Given that mLa = 36° and m

Step-by-step explanation:

An acute angle is an angle less or equal to 90 degrees.

let the other angle be x.

x + 36 = 90

grouping like terms

X = 90 - 36

X = 54 degrees

Answer:

54

Step-by-step explanation:

= ∠A + ∠B + ∠C = 180o … [∵sum of the angles of a triangle is 180]

or 180 = 36 + 90 + ∠C

C=54

Reason:

The angles x and 71 are opposite the congruent sides. We call these the base angles. The base angles are congruent for any isosceles triangle. Therefore, x = 71.

It might help to rotate the triangle so that the angles x and 71 are flat along the ground (rather than tilted).

[tex]v(x) = wlh \\ 2x {}^{3} + {x}^{2} - 16 x- 15 =(2x + 5)(x - 3)h \\ h = \frac{2x {}^{3} + x {}^{2} - 16x - 15 }{2x {}^{2} - x - 15 } \\ using \: long \: divison \\ h = x + 1[/tex]

Answer: 45%

Step-by-step explanation:

Given:

Find:

The number of groups which contains 9 is 9 and the total number of groups are 20.

So,

P = 9 / 20 = 0.45 = 45%

Therefore, the experimental probability is 45%

Based on the mean of the sample and the proportion who read below grade level, the probability that a second sample would have a proportion greater than 0.54 is 0.9772.

The probability that the second sample would be selected with a proportion greater than 0.54 can be found as:

P (x > 0.54) = P ( z > (0.54 - 0.30) / 0.12))

Solving gives:

P (x > 0.54) = P (z > 2)

P (x > 0.54) = 0.9772

In conclusion, the probability that the second sample would be selected with a proportion greater than 0.54 is 0.9772.

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Answer:

2827 m²

Step-by-step explanation:

The equation for volume of a sphere is:

[tex]V=\frac{4}{3}\pi r^3[/tex]

Using this equation and the given volume of the sphere, we can find the radius of the sphere.

[tex]V=\frac{4}{3}\pi r^3[/tex]

[tex]4500\pi = \frac{4}{3}\pi r^3[/tex]

Divide both sides by π

[tex]4500=\frac{4}{3}(r^3)[/tex]

Multiply both sides by 3

[tex]13500=4( r^3)[/tex]

Divide both sides by 4

[tex]3375=r^3[/tex]

Take the cube root of both sides

[tex]r=15[/tex]

Thus the radius of the sphere is 15m. We can use this information to find the surface area of the sphere.

The equation for surface area of a sphere is [tex]4\pi r^2[/tex]. Substituting the value we found for the radius into the equation, we find:

[tex]SA=4\pi r^2\\SA=4\pi 15^2\\SA=4\pi 225\\SA=900\pi\\SA\approx2827$ m^2[/tex]

The surface area of the sphere, rounded to the nearest square meter, is 2827 m².

Answer:35443

Step-by-step explanation:

The sales last year of the company is 5617

The given parameters are:

Company sales = 6740

Percentage increase = 20%

The sales of the company last year (x) to date is calculated as:

Company sales = (1 + Proportion) * Last year sales

Substitute the known values in the above equation

6740 = (1 + 20%) * x

Evaluate the sum

6740 = 1.2 * x

Divide both sides by 1.2

x = 5617

Hence, the sales last year is 5617

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Answer:

-2s-18 is the simplified answer

Step-by-step explanation:

s-3(s+6)

distributive property

s-3s-18

-2s-18

Answer:

B) 9:35

Step-by-step explanation:

the travel time we get by dividing the distance by the speed

distance / distance/time = time

so,

60 / 80 = 6/8 = 3/4 = 0.75 hours or 45 minutes.

since she arrived 20 minutes late, that means at 10:20 am, she left home 45 - 20 = 25 minutes before 10:00am.

and that is 9:35 am.

if any number or variable is outside a parenthesis it will go to all the number rin the parenthesis.

in this case -y is outside the parenthesis so it will got to both the sides :

-y(-6y-3) = -6y x y - 3 x y

= -6y² -3y//

The area of the garden is 74 square feet

The complete question is added as an attachment

From the attached figure, we have the following shapes and dimensions:

