9. You measure a bag of apples as weighing 85 N on a scale.

a) What is the mass of the bag of apples?

b) if you were to carry the bag of apples to the surface of the moon, where the acceleration of gravity is only 1.67 m/s2, how much would the apples weigh?

The height of the right triangular prism is 84.3 cm.

To find the height of the right triangular prism, let's use the formula for the volume of a right triangular prism.

Volume of the right triangular prism = 1/2 × base × height × length

Where, base = 25 cm and height is what we need to find.

We are given that the volume of the prism is 0.075 cubic meters. We first need to convert the volume into cubic cm as the other units are given in cm. 1 cubic meter = 100 cm × 100 cm × 100 cm= 1,000,000 cubic cm

So, 0.075 cubic meters = 0.075 × 1,000,000 = 75,000 cubic cm

Substituting all the known values into the formula, we get:

75,000 = 1/2 × 25 × height × length

Hence, height = 75,000 / (1/2 × 25 × length)

Now, we need to find the length of the prism. As the base of the right triangular prism is a right isosceles triangle, its hypotenuse is the length of the prism. Using the Pythagorean theorem to find the hypotenuse, we have:

Length² = 25² + 25²

Length² = 1,250

Length = √1,250

Length ≈ 35.36 cm

Substituting this length into the equation we found earlier, we get:

height = 75,000 / (1/2 × 25 × 35.36)height ≈ 84.3 cm

So, the height of the right triangular prism is approximately 84.3 cm.

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Answer:

-1

Step-by-step explanation:

calculator

:)

Answer:

An even function is symmetric about the y-axis. An odd function is symmetric about the origin. As the graphed function is neither symmetric about the origin, nor symmetric about the y-axis, it is neither even nor odd.

Step-by-step explanation:

Even and odd functions are special types of functions.

An even function is symmetric about the y-axis. An odd function is symmetric about the origin. As the graphed function is neither symmetric about the origin, nor symmetric about the y-axis, it is neither even nor odd.

[tex]\cfrac{89.40~~ - ~~\stackrel{ \textit{minus the tax of each} }{0.50-0.50-0.50-0.50-0.50-0.50}}{6}\implies \cfrac{86.40}{6}\implies \stackrel{ each }{14.40}[/tex]

Answer:

4

Step-by-step explanation:

Answer: w=4

Step-by-step explanation:

1).  5(3w)+2w=68

2). 15w+2w=68

3). 17w=68

4). 17w/17=68/17

5). w=4

Let's first calculate the probability of getting the question right if you eliminate one of the six choices. Since there are six options, initially you have a 1/6 chance of getting the question right. If you eliminate one of the options, you are left with five choices, so your probability of getting the question right increases to 1/5.

Now, let's calculate the expected value of the number of points you get. If you get the question right, you earn one point. If you get the question wrong, you lose 1/3 point. So, we can write the expected value as:

Expected value = Probability of getting the question right x Points for a correct answer + Probability of getting the question wrong x Points for a wrong answer

Let's plug in the values:

Expected value = (1/5) x 1 + (4/5) x (-1/3)

Expected value = 1/5 - 4/15

Expected value = -1/15

So, the expected value of the number of points you get is -1/15. This means that, on average, you can expect to lose about 0.067 points for each question you attempt.

Answer:

Any line with the slope 1 or y = x + b

Step-by-step explanation:

Parallel lines by definition have the same slope

Value of the y-intercept or b doesn't matter in this case

You have to find the slope of the line through (3,6) and (5,8)

Slope equation is

Plug in values given

[tex]\frac{8-6}{5-3}[/tex]

Evaluate

[tex]\frac{2}{2} =1[/tex]

Any line with the slope of 1 (or no co-efficient in front of x) would be parallel to the line through (3,6) and (5,8)

Answer: The point-slope form of an equation of a line is: y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope.

We are given the y-intercept at (0, -3), which means that this point is on the line and that the line crosses the y-axis at y = -3. Therefore, the y-coordinate of the y-intercept is -3.

