Can someone please please help me

Answer:

[tex]h = \bf 28.3 \space\ m[/tex]

Step-by-step explanation:

• We are given:

○ Volume = 36 m³,

○ Circumference = 4 m

• Let's find the radius of the cylinder first:

[tex]\mathrm{Circumference} = 2 \pi r[/tex]

Solving for [tex]r[/tex] :

⇒ [tex]4 = 2 \pi r[/tex]

⇒ [tex]r = \frac{4}{2\pi}[/tex]

⇒ [tex]r = \bf \frac{2}{\pi}[/tex]

• Now we can calculate the height using the formula for volume of a cylinder:

[tex]\mathrm{Volume} = \boxed{\pi r^2 h}[/tex]

Solving for [tex]h[/tex] :

⇒ [tex]36 = \pi \cdot (\frac{2}{\pi}) ^2 \cdot h[/tex]

⇒ [tex]h = \frac{36 \pi^2}{4 \pi}[/tex]

⇒ [tex]h = 9 \pi[/tex]

⇒ [tex]h = \bf 28.3 \space\ m[/tex]

Answer:

9π m ≈ 28.27m

Step-by-step explanation:

The volume of a right cylinder is given by the formula

πr²h where r is the radius of the base of the cylinder(which is a circle), h is the height of the cylinder

Circumference of base of cylinder is given by the formula 2πr

Given,

2πr = 4m

r = 2/π m

Volume given as 36 m³

So πr²h = 36
π (2/π)² h = 36

π x 4/π² h = 36

(4/π) h = 36

h = 36π/4 = 9π ≈ 28.27m

The measure of angle 1 (m ∠1) is 28° and the measure of angle 2 (m ∠2) is 62°

From the question, we are to determine the measure of angle 1 and the measure of angle 2

The given diagram is a kite and the diagonals intersect at right angles

Thus,

m ∠2 + 28° + 90° = 180°

m ∠2 = 180° - 28° - 90°

m ∠2 = 62°

Hence, the measure of angle 2 is 62°

For the measure of angle 1

Consider ΔADB

ΔADB is an isosceles triangle

Thus,

In the triangle, m ∠D = m ∠B

Then, we can write that

m ∠1 + 62° + 90° = 180°

m ∠1 = 180° - 62° - 90°

m ∠1 = 28°

∴ The measure of angle 1 is 28°

Hence, the measure of angle 1 (m ∠1) is 28° and the measure of angle 2 (m ∠2) is 62°

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The simplification of a³ - 1000b³ and 64a³ - 125b³ is (a - 10b) × (a² + 10ab + 100b²) and 4a - 5b) • (16a² + 20ab + 25b²) respectively.

Question 1: a³ - 1000b³

a³ - b³

= (a-b) × (a² +ab +b²)

So,

(a - 10b) × (a² + 10ab + 100b²)

Question 2: 64a³ - 125b³

a³ - b³

= (a-b) × (a² +ab +b²)

So,

(4a - 5b) • (16a² + 20ab + 25b²)

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Answer:

I think it's 120 dollars sorry if i get it wrong but I am sure it is right tell me when you get the right answer byeeee.

Answer:

the price of one pound of trail mix is $3

the price of one pound of jelly beans is $2.5

Step-by-step explanation :

let t be the price of one pound of trail mix.

and b be the price of one pound of jelly beans.

This statement:

“For 3 pounds of trail mix and 8 pounds of jelly beans,

the total cost is $29”

Can be translated mathematically like this :

3t + 8b = 29

This statement:

“For 5 pounds of trail mix and 2 pounds of jelly beans,

the total cost is $20”

Can be translated mathematically like this :

5t + 2b = 20

We obtain a system of two equations :

3t + 8b = 29

5t + 2b = 20

Solving the system:

3t + 8b = 29  (equation 1)

5t + 2b = 20. (equation 2)

3t + 8b = 29

20t + 8b = 80   [4×(equation 2)]

3t + 8b = 29

17t = 51  [4×(equation 2) - equation 1]

