Find the surface area of the figure. For calculations involving π give both the exact value and an approximation to the nearest hundredth of a unit. Let r = 5 and h = 5

Answer:

The number 23 is prime. Therefore, the cube root of 23 = ∛23 = 2.8439.

Step-by-step explanation:

The equation of the perpendicular line is [tex]y=-\frac{2}{3}x+4[/tex]

The equation is defined by 2y = 3x + 2

Rewrite the equation in the form y = mx + c

[tex]y=\frac{3}{2}x + 1[/tex]

The slope, m = 3/2

The y-intercept, c = 1

The equation perpendicular to the given line will be of the form:

[tex]y-y_1=\frac{-1}{m}(x-x_1 )[/tex]

Substitute [tex]x_1=3, y_1=2, and m=\frac{3}{2}[/tex] into the equation above

[tex]y-2=\frac{-2}{3} (x-3)\\\\y-2=-\frac{2}{3}x+2\\\\y=-\frac{2}{3}x+4[/tex]

Therefore, the equation of the perpendicular line is [tex]y=-\frac{2}{3}x+4[/tex]

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The zeros of the given function are x = 2 and x = -1

From the question, we are to determine the zeros of the given function

The given function is

f(x) = (x - 2)(3x + 3)

To determine the zeros of a function, we will set the function equal to zero

That is,

(x - 2)(3x + 3) = 0

Then,

x - 2 = 0 OR 3x + 3 = 0

x = 2 OR 3x = -3

x = 2 OR x = -3/3

x = 2 OR x = -1

Hence, the zeros of the given function are x = 2 and x = -1

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The equations that can be used to find the lengths of the legs of the triangle are as follows:

The equation that can be used to find the length of the leg of the right triangle uses Pythagoras theorem,

Therefore,

c²  = a² + b²

where

Therefore,

10²  = (x + 2)² + x²

100 = (x + 2)² + x²

Therefore, using the area of a triangle,

area of a triangle  = 1 / 2 bh

where

Therefore,

area of the triangle = 1 / 2 × (x + 2)(x)

24 = 0.5 × (x + 2)(x)

24 = x² + 2x / 2

cross multiply

48 = x² + 2x

x² + 2x - 48 = 0

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Answer:

[tex]\boxed {(f-g)(x) = x^{2}+3}[/tex]

Step-by-step explanation:

Given :

Now, this problem can be used using a simple formula.

[tex]\fbox {(f - g)(x) = f(x) - g(x)}[/tex]

Let's solve.

∴ The answer will be x² + 3.

The equation of the exponential function is [tex]y = 3^{-x -5} - 3[/tex]

An exponential equation is represented as:

y = b^x

Where b represents the base.

The base is 3.

So, we have:

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

When reflected over the y-axis, we have:

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

When shifted right by 5 units, we have:

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

Lastly, the function has an asymptote of y=-3

So, we have:

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

Hence, the equation of the exponential function is [tex]y = 3^{-x -5} - 3[/tex]

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The Sample space for three cards is

[tex]S=\left\{\left(B,1\right),\left(R,1\right),\left(G,1\right),\left(B,2\right),\left(R,2\right),\left(G,2\right),\left(B,3\right),\left(R,3\right),\left(G,3\right),\left(B,4\right),\left(R,4\right),\left(G,4\right)[/tex][tex]\left(B,5\right),\left(R,5\right),\left(G,5\right),\left(B,6\right),\left(R,6\right),\left(G,6\right) \right \}[/tex]

There are two cards red(R) and black(B) and a six-sided die.

The event is one card is picked randomly and then rolls the die.

So, the Sample space is as follows.

[tex]S=\left\{\left(B,1\right),\left(R,1\right),\left(B,2\right),\left(R,2\right),\left(B,3\right),\left(R,3\right),\left(B,4\right),\left(R,4\right),\left(B,5\right),\left(R,5\right),\left(B,6\right),\left(R,6\right) \right \}[/tex]

Consider there were three cards red(R), black(B) and green(G) a six-sided die.

The event is one card is picked randomly and then rolls the die.