The rectangular area is

A1 = 10 * 5

Evaluate

A1 = 50

The area of the trapezoid is

A2 = 0.5 * Sum of parallel bases * height

This gives

A2 = 0.5 * (10 + 6) * 3

Evaluate

A2 = 24

The total area is

Total = A1 + A2

This gives

Total = 50 + 24

Evaluate

Total = 74

Hence, the area of the garden is 74 square feet

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Answer: 13

Step-by-step explanation:

4x+3y=-5 solve x :    

4x = -3y + -5           | -3y

1x = -0.75y + -1.25  | : 4

-3x + 7y = 13 solve x :

-3x + 7y = 13              | -7y

-3x = -7y = 13             | : (-3)

Equalization Method Solution: -0.75y+-1.25=2.333y+-4.333

-0, 75y - 1,25 = 2,333y - 4, 333 solve y:

-0, 75y - 1,25 = 2,333y -4, 333    | -2,333y

-3, 083y - 1,25 = -4,333               | + 1, 25

-3. 083y = -3,088                         | : (-3, 083)

y = 1

Plug y = 1 into the equation 4x + 3y = -5 :

4x + 3 · 1           | Multiply 3 with 1

4x + 3 = -5        | -3

4x = -8              | : 4

x = -2

So the solution is:

y = 1, x = -2

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Answer:

3/10

Step-by-step explanation:

There are totally 5 disks. On the first pick, there are 3 red discs. So

P(red disc on first pick ) = 3/5

Assuming that a red disc has been picked the first time, there will be 2 red discs and 2 blue discs so

(P red disc on second pick given red disc on first pick) = 2/4

P(red disc on both picks) = 3/5 x 2/4 = 3/10

[tex]x > \ln(x)[/tex] for all [tex]x[/tex], so

[tex]\displaystyle \lim_{x\to\infty} (\ln(x) - x) = - \lim_{x\to\infty} x = \boxed{-\infty}[/tex]

Similarly, [tex]\displaystyle \lim_{x\to\infty} (x-e^x) = - \lim_{x\to\infty} e^x = \boxed{-\infty}[/tex]

We can of course see the limits are identical by replacing [tex]x\mapsto e^x[/tex], so that

[tex]\displaystyle \lim_{x\to\infty} (\ln(x) - x) = \lim_{x\to\infty} (\ln(e^x) - e^x) = \lim_{x\to\infty} (x - e^x)[/tex]

You can also rewrite the limands to accommodate the application of l'Hôpital's rule. For instance,

[tex]\displaystyle \lim_{x\to\infty} (x - e^x) = \ln\left(\exp\left(\lim_{x\to\infty} (x - e^x)\right)\right) = \ln\left(\lim_{x\to\infty} e^{x-e^x}\right) = \ln\left(\lim_{x\to\infty} \frac{e^x}{e^{e^x}}\right)[/tex]

Using the rule, the limit here is

[tex]\displaystyle \lim_{x\to\infty} \frac{(e^x)'}{\left(e^{e^x}\right)'} = \lim_{x\to\infty} \frac{e^x}{e^x e^{e^x}} = \lim_{x\to\infty} \frac1{e^{e^x}} = 0[/tex]

so the overall limit is

[tex]\displaystyle \lim_{x\to\infty} (x - e^x) = \ln\left(\lim_{x\to\infty} \frac{e^x}{e^{e^x}}\right) = \ln(0) = \lim_{x\to0^+} \ln(x) = -\infty[/tex]

The domain of the relation shown in the graph is -4 <= x <= 4

The relation on the graph is an ellipse function.

The domain is the set of x values the function can take

From the graph, we have the following x values

This means that the domain of x in the graph is -4 <= x <= 4

Hence, the domain of the relation shown in the graph is -4 <= x <= 4

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Answer: 250:1

Step-by-step explanation:

Divide the measurement of appearance by the actual measurement.

[tex]100,000/400=250[/tex]

The magnification scale is 250:1. The scale is used by a 250X magnifying lense.

The equation for the reflected function is:

g(x) =  (1/2)*(3)^(-x)

Here we know that function g(x) is a reflection of f(x) across the y-axis.