3t + 8b = 29

t = 51/17 = 3

9 + 8b = 29

t = 3

8b = 20

t = 3

b = 20/8 = 5/2

t = 3

b = 5/2

t = 3

Therefore ,the solution to our system is (3 , 2.5)

[tex] \qquad \qquad \bf \huge\star \: \: \large{ \underline{Answer} } \huge \: \: \star[/tex]

[tex]\textsf{ \underline{\underline{Steps to solve the problem} }:}[/tex]

Take HJ = a, GH = b and GJ = c

put the value of a from equation 1 in equation 2

[tex]\qquad❖ \: \sf \:c = (b + 2) + b - 17[/tex]

[tex]\qquad❖ \: \sf \:c = 2b - 15[/tex]

now, put the value of a and c in equation 3

[tex]\qquad❖ \: \sf \:b + 2 + b + 2b - 15 = 73[/tex]

[tex]\qquad❖ \: \sf \:4b - 13 = 73[/tex]

[tex]\qquad❖ \: \sf \:4b = 86[/tex]

[tex]\qquad❖ \: \sf \:b = 21.5 \: \: in[/tex]

Now, we need to find HJ (a)

[tex]\qquad❖ \: \sf \:a = b + 2[/tex]

[tex]\qquad❖ \: \sf \:a = 21.5 + 2[/tex]

[tex]\qquad❖ \: \sf \:23.5 \: \: in[/tex]

[tex] \qquad \large \sf {Conclusion} : [/tex]

Answer:

23.5 in

Step-by-step explanation:

To find the length of HJ in triangle GHJ, create three equations using the given information, then solve simultaneously.

Equation 1

HJ is two inches longer than GH:

⇒ HJ = GH + 2

Equation 2

GJ is 17 inches shorter than the sum of HJ and GH:

⇒ GJ + 17 = HJ + GH

Equation 3

The perimeter of ΔGHJ is 73 inches:

⇒ HJ + GH + GJ = 73

Substitute Equation 1 into Equation 2 and isolate GJ:

⇒ GJ + 17 = GH + 2 + GH

⇒ GJ + 17 = 2GH + 2

⇒ GJ = 2GH - 15

Substitute Equation 1 into Equation 3 and isolate GJ:

⇒ GH + 2 + GH + GJ = 73

⇒ 2GH + GJ = 71

⇒ GJ = 71 - 2GH

Equate the two equations where GJ is the subject and solve for GH:

⇒ 2GH - 15 = 71 - 2GH

⇒ 4GH = 86

⇒ GH = 21.5

Substitute the found value of GH into Equation 1 and solve for HJ:

⇒ HJ = 21.5 + 2

⇒ HJ = 23.5

Answer:

50°

Step-by-step explanation:

Note EFGH is an isosceles trapezoid.

∠HGF=77° (base angles of an isosceles trapezoid are congruent)

∠EGH=27° (angles in a triangle add to 180°)

∠FGE=50° (angle subtraction postulate)

The period of [tex]f(x)[/tex] is [tex]\boxed{2\pi}[/tex].

Recall that [tex]\sin(x)[/tex] and [tex]\cos(x)[/tex] both have periods of [tex]2\pi[/tex]. This means

[tex]\sin(x + 2\pi) = \sin(x)[/tex]

[tex]\cos(x + 2\pi) = \cos(x)[/tex]

Replacing [tex]x[/tex] with [tex]3x[/tex], we have

[tex]\sin(3x + 2\pi) = \sin\left(3 \left(x + \dfrac{2\pi}3\right)\right) = \sin(3x)[/tex]

In other words, if we change [tex]x[/tex] by some multiple of [tex]\frac{2\pi}3[/tex], we end up with the same output. So [tex]\sin(3x)[/tex] has period [tex]\frac{2\pi}3[/tex].