So, the sample space will be as follows,

[tex]S=\left\{\left(B,1\right),\left(R,1\right),\left(G,1\right),\left(B,2\right),\left(R,2\right),\left(G,2\right),\left(B,3\right),\left(R,3\right),\left(G,3\right),\left(B,4\right),\left(R,4\right),\left(G,4\right)[/tex][tex]\left(B,5\right),\left(R,5\right),\left(G,5\right)\left(B,6\right),\left(R,6\right),\left(G,6\right) \right \}[/tex].

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The value of x is 2 and the length of JK is 4

The given parameters from the circle are:

The lines SR and SQ are the radii of the circle P

This means that lines JK and JL are congruent

So, we have:

JK = KL

Substitute LK = 2x + 4 and JK = 4

4 = 2x + 4

Rewrite the above equation as:

2x + 4 = 8

Subtract 4 from both sides

2x + 4 - 4 = 8 - 4

Evaluate the difference

2x = 4

Divide both sides by 2

2x = 4/2

This gives

x = 2

Substitute x = 2 in LK = 2x + 4

LK = 2*2 + 4

Evaluate the product of 2 and 2

LK = 4 + 4

This gives

LK = 8

The point N divides JK into 2 equal segments

So, we have

JN = JK/2

JN= 8/2

JN = 4

Hence, the value of x is 2 and the length of JK is 4

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Answer:

y=0.5x²-4x-10.

Step-by-step explanation:

1) according to the definition of parabola:

d1=d2, then

[tex]2) \ |y+1.5|=\sqrt{(x+4)^2+(y+2.5)^2} ;[/tex]

3) after evolution:

-(2y+4)=(x+4)²; ⇔

[tex]y=-\frac{1}{2}x^2-4x-10.[/tex]

By using Pythagoran theorem, the length of the third side of the triangle ABC is 35 centimeters.

In this question we know the lengths of two sides of the right triangle. By Pythagorean theorem we get the length of the missing side:

a² + b² = c²

b² = c² - a²

b = √(c² - a²)

b = √(37² - 12²)

b = 35

By using Pythagoran theorem, the length of the third side of the triangle ABC is 35 centimeters.

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Answer:

C) The volume of the prism is equal to the volume of the cylinder.

Step-by-step explanation:

volume of triangular prism = area of base × height

volume of cylinder = area of base × height

The cross sectional area are equal, so the bases have the same area.

The heights are also equal.

The volumes must be equal.

Answer: C) The volume of the prism is equal to the volume of the cylinder.

Answer:

C) The volume of the prism is equal to the volume of the cylinder.

Step-by-step explanation:

If the cross-sectional areas of the triangular prism and the right cylinder are congruent, it means that when you look at their shapes from the top or bottom, they have the same size and shape.

Since both the triangular prism and the right cylinder have a height of 5 units, and their cross-sectional areas are the same, it implies that they have the same volume as well. So, the volume of the triangular prism is equal to the volume of the cylinder.

Please mark me as Brainliest if you're satisfied with this answer

Answer:

51

Step-by-step explanation:

Let c be the number of child tickets sold.

Let a be the number of adult tickets sold.

From the information given from the question, we can deduce:

6.10c + 9.70a = 1074.20 - Equation 1

c + a = 146 - Equation 2

From here, we can use the substitution method in solving simultaneous equations to find a and c. We will use Equation 2 as the base.

c = 146 - a - Equation 2A

We will then substitute equation 2A into Equation 1 to find a.

6.10(146-a) + 9.70a = 1074.20

890.6 - 6.10a + 9.70a = 1074.20

890.6 + 3.6a = 1074.20

3.6a = 1074.20 - 890.6

3.6a = 183.6

a = [tex]\frac{183.6}{3.6}[/tex]

a = 51

Therefore we know that 51  adult tickets were sold that day.

After that you can substitute a into equation 2 to find c.

Step-by-step explanation:

Help me am lost somewhere

Answer:

yes i help yes of course he help

Answer:

A

Step-by-step explanation:

Answer: in 5 seconds it will take the ball to the ground.

