Is 1/4 more than 7/3

Answer:

z = 71

Step-by-step explanation:

this is the answer for this equation

Answer:

z=71

Step-by-step explanation:

Answer:

3.6c + 7.5t <= 21,600

Step-by-step explanation:

Let c = number of cars, and let t = number of trucks.

c cars weigh 3.6c tons, and t trucks weigh 7.5t tons.

3.6c + 7.5t <= 21,600

Answer:

-3 ≥ b is your inequality

Step-by-step explanation:

Answer:

The answer is 7/12

Answer:

7/12

Step-by-step explanation:

[tex]\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}[/tex]

Actually Welcome to the Concept of the derivatives.

Hence, here

y = 6x^2 -4x +17

so, dy/dx = d(6x^2-4x+17)/dx

===> hence, dy/dx = 12x -4 ( used formula d(x^n) /dx = nx^n-1 )

======> dy/dx = 12x-4

Answer:

b.

Step-by-step explanation:

Step-by-step explanation:

(8+(-2)-4)

=8-2-4

=2

(3-5)

=-2

Answer:

5

Step-by-step explanation:

The interquartile range is the 3rd quartile minus the 1st quartile. The 3rd quartile is 8 and the 1st quartile is 3.

8 - 3 = 5

The interquartile range (IQR) of the data is 5.

From the box plot we can write,

Minimum = 2

First Quartile, Q1 = 3

Median = 6

Third Quartile, Q3 = 8

Maximum = 10

So, Interquartile range (IQR)

= Q3 - Q1

= 8 - 3

= 5

Thus, the interquartile range (IQR) of the data is 5.

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Answer:

Step-by-step explanation:

Two of the sides are 13.5 inches by 2.5 inches, two sides are 13.5 inches by 8 inches, and two sides are 2.5 inches by 8 inches.

Surface area = 2*13.5*2.5 + 2*13.5*8 + 2*2.5*8

Answer:

srry buddy I don't understand your language

I am from India

can ask it in English

then I could help you

Answer:

5x-5y

Step-by-step explanation:

You are using the distributive property

You can also think of it as multiplying 5 by x and 5 by y and then subtracting it just as it is subtracting in the original equation

Answer:

5x-y.

Step-by-step explanation:

so its 5 times x, minus 7. easy. 5x,-y. rephrase it. hhen a variable is next to a number, it means multiply.

Answer:

Follows are the solution to the given points:

Step-by-step explanation:

Given value:

[tex]\to f_X (x) \ \ xe^{-x} \ , \ x>0[/tex]

For point a:

Moment generating function of X=?

Using formula:

[tex]\to M(t) =E(e^{tx})= \int^{\infty}_{-\infty} \ e^{tx} f(x) \ dx[/tex]

[tex]M(t) = \int^{\infty}_{-\infty} \ e^{tx}xe^{-x} \ dx = \int^{\infty}_{0} \ xe^{(t-1)x} \ dx[/tex]

integrating the values by parts:

[tex]u = x \\\\dv = e^{(t-1)x}\\\\dx =dx \\\\v= \frac{e^{(t-1)x}}{t-1}\\\\M(t) =[\frac{e^{(t-1)x}}{t-1}]^{-\infty}_{0} -\int^{\infty}_{0} \frac{e^{(t-1)x}}{t-1} \ dx\\\\[/tex]  

        [tex]= \frac{1}{t-1} (0) - [\frac{e^{(t-1)x}}{(t-1)^2}]^{\infty}_{0}\\\\=\frac{1}{(t-1)^2}(0-1)\\\\=\frac{1}{(t-1)^2}\\\\[/tex]

Therefore, the moment value generating by the function is [tex]=\frac{1}{(t-1)^2}[/tex]

In point b:

[tex]E(X^n)=?[/tex]

Using formula: [tex]E(X^n)= M^{n}_{X}(0)[/tex]

form point (a):

[tex]\to M_{X}(t)=\frac{1}{(t-1)^2}[/tex]

Differentiating the value with respect of t

[tex]M'_{X}(t)=\frac{-2}{(t-1)^3}[/tex]

when [tex]t=0[/tex]