We are also told that the line passes through the point (4, 5), which means that this point is also on the line. Therefore, x1 = 4 and y1 = 5.

To find the slope of the line, we can use the two points (0, -3) and (4, 5):

slope = (y2 - y1) / (x2 - x1)

slope = (5 - (-3)) / (4 - 0)

slope = 8 / 4

slope = 2

Now that we have the slope and a point on the line, we can use the point-slope form to write the equation of the line:

y - y1 = m(x - x1)

y - 5 = 2(x - 4)

This is the equation of the line in point-slope form that has a y-intercept at (0, -3) and also contains the point (4, 5).

Step-by-step explanation:

Step-by-step explanation:

x = 45000 - (Discount % * 45000/100)

x = 45000 - (Discount % * 450)

Given that the selling price after adding 13% VAT is Rs. 40680, we can write:

x + (13% * x/100) = 40680

Simplifying the above equation, we get:

x = 36000

Therefore, the selling price of the TV without VAT will be x, which is Rs. 36000.

So the final answer is that the selling price of the TV without VAT is Rs. 36000

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Answer : The answer is B= -12

Answer:

21

Step-by-step explanation:

[tex]\frac{x}{6}=\frac{7}{2} \\x=21[/tex]

Answer:

------------------------------

Let the number of glasses be g.

Then g glasses can be filled with 3 1/2 bottles of water.

Set up equation to represent this and solve for g:

The value of the 7th term of the geometric progression is 223.9488

Based on the information given,

First term = 75

Second term = 90

Common ratio = 90/75 = 1.2

The nth term of a geometric progression is calculated as:

[tex]a_n=ar^{r-1}[/tex]

The 7th term will then be:

[tex]= 75 \times (1.2)^6[/tex]

[tex]=75\times2.985984[/tex]

[tex]=223.9488[/tex]

Therefore, the value of the 7th term of the geometric progression is 223.9488.

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The missing value in the table for x = –1 is Y=6 is the value that x=-1 lacks.

An equation is a mathematical statement that two expressions are equal. It consists of two expressions that contain numbers, variables, and/or operators, and is written in a symbolic form. Equations are used to explain a variety of real-world phenomena and also used to describe the behavior of physical systems.An equation can be used to solve for a single unknown variable or to find relationships between multiple variables.

How  to resolve the equation?

The equation's unknown value can be obtained by inserting known values for them.

Provided is the equation y=2-4x.

discover y when x=-1

Replace x=-1 in the given equation,

We obtain y=2-4 (-1)

y=6

Therefore, the value of y at x=-1 is 6.

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Complete questions as follows-

The table represents the equation y = 2 – 4x The x-values are the The y-values are the The missing value in the table for x = –1 is y =

USE MATH   WAY IT HELPS ALOT

When x = 3, y = 60 + 20(3) = 120. So, we can plot the point (3, 120).

An equation is a mathematical statement that uses symbols, numbers, and operations to show that two expressions are equal. It usually contains one or more variables, which represent unknown values that need to be solved for. Equations are used in a wide range of mathematical fields, such as algebra, geometry, calculus, and physics, to model real-world situations and solve problems. Some common examples of equations include:

Linear equation: y = mx + b

he total expenses for Myra's phone plan include the initial cost of buying the phone plus the monthly cost of the phone plan. Therefore, the total cost can be represented by the equation:

y = 60 + 20x

where y is the total cost and x is the number of months on the phone plan.

To graph this equation, we can plot a point at (0, 60) to represent the initial cost of buying the phone. Then, we can plot additional points by plugging in different values of x into the equation and finding the corresponding values of y. For example:

When x = 1, y = 60 + 20(1) = 80. So, we can plot the point (1, 80).

When x = 2, y = 60 + 20(2) = 100. So, we can plot the point (2, 100).

When x = 3, y = 60 + 20(3) = 120. So, we can plot the point (3, 120).

Connecting these points gives us a straight line, since the equation is linear.

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Check the picture below.

The probability that Babby gets the same number of heads that Abby gets is 1/8 + 1/8 = 1/4. To solve this problem, we need to consider all the possible outcomes of Abby and Babby's coin tosses.

