The main sail of a sailboat has the dimensions shown in the figure at the right. What is the height of the main sail?

Answer:

Step 1

According to given details

Number of delivered pizzas at the weekly meeting=5

Number of slices in one pizza =8

(a)

Now evaluate total number of slices

Number of slices in one pizza =8

Total slices =8x5

=40 slice

(b)

Yes, we can find more than one way

= 5 ÷ 1/8

= 5x8/1

=40/1

Answer:

D) all real numbers except x = 6

Step-by-step explanation:

If there is a solid line or dot at an x value, then it is included in the domain. Since there are open dots at x = 6, that means it is not included in the domain. With that being said, the domain can be written as:

(-∞, 6)∪(6, ∞)

Hope this helps!

Answer: 18,840*pi

Step-by-step explanation:

The volume of a globe would be 4/3*pi*radius^3 so that would be 4/3*14,130= 18840*pi or approximately 59,187.6055

Answer: X is 11,

Step-by-step explanation:

Hope it helps you

Answer:

[tex]x = \geqslant 34[/tex]

Step-by-step explanation:

x is greater than or equal to 34, because the illustration shows, that it can go beyond 34. The filled in dot also means that x can also be equal to 34. (i hope i explained it in a good way)

Answer:

i think it b i dont know for sure

Step-by-step explanation:

Answer:  

g(x) has the smallest possible y-value of -3

Step-by-step explanation:

f(x) = 3ˣ - 3   This is an exponential graph shifted down three units. So, it has an asymptote at y = -3, which means it approaches -3 but does not touch it.Range: y > 3   (-3, ∞)   g(x) = 7x² - 3  ⇒ g(x) = 7(x - 0)² - 3   This is a parabola with vertex at (0, -3) Range: y ≥ 3   [-3, ∞)

Answer:

D. 6(9 + 5)

Step-by-step explanation:

Let that be y

Now solve

Each youth ticket costs 7$

Answer So you can add a number with the letter t and then add and then you can 35 dollers

Answer:

B.

Step-by-step explanation:

the range is the y values of the line.

so it would be [3, 6]

Answer:

.1, 1.2, 0, 1.5, 0, 1.4 // 19.2, .8, 70.2, .7, 4791.7, .7

Step-by-step explanation:

The first set is if you meant the first number divided BY the second and the second set of numbers is if you meant it vice versa. If it was worded better I could understand better.

Answer:

A)

Step-by-step explanation:

The experimental probability on 'A' is higher than the theoretical probability  of landing on 'A'.

Answer: 4

Step-by-step explanation: how i know by guessing

Answer: 147.2

Step-by-step explanation:

(8+15) = 23

23 x 6.4= 147.2

Answer: 147.2

Step-by-step explanation: 8+15= 23 so 23*6.4 is 147.2

Answer:

13n+20 is the correct answer to ypur question

Step-by-step explanation:

The two parabolas intersect for

[tex]8-x^2 = x^2 \implies 2x^2 = 8 \implies x^2 = 4 \implies x=\pm2[/tex]

and so the base of each solid is the set

[tex]B = \left\{(x,y) \,:\, -2\le x\le2 \text{ and } x^2 \le y \le 8-x^2\right\}[/tex]

The side length of each cross section that coincides with B is equal to the vertical distance between the two parabolas, [tex]|x^2-(8-x^2)| = 2|x^2-4|[/tex]. But since -2 ≤ x ≤ 2, this reduces to [tex]2(x^2-4)[/tex].

a. Square cross sections will contribute a volume of

[tex]\left(2(x^2-4)\right)^2 \, \Delta x = 4(x^2-4)^2 \, \Delta x[/tex]

where ∆x is the thickness of the section. Then the volume would be

[tex]\displaystyle \int_{-2}^2 4(x^2-4)^2 \, dx = 8 \int_0^2 (x^2-4)^2 \, dx \\\\ = 8 \int_0^2 (x^4-8x^2+16) \, dx \\\\ = 8 \left(\frac{2^5}5 - \frac{8\times2^3}3 + 16\times2\right) = \boxed{\frac{2048}{15}}[/tex]

where we take advantage of symmetry in the first line.

b. For a semicircle, the side length we found earlier corresponds to diameter. Each semicircular cross section will contribute a volume of

[tex]\dfrac\pi8 \left(2(x^2-4)\right)^2 \, \Delta x = \dfrac\pi2 (x^2-4)^2 \, \Delta x[/tex]

We end up with the same integral as before except for the leading constant:

