A stone is thrown vertically upward at a velocity of 10 feet per second from a bridge that is 60 feet above the level of the water. The height h (in feet) of the stone at time t (in seconds) after it is thrown is given by
h = -16t2 + 10 + 60.

(a) Find the time when the stone is again 60 feet above the water. ____sec

(b) Find the time when the stone strikes the water. (Round your answer to two decimal places.) ____sec

(c) Does the stone reach a height of 90 feet? O Yes O No​

Answer:

? = 13.6

Step-by-step explanation:

Let the unknown angle be y

so

tan y= p/b

tan y =8/33

y = tan‐¹(8/33)

y = 13.62699486

y = 13.6

Answer:

Rugby lawyer

Step-by-step explanation:

Aaron is a partner in the firm’s dispute resolution division. He advises clients on a range of litigious and risk related matters, with particular expertise in the areas of corporate misconduct, white collar criminal and regulatory affairs, sports law and employment law.  Aaron leads our sports law practice, and is a member of the firm’s health and safety, public law, and organisational integrity teams.

Well regarded by clients for his ability to analyse and strategise complex situations, Aaron is internationally recognised for his ability to implement pragmatic and commercial strategies to minimise risk and create opportunity. This ability has resulted in clients avoiding significant litigation and commercial consequences.

Aaron is an experienced advocate, having argued cases in the District Court, High Court, Employment Court, the Court of Appeal and Supreme Court of New Zealand, along with numerous tribunals.

He is recognised by international legal directories including by Chambers & Partners (Asia Pacific), Who’s Who Legal, Expert Guides, Best Lawyers and Doyles.

Before joining MinterEllisonRuddWatts Aaron practiced as a barrister with Paul Davison QC, and has lectured at the University of Auckland.

Answer:

A)

[tex]k=0[/tex]

B)

[tex]\displaystyle \begin{aligned} 2k + 1& = 2\ln 20 + 1 \\ &\approx 2.3863\end{aligned}[/tex]

C)

[tex]\displaystyle \begin{aligned} k - 3&= \ln \frac{1}{2} - 3 \\ &\approx-3.6931 \end{aligned}[/tex]

Step-by-step explanation:

We are given the function:

[tex]\displaystyle h(x) = 20e^{kx} \text{ where } k \in \mathbb{R}[/tex]

A)

Given that h(1) = 20, we want to find k.

h(1) = 20 means that h(x) = 20 when x = 1. Substitute:

[tex]\displaystyle (20) = 20e^{k(1)}[/tex]

Simplify:

[tex]1= e^k[/tex]

Anything raised to zero (except for zero) is one. Therefore:

[tex]k=0[/tex]

B)

Given that h(1) = 40, we want to find 2k + 1.

Likewise, this means that h(x) = 40 when x = 1. Substitute:

[tex]\displaystyle (40) = 20e^{k(1)}[/tex]

Simplify:

[tex]\displaystyle 2 = e^{k}[/tex]

We can take the natural log of both sides:

[tex]\displaystyle \ln 2 = \underbrace{k\ln e}_{\ln a^b = b\ln a}[/tex]

By definition, ln(e) = 1. Hence:

[tex]\displaystyle k = \ln 2[/tex]

Therefore:

[tex]2k+1 = 2\ln 2+ 1 \approx 2.3863[/tex]

C)

Given that h(1) = 10, we want to find k - 3.

Again, this meas that h(x) = 10 when x = 1. Substitute:

[tex]\displaystyle (10) = 20e^{k(1)}[/tex]

Simplfy:

[tex]\displaystyle \frac{1}{2} = e^k[/tex]

Take the natural log of both sides:

[tex]\displaystyle \ln \frac{1}{2} = k\ln e[/tex]

Therefore:

[tex]\displaystyle k = \ln \frac{1}{2}[/tex]

Therefore:

[tex]\displaystyle k - 3 = \ln\frac{1}{2} - 3\approx-3.6931[/tex]

Answer:

Tandin has 16 children.

Step-by-step explanation:

Total of children:

3+5 = 8(first woman)

2(youngest wife)

1 + 1 = 2(two of the  remaining wives)

So

8 + 2 + 2 = 12

If the ratio of children of  5th wife was 1:3 with the children of other wives.

Thus the 5th wife has 12/3 = 4 children.

How many  children does Tandin have?

12 + 4 = 16

Tandin has 16 children.

Answer:

682

Step-by-step explanation:

191,919 + 262,626

454545 ÷ 666 = 682.5

Thus meaning 682 bags will have 666 berries and one bag will have 333 berries.