Remember that the reflection is written as:

g(x) = f(-x)

Here we know that:

f(x) = (1/2)*(3)^x

Then, replacing that in the equation for g(x), we get:

g(x) =  f(-x) = (1/2)*(3)^(-x)

g(x) =  (1/2)*(3)^(-x)

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a. Given the 2nd order ODE

[tex]y''(x) = 4y(x) + 4[/tex]

if we substitute [tex]z(x)=y'(x)+2y(x)[/tex] and its derivative, [tex]z'(x)=y''(x)+2y'(x)[/tex], we can eliminate [tex]y(x)[/tex] and [tex]y''(x)[/tex] to end up with the ODE

[tex]z'(x) - 2y'(x) = 4\left(\dfrac{z(x)-y'(x)}2\right) + 4[/tex]

[tex]z'(x) - 2y'(x) = 2z(x) - 2y'(x) + 4[/tex]

[tex]\boxed{z'(x) = 2z(x) + 4}[/tex]

and since [tex]y(0)=y'(0)=1[/tex], it follows that [tex]z(0)=y'(0)+2y(0)=3[/tex].

b. I'll solve with an integrating factor.

[tex]z'(x) = 2z(x) + 4[/tex]

[tex]z'(x) - 2z(x) = 4[/tex]

[tex]e^{-2x} z'(x) - 2 e^{-2x} z(x) = 4e^{-2x}[/tex]

[tex]\left(e^{-2x} z(x)\right)' = 4e^{-2x}[/tex]

[tex]e^{-2x} z(x) = -2e^{-2x} + C[/tex]

[tex]z(x) = -2 + Ce^{2x}[/tex]

Since [tex]z(0)=3[/tex], we find

[tex]3 = -2 + Ce^0 \implies C=5[/tex]

so the particular solution for [tex]z(x)[/tex] is

[tex]\boxed{z(x) = 5e^{-2x} - 2}[/tex]

c. Knowing [tex]z(x)[/tex], we recover a 1st order ODE for [tex]y(x)[/tex],

[tex]z(x) = y'(x) + 2y(x) \implies y'(x) + 2y(x) = 5e^{-2x} - 2[/tex]

Using an integrating factor again, we have

[tex]e^{2x} y'(x) + 2e^{2x} y(x) = 5 - 2e^{2x}[/tex]

[tex]\left(e^{2x} y(x)\right)' = 5 - 2e^{2x}[/tex]

[tex]e^{2x} y(x) = 5x - e^{2x} + C[/tex]

[tex]y(x) = 5xe^{-2x} - 1 + Ce^{-2x}[/tex]

Since [tex]y(0)=1[/tex], we find

[tex]1 = 0 - 1 + Ce^0 \implies C=2[/tex]

so that

[tex]\boxed{y(x) = (5x+2)e^{-2x} - 1}[/tex]

6.a) The differential equation for z(x) is z'(x) = 2z(x) + 4, z(0) = 3.

6.b) The value of z(x) is [tex]z(x) = 5e^{2x} - 2[/tex].

6.c) The value of y(x) is [tex]y(x) = \frac{5e^{2x}}{4} - \frac{1}{4e^{2x}} -1[/tex].

The given ordinary differential equation is y''(x) = 4y(x) + 4, y(0) = y'(0) = 1 ... (d).

We are also given a substitution function, z(x) = y'(x) + 2y(x) ... (z).

Putting x = 0, we get:

z(0) = y'(0) + 2y(0),

or, z(0) = 1 + 2*1 = 3.

Rearranging (z), we can write it as:

z(x) = y'(x) + 2y(x),

or, y'(x) = z(x) - 2y(x) ... (i).

Differentiating (z) with respect to x, we get:

z'(x) = y''(x) + 2y'(x),

or, y''(x) = z'(x) - 2y'(x) ... (ii).

Substituting the value of y''(x) from (ii) in (d) we get:

y''(x) = 4y(x) + 4,

or, z'(x) - 2y'(x) = 4y(x) + 4.

Substituting the value of y'(x) from (i) we get:

z'(x) - 2y'(x) = 4y(x) + 4,

or, z'(x) - 2(z(x) - 2y(x)) = 4y(x) + 4,

or, z'(x) - 2z(x) + 4y(x) = 4y(x) + 4,

or, z'(x) = 2z(x) + 4y(x) - 4y(x) + 4,

or, z'(x) = 2z(x) + 4.