[tex]M'_{X}(0)=\frac{-2}{(0-1)^3}= \frac{-2}{(-1)^3}= \frac{-2}{-1}=2\\\\M''_{X}(t)=\frac{(-2)(-3)}{(t-1)^4}\\\\M''_{X}(0)=\frac{(-2)(-3)}{(0-1)^4}= \frac{6}{(-1)^4}= \frac{6}{1}=6\\\\M''_{X}(t)=\frac{(-2)(-3)}{(t-1)^4}\\\\M''_{X}(0)=\frac{(-2)(-3)}{(0-1)^4}= \frac{6}{(-1)^4}= \frac{6}{1}=6\\\\M'''_{X}(t)=\frac{(-2)(-3)(-4)}{(t-1)^5}\\\\M''_{X}(0)=\frac{(-2)(-3)(-4)}{(0-1)^5}= \frac{24}{(-1)^5}= \frac{24}{-1}=-24\\\\\therefore \\\\M^{K}_{X} (t)=\frac{(-2)(-3)(-4).....(k+1)}{(t-1)^{k+2}}\\\\[/tex]

[tex]M^{K}_{X} (0)=\frac{(-2)(-3)(-4).....(k+1)}{(-1)^{k+2}}\\\\\therefore\\\\E(X^n) = \frac{(-2)(-3)(-4).....(n+1)}{(-1)^{n+2}}\\\\[/tex]

Step-by-step explanation:

so basically they are asking you to subtract, hope this helps, use a calculator if needed

Answer:

The answer is below

Step-by-step explanation:

The car is moving in a circular track with a radius of 1.9 miles. The distance covered by the car if the track is revolved once = 2π * radius of the track.

a) Since the car has swept out an angle of 212 degrees, the distance covered by the race car = [tex]\frac{212}{360}*2\pi(1.9)=7.03 \ miles[/tex]

b) If the car traveled 4.9 miles, the angle swept out (θ) is:

[tex]4.9= \frac{\theta}{360} * 2\pi r\\\\\theta=\frac{4.9}{2\pi (1.9)}*360 \\\\\theta=147.76^o=147.76^o*\frac{\pi}{180} \\\\\theta=2.58\ rad[/tex]

c) h = distance covered, t = time in hourss.

Hence:

h = 85.5t

The race-car's motion round the track is a repeating motion that can be

described by a sinusoidal function.

The correct responses are;

Reasons:

Known parameters are;

Radius of the track, r = 1.9 miles

Point the car starts = 3 O'clock

Speed of the car = 85.5 mph

Required:

Distance swept out when the car traveled an angle of 212°.

Solution;

Distance of one complete turn = 2·π·r

Angle of one complete turn = 360°

Therefore, at 212°, we have;

[tex]\dfrac{212^{\circ}}{360^{\circ}} \times 2 \times 1.9 \times \pi \approx 7.03[/tex]

Required:

The measure of the angle swept out by the car if the car has travelled 4.9 miles.

Solution;

Let, θ represent the angle, we have;

[tex]\dfrac{\theta}{2 \cdot \pi} \times 2 \times 1.9 \times \pi \approx 4.9 \ miles[/tex]

We get;

[tex]{\theta} = \dfrac{4.9 \ miles}{2 \times 1.9 \ miles \times \pi } \times 2\cdot \pi \approx 2.58 \ radians[/tex]

Required:

The function, h, that gives the distance of the above the horizontal

diameter of the track (in miles) in terms of the number of hours since the

race-car started driving, t.

Solution;

The function that gives the car height can be presented as follows;

The angular velocity, ω = [tex]\dfrac{v}{r}[/tex]

Therefore;

[tex]\omega = \dfrac{85.5 \ mph}{1.9 \ miles} = 45 \ rad/hour[/tex]

The general form of the sinusoidal function is h = A·sin[k·(θ - b)] + c

Therefore, we get;

h = 1.9·sin[k·(θ - b)] + c

The period, T = 2·π/ω = 2·π/k

Therefore, given that ω·t = θ, we get;

h = 1.9·sin[ω·t - b] + c = 1.9·sin[45·t - b] + c

The vertical sift, c, and the horizontal shift, b, can both be taken as zero,

given that the sin(0) = 0, therefore;

h = 1.9·sin[0] + c = c = 0, such that the at the start, t = 0, the car is at a

distance of h = 0 above the horizontal line.