Step-by-step explanation:

h=400 foot    h(t)=-16t²+h

Height a ball will take for to hit the ground  is 0 feet.

Hence,    

-16t²+h=0

400=16t²

16*25=16*t² |:16

25=t².

t²=5²

t=5 с.

Good luck an' have a nice day!

Answer:

IV 5

Step-by-step explanation:

you have to solve for x:

x-3<2

x-3+3<2+3

x=5

could you put the brainlyest Icon for me? I have 1500 points, but no brainlyest, so I can't move up the ranks.

Answer:

x = 9

Step-by-step explanation:

4x + 7 = 43

4x = 36

x = 9

[tex]order \: does \: not \:matter \\ sample \: space = 15 \: letters \\ no \: repetition\\ P(A) = 15C5 = 3003 \: ways[/tex]

I know you said "without making any assumptions," but this one is pretty important. Assuming you mean [tex]\alpha,\beta[/tex] are shape/rate parameters (as opposed to shape/scale), the PDF of [tex]X[/tex] is

[tex]f_X(x) = \dfrac{\beta^\alpha}{\Gamma(\alpha)} x^{\alpha - 1} e^{-\beta x} = \dfrac{2^8}{\Gamma(8)} x^7 e^{-2x}[/tex]

if [tex]x>0[/tex], and 0 otherwise.

The MGF of [tex]X[/tex] is given by

[tex]\displaystyle M_X(t) = \Bbb E\left[e^{tX}\right] = \int_{-\infty}^\infty e^{tx} f_X(x) \, dx = \frac{2^8}{\Gamma(8)} \int_0^\infty x^7 e^{(t-2) x} \, dx[/tex]

Note that the integral converges only when [tex]t<2[/tex].

Define

[tex]I_n = \displaystyle \int_0^\infty x^n e^{(t-2)x} \, dx[/tex]

Integrate by parts, with

[tex]u = x^n \implies du = nx^{n-1} \, dx[/tex]

[tex]dv = e^{(t-2)x} \, dx \implies v = \dfrac1{t-2} e^{(t-2)x}[/tex]

so that

[tex]\displaystyle I_n = uv\bigg|_{x=0}^{x\to\infty} - \int_0^\infty v\,du = -\frac n{t-2} \int_0^\infty x^{n-1} e^{(t-2)x} \, dx = -\frac n{t-2} I_{n-1}[/tex]

Note that

[tex]I_0 = \displaystyle \int_0^\infty e^{(t-2)}x \, dx = \frac1{t-2} e^{(t-2)x} \bigg|_{x=0}^{x\to\infty} = -\frac1{t-2}[/tex]

By substitution, we have

[tex]I_n = -\dfrac n{t-2} I_{n-1} = (-1)^2 \dfrac{n(n-1)}{(t-2)^2} I_{n-2} = (-1)^3 \dfrac{n(n-1)(n-2)}{(t-2)^3} I_{n-3}[/tex]

and so on, down to

[tex]I_n = (-1)^n \dfrac{n!}{(t-2)^n} I_0 = (-1)^{n+1} \dfrac{n!}{(t-2)^{n+1}}[/tex]

The integral of interest then evaluates to

[tex]\displaystyle I_7 = \int_0^\infty x^7 e^{(t-2) x} \, dx = (-1)^8 \frac{7!}{(t-2)^8} = \dfrac{\Gamma(8)}{(t-2)^8}[/tex]

so the MGF is

[tex]\displaystyle M_X(t) = \frac{2^8}{\Gamma(8)} I_7 = \dfrac{2^8}{(t-2)^8} = \left(\dfrac2{t-2}\right)^8 = \boxed{\dfrac1{\left(1-\frac t2\right)^8}}[/tex]

The first moment/expectation is given by the first derivative of [tex]M_X(t)[/tex] at [tex]t=0[/tex].

[tex]\Bbb E[X] = M_x'(0) = \dfrac{8\times\frac12}{\left(1-\frac t2\right)^9}\bigg|_{t=0} = \boxed{4}[/tex]

Variance is defined by

[tex]\Bbb V[X] = \Bbb E\left[(X - \Bbb E[X])^2\right] = \Bbb E[X^2] - \Bbb E[X]^2[/tex]

The second moment is given by the second derivative of the MGF at [tex]t=0[/tex].

[tex]\Bbb E[X^2] = M_x''(0) = \dfrac{8\times9\times\frac1{2^2}}{\left(1-\frac t2\right)^{10}} = 18[/tex]

Then the variance is

[tex]\Bbb V[X] = 18 - 4^2 = \boxed{2}[/tex]

Note that the power series expansion of the MGF is rather easy to find. Its Maclaurin series is

[tex]M_X(t) = \displaystyle \sum_{k=0}^\infty \dfrac{M_X^{(k)}(0)}{k!} t^k[/tex]

where [tex]M_X^{(k)}(0)[/tex] is the [tex]k[/tex]-derivative of the MGF evaluated at [tex]t=0[/tex]. This is also the [tex]k[/tex]-th moment of [tex]X[/tex].

Recall that for [tex]|t|<1[/tex],

[tex]\displaystyle \frac1{1-t} = \sum_{k=0}^\infty t^k[/tex]

By differentiating both sides 7 times, we get

[tex]\displaystyle \frac{7!}{(1-t)^8} = \sum_{k=0}^\infty (k+1)(k+2)\cdots(k+7) t^k \implies \displaystyle \frac1{\left(1-\frac t2\right)^8} = \sum_{k=0}^\infty \frac{(k+7)!}{k!\,7!\,2^k} t^k[/tex]

Then the [tex]k[/tex]-th moment of [tex]X[/tex] is

[tex]M_X^{(k)}(0) = \dfrac{(k+7)!}{7!\,2^k}[/tex]

and we obtain the same results as before,

[tex]\Bbb E[X] = \dfrac{(k+7)!}{7!\,2^k}\bigg|_{k=1} = 4[/tex]

[tex]\Bbb E[X^2] = \dfrac{(k+7)!}{7!\,2^k}\bigg|_{k=2} = 18[/tex]

and the same variance follows.

Answer:

√2² cannot be answer yesah

Using a system of equations, it is found that:

A system of equations is when two or more variables are related, and equations are built to find the values of each variable.

In this problem, the variables are:

Peter pays $50 for 1 orchestra and 4 balcony tickets, hence:

a + 4b = 50 -> a = 50 - 4b.

Justine pays $150 for 3 orchestra and 12 balcony tickets, hence:

3a + 12b = 150.

a + 4b = 50 -> a = 50 - 4b.

The equations are equal, hence we suppose that the costs are equal, that is:

a + 4b = 50 -> 5a = 50 -> a = 10.

Hence:

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Answer:

1 UNIT LEFT best describes the translation from the graph y = 6x² to the graph of y = 6(x + 1)².

Answer:

=57198.98

Step-by-step explanation:

Volume of Cylinder = pi x radius^2 x height

= pi x 17^2 x 63

=57198.9774439

=57198.98

The greatest common factor of the expression is 4m⁴.

GCF simply means the greatest common factor.

The greatest common factor (GCF) of a set of numbers is the largest factor that all the numbers share.

Hence,

56m⁵

68m⁴

The greatest common factor of the expression  is the largest positive expression that divides evenly into all numbers with zero .

Hence, the greatest common factor is 4m⁴

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The number of bats are expected to be in the colony after 11 years if the colony continues to grow at the same rate is 69,632

Let

exponential function;

y = ax^t

When t = 3

272 = ax³

when t = 5

1088 = ax^5

272/1088 = ax^3 / ax^5

1/4 = x^3-5

1/4 = x^-3

1/4 = 1/x²

1 × x² = 4 × 1

x² = 4

x = 2

Substitute x = 2 into

272 = ax³

272 = a × 2³

272 = a × 8

272 = 8a

a = 272/8

a = 34

Number of bat expected at the colony after 11 years

y = ax^t

= 34 × 2^11

= 34 × 2048

y = 69,632 bats

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Answer: 6 %

Step-by-step explanation: 6 % of 12.5 = 0.75. 12.5 plus 0.75 = 13.25