The initial value of z(0) was calculated to be 3.

6.a) The differential equation for z(x) is z'(x) = 2z(x) + 4, z(0) = 3.

Transforming z(x) = dz/dx, and z = z(x), we get:

dz/dx = 2z + 4,

or, dz/(2z + 4) = dx.

Integrating both sides, we get:

∫dz/(2z + 4) = ∫dx,

or, {ln (z + 2)}/2 = x + C,

or, [tex]\sqrt{z+2} = e^{x + C}[/tex],

or, [tex]z =Ce^{2x}-2[/tex] ... (iii).

Substituting z = 3, and x = 0,  we get:

[tex]3 = Ce^{2*0} - 2\\\Rightarrow C - 2 = 3\\\Rightarrow C = 5.[/tex]

Substituting C = 5, in (iii), we get:

[tex]z = 5e^{2x} - 2[/tex].

6.b) The value of z(x) is [tex]z(x) = 5e^{2x} - 2[/tex].

Substituting the value of z(x) in (z), we get:

z(x) = y'(x) + 2y(x),

or, 5e²ˣ - 2 = y'(x) + 2y(x),

which gives us:

[tex]y(x) = \frac{5e^{2x}}{4} - \frac{1}{4e^{2x}} -1[/tex] for the initial condition y(x) = 0.

6.c) The value of y(x) is [tex]y(x) = \frac{5e^{2x}}{4} - \frac{1}{4e^{2x}} -1[/tex].

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Answer:

d. (-6, 0)

Step-by-step explanation:

no explanation bc u need this quick

The expression that represents the perimeter of the rectangle is 30h - 2

The dimensions of the rectangle are given as:

Length = 7h + 3

Width = 8h - 4

The perimeter of the rectangle is given as:

P = 2 * (Length + Width)

Substitute the known values in the above equation

P =2 * (7h + 3 + 8h - 4)

Evaluate the like terms

P = 2 * (15h - 1)

Evaluate the product

P = 30h - 2

Hence, the expression that represents the perimeter of the rectangle is 30h - 2

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Answer:

  1∠22.5°, 1∠112.5°, 1∠202.5°, 1∠292.5°

Step-by-step explanation:

A root of a complex number can be found using Euler's identity.

For some z = a·e^(ix), the n-th root is ...

  z = (a^(1/n))·e^(i(x/n))

Here, we have z = i, so a = 1 and z = π/2 +2kπ.

Using r∠θ notation, this is ...

  i = 1∠(90° +k·360°)

and

  i^(1/4) = (1^(1/4))∠((90° +k·360°)/4)

  i^(1/4) = 1∠(22.5° +k·90°)

For k = 0 to 3, we have ...

 for k = 0, first root = 1∠22.5°

  for k = 1, second root = 1∠112.5°

  for k = 2, third root = 1∠202.5°

  for k = 3, fourth root = 1∠292.5°

The Young Modulus of the brass rod ≈ 8.5 × 10¹⁰ Pa or N/m².

Given: Physical Properties of the Brass rod exist

Length, L = 250 mm

Radius, R = Diameter / 2 = 25/2 mm

Load of 50 kN hanging on the brass rod which results in elongation of the rod to 0.3 mm.

To estimate the Young Modulus of the brass rod we will require Stress on the rod, Ratio of elongation to the total length.

To find the stress on the rod,

Stress(Pressure) = Force / Area

⇒ Stress = 50,000 / (Area of cross-section)

The rod exists shaped like a cylinder then the area of its cross-section will be πR² ( R = radius)

⇒ Stress = 50000 / (π (25/2 mm)² )

The force exists given in newton, then we must convert the unit of the area into m² to obtain the solution in pascal.