The function, h, that gives the distance of the car above the

horizontal track, in terms of the number of hours since the car started

driving, t, is therefore;

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Answer:

1.) 7x+5y

2.)-19m+14r

3.)22b+40c

4.)10x+25

5.)7m+7

Step-by-step explanation:

ez

First one: 7

Second one: 22

Third one: 137

Hope this helps.

Answer:

123

Step-by-step explanation:

213

Answer:

no

Step-by-step explanation:

5/8=0.6250 so its rational

Answer: no

Step-by-step explanation: 5/8 is rational because it is another way of writing the ratio 5:8. So no, 5/8 is not an irrational number.

Answer:

the surface area of the wedge is 360ft²

Step-by-step explanation:

6x5x3x4 = 360ft²

Answer:

The value for x should be 35

Step-by-step explanation:

Answer:

32 degrees

Step-by-step explanation:

35 + x + 29 + 84 = 180

148 + x = 180

x = 32

Answer:

The quadratic equation is [tex]h(t) = -318.22t^2 + 954.67t + 12728.89[/tex]

Step-by-step explanation:

Vertex of a quadratic function:

Suppose we have a quadratic function in the following format:

[tex]f(x) = ax^{2} + bx + c[/tex]

It's vertex is the point [tex](x_{v}, y_{v})[/tex]

In which

[tex]x_{v} = -\frac{b}{2a}[/tex]

[tex]y_{v} = -\frac{\Delta}{4a}[/tex]

Where

[tex]\Delta = b^2-4ac[/tex]

If a<0, the vertex is a maximum point, that is, the maximum value happens at [tex]x_{v}[/tex], and it's value is [tex]y_{v}[/tex].

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

[tex]ax^{2} + bx + c, a\neq0[/tex].

This polynomial has roots [tex]x_{1}, x_{2}[/tex] such that [tex]ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})[/tex], given by the following formulas:

[tex]x_{1} = \frac{-b + \sqrt{\Delta}}{2*a}[/tex]

[tex]x_{2} = \frac{-b - \sqrt{\Delta}}{2*a}[/tex]

[tex]\Delta= b^{2} - 4ac[/tex]

Quadratic equation:

[tex]ax^2 + bx + c = 0[/tex]

The rock reaches a maximum height of 676 feet above the ground after 1.5 seconds.

This means that:

[tex]-\frac{b}{2a} = 1.5[/tex]

[tex]y_{v} = -\frac{b^2-4ac}{4a} = 676[/tex]

If the rock hits the ground below the ledge 8 seconds after it is thrown

This means that:

[tex]64a + 8b + c = 0[/tex]

Relation of b and a:

[tex]-\frac{b}{2a} = 1.5[/tex]

This means that:

[tex]b = -3a[/tex]

Relationship of c and a:

[tex]64a + 8b + c = 0[/tex]

[tex]64a - 24a + c = 0[/tex]

[tex]c = -40a[/tex]

Finding a:

[tex]y_{v} = -\frac{(-3a)^2-4a(-40a)}{4a} = 676[/tex]

[tex]9a^2+160a = -2704a[/tex]

[tex]9a^2 + 2864a = 0[/tex]

[tex]a(9a + 2864) = 0[/tex]

Since it is a quadratic equation, a cannot be 0.

[tex]9a + 2864 = 0[/tex]

[tex]9a = -2864[/tex]

[tex]a = -\frac{2864}{9}[/tex]

[tex]a = -318.22[/tex]

Finding b and c:

[tex]b = -3a = -3(-318.22) = 954.67[/tex]

[tex]c = -40a = -40(-318.22) = 12728.89[/tex]

The quadratic equation is given by:

[tex]h(t) = -318.22t^2 + 954.67t + 12728.89[/tex]

The quadratic equation is [tex]h(t) = -318.22t^2 + 954.67t + 12728.89[/tex]

Complete Question:

If a 100kg bag of rice cost #120, how many kilogram of rice can one buy for #40

Answer:

33 kilograms.