Similarly, [tex]\cos(5x)[/tex] has a period of [tex]\frac{2\pi}5[/tex],

[tex]\cos(5x + 2\pi) = \cos\left(5 \left(x + \dfrac{2\pi}5\right)\right) = \cos(5x)[/tex]

We want to find the period [tex]p[/tex] of [tex]f(x)[/tex], such that

[tex]f(x + p) = f(x)[/tex]

[tex] \implies \sin(3x + p) + \cos(5x + p) = \sin(3x) + \cos(5x)[/tex]

On the left side, we have

[tex]\sin(3x + p) = \sin(3x + 2\pi + p - 2\pi) \\\\ ~~~~~~~~ = \sin(3x+2\pi) \cos(p-2\pi) + \cos(3x+2\pi) \sin(p-2\pi) \\\\ ~~~~~~~~ = \sin(3x) \cos(p-2\pi) + \cos(3x) \sin(p - 2\pi)[/tex]

and

[tex]\cos(5x + p) = \cos(5x + 2\pi + p - 2\pi) \\\\ ~~~~~~~~ = \cos(5x+2\pi) \cos(p-2\pi) - \sin(5x+2\pi) \sin(p-2\pi) \\\\ ~~~~~~~~ = \cos(5x) \cos(p-2\pi) - \sin(5x) \sin(p-2\pi)[/tex]

So, in terms of its period, we have

[tex]f(x) = \sin(3x) \cos(p - 2\pi) + \cos(3x) \sin(p - 2\pi) \\\\ ~~~~~~~~ ~~~~+ \cos(5x) \cos(p - 2\pi) - \sin(5x) \sin(p - 2\pi)[/tex]

and we need to find the smallest positive [tex]p[/tex] such that

[tex]\begin{cases} \cos(p - 2\pi) = 1 \\ \sin(p - 2\pi) = 0 \end{cases}[/tex]

which points to [tex]p=2\pi[/tex], since

[tex]\cos(2\pi-2\pi) = \cos(0) = 1[/tex]

[tex]\sin(2\pi - 2\pi) = \sin(0) = 0[/tex]

Since [tex]a,b,c[/tex] are in geometric progression, if [tex]r[/tex] is the common ratio between consecutive terms, then

[tex]a=a[/tex]

[tex]b = ar[/tex]

[tex]c=ar^2[/tex]

Since [tex]a,b,c[/tex] are also in arithmetic progression, if [tex]d[/tex] is the common difference between consecutive terms, then

[tex]a = a[/tex]

[tex]b = a + d \implies d = b-a[/tex]

[tex]c = b + d = a + 2d \implies c = a + 2(b-a) = 2b-a[/tex]

Given that [tex]abc=27[/tex], we have

[tex]abc = a\cdot ar\cdot ar^2 = (ar)^3 = 27 \implies ar = 3 \implies a = \dfrac3r[/tex]

[tex]b = \dfrac3r \cdot r = 3[/tex]

[tex]c = \dfrac3r \cdot r^2 = 3r[/tex]

It follows that

[tex]c = 2b-a \iff 3r = 6 - \dfrac3r[/tex]

Solve for [tex]r[/tex].

[tex]3r - 6 + \dfrac3r = 0[/tex]

[tex]3r^2 - 6r + 3 = 0[/tex]

[tex]r^2 - 2r + 1 = 0[/tex]

[tex](r-1)^2 = 0[/tex]

[tex]\implies r=1 \implies a=b=c=3[/tex]

so the only possible sequence is {3, 3, 3, …}.

f(x) = x³ + 2x² - 15x - 36 is the equation of a degree 3 polynomial (in factored form) with the given zeros of f(x) are − 3 , 4 , − 3 assuming that the leading coefficient is 1. This can be obtained by formula of polynomial function.

⇒ f(x) =  a(x - r₁)(x - r₂)(x - r₃)

where a is the leading coefficient

Here in the question it is given that,

By using the formula of polynomial function we get,

⇒ f(x) =  a(x - r₁)(x - r₂)(x - r₃)

⇒ f(x) = 1(x - (-3))(x - (4))(x - (-3))

⇒ f(x) = 1(x + 3)(x - 4)(x + 3)

⇒ f(x) = (x + 3)(x² - x - 12)

⇒ f(x) = x³ - x² - 12x + 3x² - 3x - 36

⇒ f(x) = x³ + 2x² - 15x - 36

Hence f(x) = x³ + 2x² - 15x - 36 is the equation of a degree 3 polynomial (in factored form) with the given zeros of f(x): − 3 , 4 , − 3 assuming that the leading coefficient is 1.