1 mm = 0.001 m

⇒ Stress = 50,000 / 22/7 ( 25/2 × 10⁻³ m)²

take, π = 22/7

⇒ Stress = 50000 / ( 22/7 × 625 / 4 × 10⁻⁶ )

⇒ Stress = 50000 / ( 11/14 × 625 × 10⁻6 )

⇒ Stress = 50000 × 14 × 10⁶ / 625×11

⇒ Stress = 80 × 14 × 10⁶ / 11

⇒ Stress = 1120/11 × 10⁶ Pa

Now, that we maintain Stressed, to estimate Strain (ratio of elongation to original length),

⇒ Strain = (Elongation) / (Original length)

⇒ Strain = 0.3 / 250

⇒ Strain = 3 / 2500

Therefore, Young Modulus, Y = Stress / Strain

⇒ Y = 1120/11 × 10⁶ / 3/2500

⇒ Y = 1120/11 × 2500/3 × 10⁶

⇒ Y = 101.8 × 833.3 × 10⁶

⇒ Y = 84829.94 × 10⁶

⇒ Y ≈ 8.5 × 10¹⁰ Pa

Therefore, the Young Modulus of the brass rod ≈ 8.5 × 10¹⁰ Pa or N/m².

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Answer:

The most reasonable unit to measure the time spent at school on an average school day is in hours.

In anything we are going to measure, we have to pay special attention to the unit. When doing this, we want to keep the absolute value of the measure, that is, the number small, hence the do this conversion of units may be used.

For example, it is easier and more practical to say one day instead of 87,840 seconds. The same logic is applied to this problem, in which the most reasonable unit to measure the time spent at school on an average school day is in hours, as it keeps the numerical measure smaller than it would be in minutes or seconds.

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The correct option regarding whether the relation is a function is:

B. The relation is a function. Each input value is paired with one output value.

A relation represent a function if each value of the input is paired with one value of the output.

In this problem, when the input - output mappings are given by:

{(–2, 3), (–1, 3), (0, 2), (1, 4), (5, 5)}.

Which means that yes, each input value is paired with one output value, hence the relation is a function and option B is correct.

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(a) We want to find a scalar function [tex]f(x,y,z)[/tex] such that [tex]\mathbf F = \nabla f[/tex]. This means

[tex]\dfrac{\partial f}{\partial x} = 2xy + 24[/tex]

[tex]\dfrac{\partial f}{\partial y} = x^2 + 16[/tex]

Looking at the first equation, integrating both sides with respect to [tex]x[/tex] gives

[tex]f(x,y) = x^2y + 24x + g(y)[/tex]

Differentiating both sides of this with respect to [tex]y[/tex] gives

[tex]\dfrac{\partial f}{\partial y} = x^2 + 16 = x^2 + \dfrac{dg}{dy} \implies \dfrac{dg}{dy} = 16 \implies g(y) = 16y + C[/tex]

Then the potential function is

[tex]f(x,y) = \boxed{x^2y + 24x + 16y + C}[/tex]

(b) By the FTCoLI, we have

[tex]\displaystyle \int_{(1,1)}^{(-1,2)} \mathbf F \cdot d\mathbf r = f(-1,2) - f(1,1) = 10-41 = \boxed{-31}[/tex]

[tex]\displaystyle \int_{(-1,2)}^{(0,4)} \mathbf F \cdot d\mathbf r = f(0,4) - f(-1,2) = 64 - 41 = \boxed{23}[/tex]

[tex]\displaystyle \int_{(0,4)}^{(2,3)} \mathbf F \cdot d\mathbf r = f(2,3) - f(0,4) = 108 - 64 = \boxed{44}[/tex]

Based on the task content given; the solution to the equation for x 3 1/2(2 1/5x - 4 2/3) - 5 5/6 = 0 is 2 203/231

3 1/2(2 1/5x - 4 2/3) - 5 5/6 = 0

(3 1/2 × 2 1/5x) - (3 1/2 × 4 2/3) - 5 5/6 = 0

(7/2 × 11/5x) - (7/2 × 14/3) - 35/6 = 0

77/10x - 98/6 - 35/6 = 0

77/10x = 98/6 + 35/6

77/10x = 98+35 / 6

77/10x = 133/6

x = 133/6 ÷ 77/10

= 133/6 × 10/77

= 1330/462

x = 2 406/462

= 2 203/231

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Based on the given parameters in the question, the value of the probability of A and D is 0.30

The given parameters from the tree diagram are:

The probability P(A and D) is calculated using the following probability formula

P(A and D) = P(A) × P(D)

Substitute the known values in the above equation

P(A and D) = 0.5 × 0.6

Evaluate the product of 0.5 and 0.6

P(A and D) = 0.30

Hence, the value of the probability of A and D is 0.30 based on the given parameters in the question

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