[tex]\displaystyle \int_{-2}^2 \frac\pi2 (x^2-4)^2 \, dx = \pi \int_0^2 (x^2-4)^2 \, dx[/tex]

Using the result of part (a), the volume is

[tex]\displaystyle \frac\pi8 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{256\pi}{15}}}[/tex]

c. An equilateral triangle with side length s has area √3/4 s², hence the volume of a given section is

[tex]\dfrac{\sqrt3}4 \left(2(x^2-4)\right)^2 \, \Delta x = \sqrt3 (x^2-4)^2 \, \Delta x[/tex]

and using the result of part (a) again, the volume is

[tex]\displaystyle \int_{-2}^2 \sqrt 3(x^2-4)^2 \, dx = \frac{\sqrt3}4 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{512}{5\sqrt3}}[/tex]

Answer:

192 liters

Step-by-step explanation:

Tank A

w = 100 + 80.5m

where:

Therefore, when m = 4:

⇒ w = 100 + 80.5(4) = 422 liters

Tank B

Given ordered pairs:  (0, 154)  (2, 384)  (6, 844)

[tex]\sf slope\:(m)=\dfrac{change\:in\:y}{change\:in\:x}=\dfrac{384-154}{2-0}=115[/tex]

Point-slope form of linear equation: [tex]\sf y-y_1=m(x-x_1)[/tex]

(where m is the slope and (x₁, y₁) is a point on the line)

[tex]\sf \implies y-154=115(x-0)[/tex]

[tex]\sf \implies y=115x+154[/tex]

Therefore, the equation for Tank B is:

w = 115m + 154

Therefore, when m = 4:

⇒ w = 115(4) + 154 = 614 liters

Difference

614 - 422 = 192 liters

Find equation for tank B

Slope:-

Equation in point slope form

For tank B

Put 4on both

Tank A:-

TankB

Difference:-

B

&

D

Step-by-step explanation:

Answer:

The Answers are A and D

Step-by-step explanation:

I took the test and got em right!

○ 2,350 square feet

○ 2,450 square feet

○ 2,500 square feet

● 2,400 square feet answer

Step-by-step explanation:

wellcome

rip

The sequence is geometric, so

[tex]a_n = r a_{n-1}[/tex]

for some constant r. From this rule, it follows that

[tex]a_3 = r a_2 \implies 20 = 2r \implies r = 10[/tex]

and we can determine the first term to be

[tex]a_2 = r a_1 \implies 2 = 10 a_1 \implies a_1 = \dfrac15[/tex]

Now, by substitution we have

[tex]a_n = r a_{n-1} = r^2 a_{n-2} = r^3 a_{n-3} = \cdots[/tex]

and so on down to (D)

[tex]a_n = r^{n-1} a_1 = 10^{n-1} \cdot \dfrac15[/tex]

(notice how the exponent on r and the subscript on a add up to n)

[tex] \sf\longmapsto \: \frac{371}{36} [/tex]

Perimeter of a triangle = Sum of the length of the three sides

perimeter of the given triangle = 11/3 + 15/4 + 26/9

( LCM of 3 , 4 and 9 is 36 )

[tex]\sf\longmapsto \: \frac{132 \: + \: 135 \: + \: 104}{36} \: \: \: [/tex]

[tex]\sf\longmapsto \: \frac{371}{36} \: cm \: [/tex]

♧ Therefore, the perimeter of a triangle is 371/36 cm

Answer:

11 feet

Explanation:

Here given:

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

Solving steps:

⇒ 1/2 * (15 + b) * 7.6 = 98.8

⇒ (15 + b) * 7.6 = 197.6

⇒ (15 + b) = 26

⇒ b = 11 ft

Let unknown be x

Answer:

1/2, 7/12, 5/8, 3/4.

Step-by-step explanation:

I did that in ascending order.

Answer:

1/89

Step-by-step explanation:

Two digit numbers include all numbers from 10 to 99. 99-10 = 89. Therefore the probability that Simon guesses the number correctly on the first try is 1 out of 89 which can be written as 1/89.

Answer:

[tex]\displaystyle k = \frac{1}{32}\ln\frac{1}{2} \approx -0.02166[/tex]

Step-by-step explanation:

The decay formula is given by:
[tex]\displaystyle P(t) = P_0e^{kt}[/tex]

Where k is some constant, P₀ is the initial population, and t is the number of years.