There are four possible outcomes for Abby's toss: heads or tails, and two outcomes for Babby's toss: heads or tails. Therefore, there are a total of 4 x 4 = 16 possible outcomes for the two coin tosses.

To determine the probability that Babby gets the same number of heads as Abby, we need to identify the outcomes where this happens. Abby can get either 0 or 1 heads, so we need to consider the outcomes where Babby gets 0 or 1 heads as well.

The possible outcomes where Abby gets 0 heads are:

Abby: tails, Babby: tails tails

Abby: tails, Babby: heads tails

In both of these outcomes, Babby also gets 0 heads, so the probability of this happening is 2/16 = 1/8.

The possible outcomes where Abby gets 1 head are:

Abby: heads, Babby: tails heads or heads tails

Abby: tails, Babby: heads heads

In two of these outcomes, Babby also gets 1 head, so the probability of this happening is 2/16 = 1/8. Therefore, the probability that Babby gets the same number of heads that Abby gets is 1/8 + 1/8 = 1/4.

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The equation would be: ΔT = 2 - (-12) = 14 degrees Celsius, ΔT = T2 - T1, where ΔT is the change in temperature, T2 is the final temperature (-12 degrees Celsius)

A. d = rt, where d = 12 meters, r = 2 meters per second (rate), and t is the time it takes for the whale to get to an elevation of 12 meters. The equation would be:

12 = 2t

B. r = d/t, where r is the rate of diving, d = 12 meters (depth), and t = 2 minutes (time). The equation would be:

r = 12/(2 x 60) = 0.1 meters per second

C. ΔT = T2 - T1, where ΔT is the change in temperature, T2 is the final temperature (2 degrees Celsius), and T1 is the initial temperature (-12 degrees Celsius). The equation would be:

ΔT = 2 - (-12) = 14 degrees Celsius

D. ΔT = T2 - T1, where ΔT is the change in temperature, T2 is the final temperature (-12 degrees Celsius), and T1 is the initial temperature (2 degrees Celsius). The equation would be:

ΔT = -12 - 2 = -14 degrees Celsius

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Answer:

60 and 100 degrees.

Step-by-step explanation:

Sum of the other 2 angles = 180-20 = 160 degrees.

3 + 5 = 8

160 / 8 = 20

- so the other 2 angles are:

3*20 = 60 and

5*20 = 100 degrees.

Answer:

x < -7

Step-by-step explanation:

3(x-4)-2>5x

3x - 12 - 2 > 5x

3x - 14 > 5x

-2x > 14

x < -7

Answer: x<-7

Step-by-step explanation: 3(x-4)= 3x-12

3x-14>5x

5x-3x=2x

-14>2x

-14/2=-7

-7>x

a. Φ(t)= P.D.[tex]P^{-1}[/tex] is the fundamental matrix of the homogeneous system.

b. [tex]P^{-1}[/tex]= 1/√5 || 1 2 || || -2 1 || is the inverse

c. x(t)= Φ(t)∫ Φ(t)-1.b(t) dt + Φ(t).[tex]x_0[/tex]

d. The solution to the system is x(t)= || e-t/√5 (1-e2t)/2√5 - (2e2t -3)/2√5 || || 1/√5 e-t/√5 +2/√5 (e2t-1)/2√5 ||

In this problem, we will solve the nonhomogeneous system.

Firstly, we will write a fundamental matrix for the associated homogeneous system 1 1 2 0.

a) Fundamental matrix for the homogeneous system

The homogeneous system of equations can be written as:

X'=AX

Here, X is the vector of unknowns, and A is a matrix of constants.