Step-by-step explanation:

Given the following data;

100 kg of rice = #120

1 kg of rice = #x

Cross-multiplying, we have;

100x = 120

x = 120/100

x = 1.2

Now, to find how much #40 can buy;

[tex] Quantity = \frac {Cost}{Cost \; of \; a \; bag} [/tex]

Substituting into the equation, we have;

[tex] Quantity = \frac {40}{1.2} [/tex]

Quantity of rice = 33.33 ≈ 33 kilograms.

Method II

100 kg of rice = #120

x kg of rice = #40

Cross-multiplying, we have;

120x = 40 * 100

x = 4000/120

x = 33.33 ≈ 33 kilograms.

Answer:

6x=18

x=18/6

x=3.

-3x=15

x=15/-3

x=-5

7x=-63

x=-63/7

x=-9

5x-3=12

5x=12+3

5x=15

x=15/5

x=3

-4x+1=13

-4x=13-1

-4x=12

x=12/-4

x=-3

-x+9=16

-x=16-9

-x=7

x=7/-1

x=-7

Answer:

A

Step-by-step explanation:

A is the answer for the first question I did the math on edge

The second one after is B

We are Simplifying Rational Expressions with Negative Exponents

Answer:

A.) Afriv

Step-by-step explanation:

Heat oven to 350°F. Generously grease 12-cup fluted tube pan with shortening or cooking spray. In large 1-gallon plastic food storage bag, mix granulated sugar and cinnamon.

2

Separate dough into 16 biscuits; cut each into quarters. Shake in bag to coat. Arrange in pan, adding walnuts and raisins among the biscuit pieces. Sprinkle any remaining sugar over biscuits.

3

In small bowl, mix brown sugar and butter; pour over biscuit pieces.

4

Bake 30 to 40 minutes or until golden brown and no longer doughy in center. Loosen edges of pan with metal spatula. Cool in pan 5 minutes. Turn upside down onto serving plate; replacing any biscuit pieces and caramel from pan. Pull apart to serve. Serve warm.

Answer: 400 because when they test 200 they get that 4 are dead wich is 1/50 of 200 so 1/50 times 20,000 is 400

Step-by-step explanation:

Answer:

83.8

Step-by-step explanation:

first you add the grades together: 75 + 87 + 90 + 88 + 79 = 419

then you divide by the amount of grades: 419 / 5 = 83.8

Answer:it’s c y=|x-5|-4

Step-by-step explanation:

The correct function which is graphed is,

D. [tex]y = |x- 5| - 4[/tex]

We have to give that,

A graph of a function is shown in the image.

Now, From the graph;

Two points on the function are (0, 1) and (5, 4).

Since both points are lies on the graph of a function, hence it satisfies the equation of the function.

From the given function,

A. [tex]y = |x + 5|+ 4[/tex]

Substitute x = 0, y = 1;

1 = |0 + 5| + 4

1 = 5 + 4

1 = 9

Which is not true.

B. [tex]y = |x -5| + 4[/tex]

Substitute x = 0, y = 1;

1 = |0 - 5| + 4

1 = 5 + 4

1 = 9

Which is not true.

C. [tex]y = |x + 5| - 4[/tex]

Substitute x = 0, y = 1;

1 = |0 + 5| - 4

1 = 5 - 4

1 = 1

Which is true.

And, Substitute x = 5, y = 4;

4 = |5 + 5| + 4

4 = 10 + 4

4 = 14

Which is not true.

D. [tex]y = |x- 5| - 4[/tex]

Substitute x = 0, y = 1;

1 = |0 - 5| - 4

1 = 5 - 4

1 = 1

Which is true.

And, Substitute x = 5, y = 4;

4 = |5 - 5| + 4

4 = 0 + 4

4 = 4

Which is true.

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Answer:

a. area of triangle: A=1/2(b)(h)

                               A= 1/2 (3)(2.4)

                               A= 3.6cm^2

All the triangles have the same area, so Total area of the triangles is

A= 4(3.60)

A= 14.4 cm^2

Area of the square: A= s^2 (3)^2= 9cm^2

Total surface area o the Pyramid:  14.4+ 9=23.4cm^2

b. Lateral surface area of the Pyramid: A= 1/2 P(l)              

                                                                 A= 1/2 (12)(4)

                                                                  A= 24cm^2

 P is perimeter of the base= (3)(4)=12cm

 l s the slant height= 4cm

Step-by-step explanation: