{3}}}[/tex]

[tex]b = \frac{\frac{5}{3}}{\sqrt{\frac{5}{3}}}[/tex]

Apply law of indices

[tex]b = (\frac{5}{3})^{1 - \frac{1}{2}}[/tex]

[tex]b = (\frac{5}{3})^{\frac{1}{2}}[/tex]

[tex]b = \sqrt{\frac{5}{3}}[/tex]

So:

[tex]V = a^2b[/tex]

[tex]V =\sqrt{(\frac{5}{3})^2} * \sqrt{\frac{5}{3}}[/tex]

[tex]V =\frac{5}{3} * \sqrt{\frac{5}{3}}[/tex]

[tex]V =\frac{5}{3}\sqrt{\frac{5}{3}}[/tex]

The maximum volume of the box which has a 10 m² surface area is given below.

[tex]\rm V_{max} = \dfrac{5}{3} *\sqrt{\dfrac{5}{2}}[/tex]

The rate of change of a function with respect to the variable is called differentiation. It can be increasing or decreasing.

We want to construct a box with a square base and we currently only have 10 m² of material to use in the construction of the box.

The surface area = 10 m²

Let a be the base length and b be the height of the box.

Surface area = 2(a² + 2ab)

  2(a² + 2ab) = 10

      a² + 2ab = 5

Then the value of b will be

[tex]\rm b = \dfrac{5-a^2}{2a}[/tex]

The volume of the box is given as

V = a²b

Then we have

[tex]\rm V = \dfrac{5-a^2 }{2a}* a^2\\\\V = \dfrac{5a - a^3}{2}\\\\V = \dfrac{5a}{2} - \dfrac{a^3}{2}[/tex]

Differentiate the equation with respect to a, and put it equal to zero for the volume to be maximum.

[tex]\begin{aligned} \dfrac{dV}{da} &= \dfrac{d}{da} ( \dfrac{5a}{2} - \dfrac{a^3}{2} ) \\\\\dfrac{dV}{da} &= 0 \\\\\dfrac{5}{2} - \dfrac{3a^2 }{2} &= 0\\\\a &= \sqrt{\dfrac{5}{2}} \end{aligned}[/tex]

Then the value of b will be

[tex]b = \dfrac{5-\sqrt{\dfrac{5}{2}} }{2*\sqrt{\dfrac{5}{2}} }\\\\\\b = \sqrt{\dfrac{5}{2}}[/tex]

Then the volume will be

[tex]\rm V = (\sqrt{\dfrac{5}{2}} )^2*\sqrt{\dfrac{5}{2}} \\\\V = \dfrac{5}{3} *\sqrt{\dfrac{5}{2}}[/tex]

More about the differentiation link is given below.

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4

Hope this helps! :)

______________

Answer:

81=1/2h×9,

81=1/18h

1458h=1

h=1/1458

Answer:

h=1/ 1458

hope it is helpful to you

Answer:

(0.5X) - 3 = Distance from the finish line

Step-by-step explanation:

Given that a student is running a 3-kilometer race, and runs 1 kilometer every 2 minutes, to determine the function that describes the distance from the finish line after X minutes, the following calculation must be performed:

1 = kilometers for every 2 minutes

X = every number of minutes

1/2 = 0.5 = kilometers per minute

(0.5X) - 3 = Distance from the finish line

Thus, if the student runs for 4 minutes, the equation would operate as follows:

0.5 x 4 - 3 = X

2 - 3 = X

-1 = X

Answer:

202 m

Step-by-step explanation:

Given :

Angle of depression, θ = 36°45' ; 45/60 = 0.75 = 36 + 0.75 = 36.75°

The height, h = 270

The distance of park bench from the building, d is given by :

Tan θ = opposite / Adjacent

Tan 36.75 = d / 270

d = 270 * Tan 36.75

d = 201.61856

d, distance of Park bench from building is 202 m

Answer:

a) The 98% Confidence Interval for the proportion of all students that prefer ebooks is (0.55, 0.65).

b) The margin of error is of 0.05.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

z is the z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].

The margin of error is of:

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In a sample of 400 students, 60% of them prefer eBooks.

This means that [tex]n = 400, \pi = 0.6[/tex]

98% confidence level

So [tex]\alpha = 0.02[/tex], z is the value of Z that has a p-value of [tex]1 - \frac{0.02}{2} = 0.99[/tex], so [tex]Z = 2.054[/tex].

Margin of error -> Question b:

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

[tex]M = 2.054\sqrt{0.6*0.4}{400}}[/tex]

[tex]M = 0.05[/tex]

The margin of error is of 0.05.

A.Find 98% Confidence Interval for the proportion of all students that prefer ebooksb.

Sample proportion plus/minus the margin of error.

0.6 - 0.05 = 0.55

0.6 + 0.05 = 0.65

The 98% Confidence Interval for the proportion of all students that prefer ebooks is (0.55, 0.65).

Answer and Step-by-step explanation:

This is for your other question in case you don't see it.

1. 2 pairs (aka 4 sides) of Opposite Sides are equal

2. AB and DC are parallel, and AD and BC are parallel  

3. Angles BDC and ACD are equal, angles DAC and DBC are equal, the angles ADB and BCD are equal, angles CAB and DBA are equal

4. 4 right angles

5. AB and DC are equal, and AD and BC are equal

6. 4 Triangles

7. False

8. Diagonals of a rectangle.

#teamtrees #PAW (Plant And Water)

Answer:20, 40, 40,100

Step-by-step explanation:

5x+2x+2x=180 9x=180 x=20

Answer:

2,160

Step-by-step explanation:

432-324=108 earnings based on sales

sales x 5%=108

sales=108/.05

sales=2,160

Answer:

[tex]B' = (-3,-3)[/tex]

[tex]C' = (-4.5,-7.5)[/tex]

Step-by-step explanation:

Given

[tex]k = 1.5[/tex]

[tex]B = (-2,2)[/tex]

[tex]C =(-3,-5)[/tex]

Required

B' and C'

This is calculated as;

[tex]B' = k * B[/tex]

[tex]B' = 1.5 * (-2,-2)[/tex]

[tex]B' = (-3,-3)[/tex]

and

[tex]C' =k * C[/tex]

[tex]C' = 1.5 * (-3,-5)[/tex]

[tex]C' = (-4.5,-7.5)[/tex]

Answer:

3.75

Step-by-step explanation:

DF = 6/24 × 15 = 3.75

________________

Answer:

18.0167≤x≤21.9833

Step-by-step explanation:

Given the following

sample size n = 10

standard deviation s = 3.2

Sample mean = 20

z-score at 95% = 1.96

Confidence Interval = x ± z×s/√n

Confidence Interval = 20 ± 1.96×3.2/√10

Confidence Interval = 20 ± (1.96×3.2/3.16)

Confidence Interval = 20 ± (1.96×1.0119)

Confidence Interval = 20 ±  1.9833

CI = {20-1.9833, 20+1.9833}

CI = {18.0167, 21.9833}

Hence the required confidence interval is 18.0167≤x≤21.9833

Answer:

Option A: [tex]5\frac{1}{4} *4\frac{1}{5}[/tex]

Step-by-step explanation:

Area of a parallelogram is base times height so in this case it would be 5 1/4 times 4 1/5. 5 1/4 being the base and 4 1/5 being the height.

Answer:

The length of the three sides [tex]5, \sqrt{58} , \sqrt{65}[/tex]

The triangle is not a right triangle

Step-by-step explanation:

A = (3, 2) , B = ( 6, 9) , C = (10, 6)

Find the lengths using distance formula.

[tex]distance = \sqrt{(x_2 -x_1)^2 + (y_2 - y_1)^2}[/tex]

[tex]AB = \sqrt{(6-3)^2 + (9-2)^2} = \sqrt{9 + 49 } = \sqrt{58}[/tex]

[tex]BC = \sqrt{(10-6)^2 + (6-9)^2} = \sqrt{16 + 9} = \sqrt{25} = 5[/tex]

[tex]AC = \sqrt{(3-10)^2+(2-6)^2} = \sqrt{49 + 16} = \sqrt{65}[/tex]

Using Pythagoras theorem :

[tex](Longer \ side)^2 = sum \ of \ square \ of \ two \ other \ sides[/tex]

Longest side is AC . So we will check if it satisfies Pythagoras theorem :

[tex]\sqrt{65} = \sqrt{58} + 5^2\\65 = 58 + 25\\[/tex]

65 ≠ 58 + 25

So the sides does not satisfy Pythagoras theorem. Hence the triangle is not a right triangle.

Answer:

20$ i think

Step-by-step explanation:

A fraction is a way to describe a part of a whole. If a car is parked in the garage for 5 hours and 30 minutes the charge will be $10.

A fraction is a way to describe a part of a whole. such as the fraction ¼ can be described as 0.25.

Given that for the first hour that a car is parked in the garage, there is no charge. After the first hour, for the next two hours that a car is parked in the garage, there is a $5 charge. After the third hour, the garage charges $2 for each additional hour that the car is parked in the garage.

1.) Since the car is parked for 30 minutes only.

Charge = $0

Hence, No charge will be charged for 30 minutes.

2.) If a car is parked in the garage for 2 hours and 30 minutes.

Charge for 1 hour = $0

Charge for the next 1 hour and 30 minutes = $5

Hence, the charge will be $5.

3.)  If a car is parked in the garage for 5 hours.

Charge for 1 hour = $0

Charge for the next 2 hour = $5

Charge for the next 2 hour = 2×$2 = $4

Charge = $5 + $4 = $9

Hence, the charge will be $9.

4.) If a car is parked in the garage for 5 hours and 30 minutes.

Charge for 1 hour = $0

Charge for the next 2 hour = $5

Charge for the next 2 hour 30 minutes = 2.5 ×$2 = $5

Charge = $5 + $5 = $10

Hence, the charge will be $10.

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Answer:

the last one is correct.............

Answer:

To be able to approximate the sampling distribution with a normal model, it is needed that [tex]np \geq 10[/tex] and [tex]n(1-p) \geq 10[/tex], and both conditions are satisfied in this problem.

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they will make payments on time, or they won't. The probability of a person making the payment on time is independent of any other person, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

Probability of exactly x successes on n repeated trials, with p probability.

The sampling distribution can be approximated to a normal model if:

[tex]np \geq 10[/tex] and [tex]n(1-p) \geq 10[/tex]

Based on past experience, a bank believes that 8.9 % of the people who receive loans will not make payments on time.

This means that [tex]p = 0.089[/tex]

The bank has recently approved 220 loans.

This means that [tex]n = 220[/tex]

What must be true to be able to approximate the sampling distribution with a normal model?

[tex]np = 220*0.089 = 19.58 \geq 10[/tex]

[tex]n(1-p) = 220*0.911 = 200.42 \geq 10[/tex]

To be able to approximate the sampling distribution with a normal model, it is needed that [tex]np \geq 10[/tex] and [tex]n(1-p) \geq 10[/tex], and both conditions are satisfied in this problem.

243 = the pages read on Saturday

243 - 53 = 190 = the pages read on Sunday

190 + 243 =  433 = total pages read

Answer = 433

Answer:

The rate of change is 13 new participants every year.

Step-by-step explanation:

y=13x+50

This is in the form y = mx+b  where m is the slope and b is the y intercept

x is the number of years so for every year increase we gain 13 new participants