Answer:

Step-by-step explanation:

[tex](a^2-1)x^2+(a-1)x+a^2-4a+3=0\\\\Calculate\ and\ identify\ the\ polynomials\\\\\Longleftrightarrow\ a^2x^2-x^2+ax-x+a^2-4a+3=0\\\\\Longleftrightarrow\ a^2x^2+ax+a^2-4a+3=x^2+x+0\\\\\Longleftrightarrow\ \left\{\begin{array}{ccc}a^2&=&1\\a&=&1\\a^2-4a+3&=&0\\\end{array} \right.\\\\\Longleftrightarrow\ \left\{\begin{array}{ccc}(a-1)(a+1)&=&0\\a-1&=&0\\(a-1)(a-3)&=&0\\\end{array} \right.\\\\\\We\ must\ exclude\ a=-1\ and\ a=3\ (not\ solution)\\\Longrightarrow\ a=1\\[/tex]

Answer:

true. A rational number can be expressed as the quotient a/b where b ≠ 0

Answer:

The probability tree is;

                            0.95        [tex](+)[/tex]

                   [tex](S)[/tex]

          0.1              0.05        [tex](-)[/tex]

[  P  ]

          0.9             0.15          [tex](+)[/tex]                                                        

                  [tex](S_{no})[/tex]    

                             0.85         [tex](-)[/tex]        

Step-by-step explanation:

Given the data in the question;

10% of the rugby team members use steroids

so Probability of using steroid; P( use steroid ) = 10% = 0.10

Probability of not using steroid; P( no steroid use ) = 1 - 0.10 = 0.90

Since the test show positive for an athlete who uses steroids, 95% of the time.

Probability of using steroids and testing positive = 95% = 0.95

Probability of using steroids and testing Negative = 1 - 0.95 = 0.05

Also from the test, 15% of all steroid-free individuals also test positive.

so

Probability of not using steroids and testing positive = 15% = 0.15

Probability of not using steroids and testing negative = 1 - 0.15 = 0.85

To set up the probability tree, Let;

[tex](S)[/tex] represent steroid use

[tex](S_{no})[/tex] represent no steroid use

[tex](+)[/tex] represent test positive

[tex](-)[/tex] represent test negative

so we have;

                            0.95        [tex](+)[/tex]

                   [tex](S)[/tex]

          0.1              0.05        [tex](-)[/tex]

[  P  ]

          0.9             0.15          [tex](+)[/tex]                                                        

                  [tex](S_{no})[/tex]    

                             0.85         [tex](-)[/tex]        

Answer:

6'2

Step-by-step explanation:

Consecutive terms in this sequence are separated by a constant c, so if the 4th term is 15, then the next terms would be

5th: 15 + c

6th: (15 + c) + c = 15 + 2c

7th: (15 + 2c) + c = 15 + 3c

and so on. More generally, since any given number in the sequence depends on the number that came before it, we can write the n-th term in terms of the 4th term,

n-th: 15 + (n - 4) c

Then the 9th term in the sequence is

15 + (9 - 4) c = 35

and solving for c gives

15 + 5c = 35   ==>   5c = 20   ==>   c = 4

Then the 15th term would be

15 + (15 - 4)×4 = 15 + 11×4 = 15 + 44 = 59

We can seperate (3b³) into two different parts, the constant and the variable.

The constant (3) and the variable (b) can both be squared and multiplied to get the correct answer, so:

3² = 9

(b³)² = [tex]b^{6}[/tex]

So, [tex](3b^{3})^{2} = 9b^{6}[/tex]

Complete Question

Polymer composite materials have gained popularity because they have high strength to weight ratios and are relatively easy and inexpensive to manufacture. However, their nondegradable nature has prompted development of environmentally friendly composites using natural materials. An article reported that for a sample of 10 specimens with 2% fiber content, the sample mean tensile strength (MPa) was 51.1 and the sample standard deviation was 1.2. Suppose the true average strength for 0% fibers (pure cellulose) is known to be 48 MPa. Does the data provide compelling evidence for concluding that true average strength for the WSF/cellulose composite exceeds this value? (Use α = 0.05.)  

t=8.169  

P-value= ?

Answer:

a)  [tex]P-value=0[/tex]

b)  Hence,We FAil to reject the alternative hypothesis and accept that the true average strength for the WSF/  cellulose composite exceeds 48 MPa.