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If a kite is flying 95 ft. off the ground, and its string is pulled taut. The angle of elevation of the kite is 59 degrees. Then the length of the string will be 110.8 ft.

Given information constitutes the following,

The distance of the flying kite from the ground, length AB (refer the figure) = 95 ft.

The angle of elevation of the kite, ∠ACB = 59°

We have to find the length of the string, that is the length AC. For that, we can apply Trigonometry as shown in the next steps of the solution.

In ΔABC, as shown in the attached figure,

sin (∠ACB ) = AB / AC

⇒ sin (59°) = 95 / AC

0.8572 = 95 / AC

AC = 95 / 0.8572

AC = 110.814

AC ≈ 110.8 ft.                [After rounding off to the nearest tenth]

Hence, the length of the string comes out to be 110.8 ft.

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Using the normal distribution, the value of z is of z = 1.5.

The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

For this problem, considering the symmetry of the normal distribution, the area above the mean is given by:

0.8664/2 = 0.4332.

Hence z has a p-value of 0.5  + 0.4332 = 0.9332, hence the value of z is z = 1.5.

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Answer:

The shortest piece is the first piece and it is 14 cm long.

Step-by-step explanation:

We have three unknowns so we need 3 equations.

Let x = the length of the first piece

Let y = the length of the second piece

Let z = the length of the third piece.

x + y + z = 74            y = 2x          z = y + 4

There are a number of ways to solve this.  I am going to plug in 2x for y into the first and the third equation to get:

x + y + z = 74

x + 2x + z = 74 Combine the x terms

3x + z = 74

Next, I am going to substitute 2x in for y in the third equation above.

z = y + 4

z = 2x + 4  I am going to put both variable on the left side of the equation

z - 2x = 4

I can know take the two bold equations that I have above and solve for the either x or z.  I am going to solve for z.  I need one of the equation to have a z and the other equation to have -z so that they will cancel one another out.  I am going to multiple z - 2x = 4 all the way through by -1 to get:

z - 2x = 4

-1(z - 2x) = 4(-1)

-z +2x = -4  

I am going to rearrange 3x + z = 74 so that the z term is first and add it to -z + 2x = -4

z + 3x = 74

-z + 2x = -4

      5x = 70 divide both sides by 5

x = 14  This is the length of the first piece.

y = 2x

y = 2(14) = 28

y = 28 This is the length of the second piece.

z = y+4

z = 28 + 4 = 32

or

x + y + z = 74

14 + 28 + z = 74

42 + z = 74  Subtract 42 from both sides.

z = 32

The general solution of the logistic equation is y = 14 / [1 - C · tⁿ], where a = - 14² / 3 and C is an integration constant. The particular solution for y(0) = 10 is y = 14 / [1 - (4 / 10) · tⁿ], where n = - 14² / 3.

Herein we have a kind of ordinary differential equation with separable variables, that is, that variables t and y can be separated at each side of the expression prior solving the expression:

dy / dt = 3 · y · (1 - y / 14)

dy / [3 · y · (1 - y / 14)] = dt

dy / [- (3  / 14) · y · (y - 14)] = dt

By partial fractions we find the following expression:

- (1 / 14) ∫ dy / y + (1 / 14) ∫ dy / (y - 14) = - (14 / 3) ∫ dt

- (1 / 14) · ln |y| + (1 / 14) · ln |y - 14| = - (14 / 3) · ln |t| + C, where C is the integration constant.

y = 14 / [1 - C · tⁿ], where n = - 14² / 3.

If y(0) = 10, then the particular solution is:

y = 14 / [1 - (4 / 10) · tⁿ], where n = - 14² / 3.

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Answer:

40°, 50°, and 90°

Step-by-step explanation:

Let the smallest angle be x.

Then, the other small angle is x+10.

The acute angles of a right triangle are complementary, so x+x+10=90, and thus x=40.

So, the acute angles measure 40° and 50°.

Therefore, the three angles are 40°, 50°, and 90°.

The frequency distribution based on the information given is illustrated below.

A frequency distribution table is the

chart that summarizes all the data under two columns - variables/categories, and their frequency.

It should be noted that the distribution table has two or three columns and the first column lists all the outcomes as individual values or in the form of class intervals, depending upon the size of the data set.

Given the above information the frequency distribution table is:

Cost of living Number of states

85.0 - 94.9 9

95.0 - 104.9 5

105.0 - 114.9 0

115.0 - 124.9 2

125.0 - 134.9 2

135.0 - 144.9 1

145.0 - 154.9 1

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The value of the composite functions (g∘f)(x) and (f∘g)(x)   are 32x^2 + 16x + 3 and 8x^2 + 5 respectively

Composite function is also known as function of a function. They are determined by representing x with the other function.

Given the following functions

f(x)=4x+1

g(x)=2x^2+1

(f∘g)(x) = f(g(x))

(f∘g)(x) = f(2x^2+1)

(f∘g)(x) = 4(2x^2+1) + 1

(f∘g)(x)  =8x^2 + 5

For the composite function (g∘f)(x)

(g∘f)(x) = g(f(x))

(g∘f)(x) = g(4x+1)

Replace x wit 4x+1  to have:

(g∘f)(x) = 2(4x+1)^2 + 1

(g∘f)(x)= 2(16x^2+8x+1) + 1

(g∘f)(x) = 32x^2 + 16x + 3

Hence the value of the composite functions (g∘f)(x) and (f∘g)(x)   are 32x^2 + 16x + 3 and 8x^2 + 5 respectively

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The focal length of the given ellipse is given as (±6, 0)

An ellipse is defined as a regular oval shape, traced by a point moving in a plane so that the sum of its distances from two other points (the foci) is constant or when a cone is cut by an oblique plane which does not intersect the base.

The standard equation of an ellipse is expressed as;

x^2/a^2 + y^2/b^2 = 1

The formula for calculating the focus of the ellipse is given as:

c^2 = b^2 - a^2

Given the equation of an ellipse

(x-7)^2/64 + (y-5)^2/100 = 1

This can also be expressed as:

(x-7)^2/8^2 + (y-5)^2/10^2 = 1

Comparing with the general equation

a = 8 and b = 10

Substitute

c^2 = 10^2 - 8^2

c^2 = 100 - 64

c^2 = 36

c = 6

Hence the focal length of the given ellipse is given as (±6, 0)

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The system of inequalities y < -x+4 and y >-x 2 do not have a solution

Inequalities are expressions that have unequal values when compared or evaluated

The system of inequalities is given as

y < -x+4

y >-x 2

Next, we plot the inequalities on a graphing tool

See attachment for the graph

From the attached graph, the lines of the inequalities do not intersect

This means that the system of inequalities do not have a solution

Hence, the system of inequalities y < -x+4 and y >-x 2 do not have a solution

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The limit does not exist. There are infinitely many infinite discontinuities at [tex]x=n\pi[/tex], where [tex]n\in\Bbb N[/tex]. The function oscillates wildly between negative and positive infinity.

13) 120 + 1.5 + 0.25 = 121.75 pounds

14) 4 - 1.5 = 2.5 pints

15) (2.25)(32)= $72

As per the unitary method, Mrs. Smith's current weight is 121 pounds and 3 ounces.

To find Mrs. Smith's current weight, we need to add the weight she gained over the last two months to her initial weight. First, we will convert the mixed fractions to improper fractions for easier calculations.

1½ pounds can be written as (2 * 1) + 1/2 = 3/2 pounds.

1/4 pound remains as it is.

Now, let's add the weight gained in the last two months:

3/2 pounds + 1/4 pound = (3/2) + (1/4) = (6/4) + (1/4) = 7/4 pounds.

Next, we add the total weight gained to Mrs. Smith's initial weight:

120 pounds + 7/4 pounds = (120 * 4/4) + (7/4) = (480/4) + (7/4) = 487/4 pounds.

To express the answer in pounds, we convert the improper fraction back to a mixed fraction:

487/4 pounds can be written as (4 * 121) + 3 pounds.