Because the substance has a half-life of 32 years, P(t) = 1/2P₀ when t = 32. Substitute:
[tex]\displaystyle \frac{1}{2}P_0 = P_0 e^{k(32)}[/tex]

Solve for k:

[tex]\displaystyle \begin{aligned} \frac{1}{2} & = e^{32k} \\ \\ \ln\left(e^{32k}\right) & = \ln\left(\frac{1}{2}\right) \\ \\ 32k & = \ln\frac{1}{2} \\ \\ k & = \frac{1}{32}\ln\frac{1}{2} \\ \\ &\approx -0.02166\end{aligned}[/tex]

In conclusion, the value of k is about -0.02166.

The value of k in the exponential decay formula is k = 0.022

To find the constant k in the decay formula for the radioactive substance, we can use the formula for exponential decay, which is given by:

N(t) = N₀*exp(-kt)

where:

The half-life of the substance is the time it takes for half of the substance to decay, which is 32 years in this case. So, after one half-life, N(t) = N₀ / 2.

Now, let's set up the equation for one half-life:

N(t) = N₀ * exp(-k * 32)

Since N(t) = N₀ / 2 after one half-life, we can write:

N₀ / 2 = N₀ * e^(-k * 32)

Now, divide both sides by N₀:

1/2 = exp(-k * 32)

To solve for k, take the natural logarithm (ln) of both sides:

ln(1/2) = -k * 32

k = ln(1/2) / -32

k = 0.022

Learn more about half-life at:

https://brainly.com/question/1160651

#SPJ3

Answer:

the correct answers are A, C, and D

Step-by-step explanation:

Those are the only ones that make sense enough!

What is the average height of 6th grade students?

And

What was the low temperature...?

Those responses will be given as numerical values

E.G. 162cm or 2 degrees Celsius (2*C)

(Edit: the first & third options are the correct ones)

Hope this helps!

Using sampling concepts, it is found that this is an example of a random sample.

Samples may be classified as:

In this problem, one winner is chosen at random from all the participants, hence random sampling is used.

More can be learned about sampling concepts at https://brainly.com/question/25122507

Step-by-step explanation:

We know the quadratic formula, which is :

[tex] \\ {\longrightarrow \qquad{ \sf{x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a} }}} \\ \\ [/tex]

Here,

So,

[tex] \\ {\longrightarrow \qquad{ \sf{x = \frac{ - ( - 6) \pm \sqrt{ {( - 6)}^{2} - 4(1)( - 5) } }{2(1)} }}} \\ \\ [/tex]

[tex] {\longrightarrow \qquad{ \sf{x = \frac{ 6 \pm \sqrt{ 36 - 4( - 5) } }{2} }}} \\ \\ [/tex]

[tex]{\longrightarrow \qquad{ \sf{x = \frac{ 6 \pm \sqrt{ 36 + 20 } }{2} }}} \\ \\ [/tex]

[tex]{\longrightarrow \qquad{ \sf{x = \frac{ 6 \pm \sqrt{ 56 } }{2} }}} \\ \\ [/tex]

[tex]{\longrightarrow \qquad{ \sf{x = \frac{ 6 \pm \sqrt{ 2(28) } }{2} }}} \\ \\ [/tex]

[tex]{\longrightarrow \qquad{ \sf{x = \frac{ 6 \pm \sqrt{ 2(2)(14) } }{2} }}} \\ \\ [/tex]

[tex]{\longrightarrow \qquad{ \sf{x = \frac{ 6 \pm \sqrt{ 2(2)(2)(7) } }{2} }}} \\ \\ [/tex]

[tex]{\longrightarrow \qquad{ \sf{x = \frac{ 6 \pm \sqrt{ 2} . \sqrt{2} . \sqrt{2 . 7} }{2} }}} \\ \\ [/tex]

[tex]{\longrightarrow \qquad{ \sf{x = \frac{ 6 \pm 2 \sqrt{14} }{2} }}} \\ \\ [/tex]

Now, Separating the solutions :

[tex] \\ {\longrightarrow \qquad{ \sf{x = \frac{ 6 + 2 \sqrt{14} }{2} }}} \\ \\ [/tex]

[tex] {\longrightarrow \qquad{ \sf{x = 3 + \sqrt{14}}}}\\ \\ [/tex]

[tex]\\ {\longrightarrow \qquad{ \sf{x = \frac{ 6 - 2 \sqrt{14} }{2} }}} \\ \\ [/tex]

[tex] {\longrightarrow \qquad{ \sf{x = 3 - \sqrt{14}}}}\\ \\ [/tex]