Step-by-step explanation:

From the question we are told that:

Sample size [tex]n=10[/tex]

Mean [tex]\=x= 51.3[/tex]

Standard deviation [tex]\sigma=1.2[/tex]

Significance level is taken as [tex]\alpha=0.05[/tex]

t test statistics

[tex]t=8.169[/tex]

Therefore

[tex]P-Value=P(t>8.169)[/tex]

Critical point

[tex]t_{\alpha,df}[/tex]

[tex]\alpha=0.05[/tex]

[tex]df=10-1=>9[/tex]

Therefore

P-value from T distribution table

[tex]P-value=0[/tex]

Conclusion

[tex]P-value (0)< \alpha(0.05)[/tex]

We Reject the Null Hypothesis [tex]H_0[/tex]

Hence,We FAil to reject the alternative hypothesis and accept that the true average strength for the WSF/  cellulose composite exceeds 48 MPa.

Answer:

Step-by-step explanation:

[tex]\tt{ \green{P} \orange{s} \red{y} \blue{x} \pink{c} \purple{h} \green{i} e}[/tex]

Answer:

[tex]88 \frac{19}{32} [/tex]

Answer:

Here we have the matrix:

[tex]M = \left[\begin{array}{ccc}1&0\\0&3\end{array}\right][/tex]

And we want to find its inverse.

The inverse of a 2x2 matrix A is:

(1/det(A))*adj(A)

where det(A) is the determinant of the matrix.

Such that for a matrix:

[tex]A = \left[\begin{array}{ccc}a_{11}&a_{12}\\a_{21}&a_{22}\end{array}\right][/tex]

The determinant is:

det(A) = a₁₁*a₂₂ - a₁₂*a₂₁

in the case of our matrix M, the determinant is:

det(M) = 1*3 - 0*0 = 3

and adj(A) is a transposition along the diagonal, and for the other elements, we just change its sign.

Then for our matrix A we would have:

[tex]adj(A) = \left[\begin{array}{ccc}a_{22}&-a_{12}\\-a_{21}&a_{11}\end{array}\right][/tex]

Then for our matrix M, we have:

[tex]adj(M) = \left[\begin{array}{ccc}3&-0\\-0&1\end{array}\right][/tex]

Then the inverse of the matrix M is:

[tex]M^{-1} = \frac{1}{det(M)} *adj(M) = \frac{1}{3}\left[\begin{array}{ccc}3&0\\0&1\end{array}\right] = \left[\begin{array}{ccc}1&0\\0&1/3\end{array}\right][/tex]

Answer:

z(max) = 16

z(min) = -24

Step-by-step explanation:

2x - y = 12  multiply by 2

4x - 2y = 24  (1)

4x + 2y = 0  add equations

8x = 24

x = 3

4(3) + 2y = 0

y = -6

so (3, -6) is a common point on these two lines

z = 2(3) + 5(-6) = -24

4x - 2y = 24   (1)

x + 2y = 6   add equations

      5x = 30

        x = 6

6 + 2y = 6

         y = 0

so (6, 0) is a common point on these two lines

z = 2(6) + 5(0) = 12

4x + 2y = 0    multiply by -1

-4x - 2y = 0

  x + 2y = 6  add equations

       -3x = 6

         x = -2

-2 + 2y = 6

         y = 4

so (-2, 4) is a common point on these two lines

z = 2(-2) + 5(4) = 16

Answer:

????

Step-by-step explanation:

as in y = 4.95 + 3.99 or points? if so just draw a horizontal line at 8.94

Answer:

Substitute the functions and the value of the functions.

Step-by-step explanation:

Doing all will be long, so i'll present a and d

Here,(no a)

f(x)=3x-1, g(x)=x^2+2

Now,

f(g(x))=f(x^2+2)=3(x^2+2)-1=3x^2+6-1=3x^2+5

g(f(x))=g(3x-1)=(3x-1)^2+2=9x^2-6x+1+2=9x^2-6x+3

Here, (no d)

f(x)=x^2-9, g(x)=√(x+4)

Now,

f(g(x))=f(√(x+4))=(√(x+4))^2-9=x+4-9=x-5

g(f(x))=g(x^2-9)=√(x^2-9+4)=√(x^2-5)

first we shall learn the rules.when numbers with same sign are divided it gives pisitive sign but, when numbers of different signs are divided it gives negetive sign.

here,

7. (-154) ➗ (-14) =11

11. (-40) ➗10=-4

15. 90 ➗ (-15)=-6

16. 108 ➗ (-9)=-12

17. (-48) ➗ (-6)=8

18. (-105) ➗ 7=-15

hope it helps you......

Answer:

5765865876+5737555586=11503421462

Answer:

[tex]f(10) = 1147.25[/tex]

Step-by-step explanation:

Given

[tex]f(-1.5) = 26[/tex]

[tex]f(5.5) = 7[/tex]

Required

f(10)

An exponential function is represented as:

[tex]f(x) = ab^x[/tex]

[tex]f(-1.5) = 26[/tex] impleies that:

[tex]26 = ab^{-1.5}[/tex] --- (1)

[tex]f(5.5) = 7[/tex] implies that

[tex]7 = ab^{5.5}[/tex] --- (2)

Divide (2) by (1)

[tex]26/7 = ab^{-1.5}/ab^{5.5}[/tex]

[tex]3.71429 = b^{-1.5+5.5}[/tex]

[tex]3.71429 = b^{4}[/tex]

Take 4th root

[tex]b = 1.39[/tex]

Substitute [tex]b = 1.39[/tex] in [tex]26 = ab^{-1.5}[/tex]

[tex]26 = a * 1.39^{-1.5}[/tex]

[tex]26 = a * 0.6102[/tex]

Solve for (a)

[tex]a = 26/0.6102[/tex]

[tex]a = 42.61[/tex]

f(10) is calculated as:

[tex]f(10) = ab^{10}[/tex]

[tex]f(10) = 42.61 * 1.39^{10}[/tex]

[tex]f(10) = 1147.25[/tex]

answer is 6

Step-by-step explanation:

3x+11=y

y=29

3x+11=29

3x=29-11

3x=18

x=18÷3

x=6

Answer:6

Step-by-step explanation:

3x+11=29

3x=29-11

3x=18

X=18/3

X=6

Answer:

No

Step-by-step explanation:

4:1

Multiply each side by 3

4*3 : 1*3

12: 3

This is not equal to 12:9

Answer:

no

Step-by-step explanation:

12 : 9

4 : 3 ≠ 4 : 1

Let the number be x

ATQ

[tex]\\ \sf\twoheadrightarrow 6x-18=96[/tex]

[tex]\\ \sf\twoheadrightarrow 6x=96+18[/tex]

[tex]\\ \sf\twoheadrightarrow 6x=112[/tex]

[tex]\\ \sf\twoheadrightarrow x=\dfrac{112}{6}[/tex]

[tex]\\ \sf\twoheadrightarrow x=7[/tex]

Answer:

4 = 6

5 = -17

Step-by-step explanation:

4. a² - b / b² - c

a² = 2² = 4

b = -2

4 - (-2) = 4 + 2 = 6

6 / b² - c = 6/4 - 3 = 6/1 = 6

4 = 6

5. -3x² + 2xy + 7

-3x² = -3 * -2² = -3 * 4 = -12

2xy = 2 * -2 * 3 = -4 * 3 = -12

-12 + -12 + 7 = -24 + 7 = -17

5 = -17

If my answer is incorrect, pls correct me!

If you like my answer and explanation, mark me as brainliest!

-Chetan K

The final amount is $7,615.27

A = P(1 + r/n)^t

Where,

A = Final amount

P = principal = $7, 200

r = interest rate = 2.5% = 0.025

n = number of periods = 4

t = time = 9 years

A = P(1 + r/n)^t

= 7,200(1 + 0.025/4)^9

= 7,200(1 + 0.00625)^9

= 7,200(1.00625)^9

= 7,200(1.0576769512798)

= 7,615.2740492152

Approximately,

A = $7,615.27

https://brainly.com/question/14003110

Answer:

16, 16(sqrt3)

Step-by-step explanation:

30-60-90 triangles follow a rule where the hypotenuse is 2 times the shortest leg, and the longer leg is sqrt3 times the shorter leg.

So, the sides are x, 2x, and (sqrt3)x.

If 2x = 32, then x = 16.

Therefore the two legs are 16 and 16(sqrt3)

you can search up 30-60-90 triangle for more information

What do you personaly feel like is most dificult.

For me its rembering minus signs

It sounds like R is the region (in polar coordinates)

R = {(r, θ) : 2 ≤ r ≤ 3 and 0 ≤ θ ≤ π/2}

Then the integral is

[tex]\displaystyle \iint_R\frac{\mathrm dx\,\mathrm dy}{\sqrt{x^2+y^2}} = \int_0^{\pi/2}\int_2^3 \frac{r\,\mathrm dr\,\mathrm d\theta}{\sqrt{r^2}} \\\\ = \int_0^{\pi/2}\int_2^3 \mathrm dr\,\mathrm d\theta \\\\ = \frac\pi2\int_2^3 \mathrm dr \\\\ = \frac\pi2r\bigg|_2^3 = \frac\pi2 (3-2) = \boxed{\frac\pi2}[/tex]