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If y = 3 cot x/8, the double differentiation of y gives,

y'' = (3/32) (cosec x/8)(cot x/8)

Given value of y is,

y = 3 cot x/8

Differentiation of the above equation will give us the following,

y' = d(3 cot x/8) / dx

y' = 3d(cot x/8) / dx ........... (1)

Now, we know that the differentiation of cot x is -cosec²x

Therefore, d(cot x/8) / dx = -(cosec²x/8)/8

Thus, equation (1) transforms as follows,

y' = -3(cosec²x/8)/8

Taking differentiation of the above equation, we get,

y'' = -3d((cosec²x/8)/ 8dx

y'' = -6d(cosec x/8) / 8dx   [Using chain rule] .......... (2)

As we know, the differentiation of cosec x is -(cosec x)(cot x),

d(cosec x/8) / dx = -(cosec x/8)(cot x/8) / 8 [Using chain rule]

Therefore, equation (2) can be written as,

y'' = -(-6(cosec x/8)(cot x/8) / (8×8)

∴ y '' = 3(cosec x/8)(cot x/8) / 32

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If the length of hypotenuse is 95 m ,perpendicular is 57 m then the length of missing leg is 76m.

Given that the length of hypotenuse is 95 m ,the length of perpendicular is 57 m.

We are required to find the length of base or missing leg.

The given triangle is a right angled triangle. We can easily find out the length of the base of the triangle by using pythagoras theorem.

Pythagoras theorem says that the square of hypotenuse of a right angled triangle is equal to the sum of squares of the base and perpendicular of that triangle.

[tex]H^{2} =P^{2} +B^{2}[/tex]

We have to find the base of the triangle.

B=[tex]\sqrt{H^{2} -P^{2} }[/tex]

=[tex]\sqrt{(95)^{2} -(57)^{2} }[/tex]

=[tex]\sqrt{9025-3249}[/tex]

=[tex]\sqrt{5776}[/tex]

=76 m.

Hence if the length of hypotenuse is 95 m ,perpendicular is 57 m then the length of missing leg is 76m.

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Answer:

Point C will be at (3,1), Point B will be at (7,1) and Point A will be at (7,5)

Step-by-step explanation:

The probability that the student plays football is 20/33.

Probability determines the chances that an event would happen. The probability the event occurs is 1 and the probability that the event does not occur is 0.

The more likely the event is to happen, the closer the probability value would be to 1. The less likely it is for the event not to happen, the closer the probability value would be to zero.

The probability that the student plays football = total number of students who play football / total number of students

The probability that the student plays football = 40/66 = 20/33

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the area of the minor segment is 0. 29 cm^2

From the information given, we have the following parameters;

It is important to note the formula for area of a sector is given as;

Area = πr² + θ/360° - 1/ 2 r² sin θ

The value for π = 3.142

θ = 30°

Now, let's substitute the values

Area = 3. 142 × 5² × 30/ 360 - 1/ 2 × 5² × sin 30

Find the difference

Area = 3. 142 × 25 × 1/ 12 - 1/ 2 × 25 × 1/2

Multiply through

Area = 6. 54 - 6. 25

Area = 0. 29 cm^2

The area of the minor segment is given as 0. 29 cm^2

Thus, the area of the minor segment is 0. 29 cm^2

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Using proportions, it is found that option A gives a figure that is a scale drawing of the same courtyard using the scale 1 centimeter = 3 feet.

A proportion is a fraction of a total amount, and the measures are related using a rule of three. Due to this, relations between variables, either direct or inverse proportional, can be built to find the desired measures in the problem.

Researching this problem on the internet, the figure with a scale of 1 cm = 4 feet has the dimensions of:

51 cm, 75 cm, 30 cm and 72cm.

For a scale of 1 centimeter = 3 feet, these measures will be multiplied by 4/3, hence the figure is given in option A, as:

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Answer:

Step-by-step explanation:

in the smallest triangle (BCD) you have an angle of 90° and one of 63°, the sum of the internal angles in a triangle is 180°, remove the known angles from 180 ° and you will have the measure of the CBD angle

180 - 63 - 90 =

27°

Answer:

27°

Step-by-step explanation:

180°-90°-63° = 27°