For the given homogeneous system of equations, we have:

1 1 2 0X1' X2'= 1 2 X1X2

X'= AX

Thus, the matrix A can be written as:

A= 1 2 1 0

Now, let us solve the eigenvalue problem for matrix A as follows:

| A - λI |= 0

⇒| 1 - λ 2 || 1 || λ - 0 || 1 || 2 || 0 - λ || 0 |

⇒ (1-λ)(0-λ)-2⋅1=0

⇒ λ2 - λ - 2 = 0

On solving, we get:

λ1 = -1 and λ2 = 2

Thus, the eigenvectors corresponding to these eigenvalues are:

For λ1 = -1, A - (-1)I= 2 2 || x || = 0 1 || y ||

⇒2x+2y = 0y = -2x

Thus, the eigenvector corresponding to λ1 is:

x1= || 1 || and x2 = || -2 ||

The normalized eigenvector corresponding to λ1 is:

x1= || 1/√5 || and x2 = || -2/√5 ||

For λ2 = 2, A - 2I= -1 2 || x || = 0 -2 || y ||

⇒-x+2y = 0y = 1/2x

Thus, the eigenvector corresponding to λ2 is:

x1= || 2 || and x2 = || 1 ||

The normalized eigenvector corresponding to λ2 is:

x1= || 2/√5 || and x2 = || 1/√5 ||

Thus, the matrix P of eigenvectors is:

P= || 1/√5 -2/√5 || || 2/√5 1/√5 ||

Hence, the fundamental matrix of the homogeneous system is:

Φ(t)= P.D.[tex]P^{-1}[/tex]

where D= diagonal matrix of eigenvalues= || -1 0 || || 0 2 ||and [tex]P^{-1}[/tex] is the inverse of matrix P.

We will now find the inverse of matrix P.

b) Computation of inverse of matrix P

The inverse of matrix P is:

[tex]P^{-1}[/tex]= 1/√5 || 1 2 || || -2 1 ||

c) Multiplication by and integration

Multiplying Φ(t) by e

At, we have:e AtΦ(t)= PeDt.

[tex]P^{-1}[/tex].eAt= P.|| e-1t 0 || || 0 e2t ||.[tex]P^{-1}[/tex]

The solution to the system is:

x(t)= Φ(t)∫ Φ(t)-1.b(t) dt + Φ(t).[tex]x_0[/tex]

where b(t) is the vector of inhomogeneous terms, and [tex]x_0[/tex] is the initial condition.

d) Solution to the system

The nonhomogeneous system of equations can be written as:

X'=AX + b

where b= 1 [tex]e^{-t}[/tex]

The solution to the system is:

x(t)= Φ(t)∫ Φ(t)-1.b(t) dt + Φ(t).x0= P.|| e-t/√5 (e2t -1)/2√5 || || -2(e2t-1)/2√5 e-t/√5 ||.P-1 .|| 0 || || 1 ||+ P.|| 1/√5 -2/√5 || || e-t/√5 (e2t -1)/2√5 || || -2(e2t-1)/2√5 e-t/√5 ||.P-1 .|| -1/√5 || || 2/√5 ||= || 1/√5 -2/√5 || || e-t/√5 (1-e2t)/2√5 || || (e2t-3)/2√5 e-t/√5 ||.|| 0 || || 1 ||+ || 1/√5 -2/√5 || || e-t/√5 (e2t -1)/2√5 || || -2(e2t-1)/2√5 e-t/√5 ||.|| -1/√5 || || 2/√5 ||
The final answer is: x(t)= || e-t/√5 (1-e2t)/2√5 - (2e2t -3)/2√5 || || 1/√5 e-t/√5 +2/√5 (e2t-1)/2√5 ||

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Answer:

4 if x < 3

Step-by-step explanation:

First, find the equation of the line. From the graph we can tell the y-intercept is 4 and the slope is 0. Using y=mx+b, we find that y=(0)x+4 (same as y=4). f(x) is the same as y, so all you have to write is 4.

To find the domain, you must look at where the function is defined. It is only defined to the left of the point (3,4), where x < 3. We know its only defined there because the arrow and line only go in that direction. We know that it should be a < not a ≤ because the point at (3,4) is hollow, so (3,4) is not a point in the function and not included in the domain.

Answer:

Step-by-step explanation:

smelly a becasue

The estimated total population size of stickleback fish in the lake is approximately 530 individuals.

The mark and recapture method can be used to estimate the total population size of a species in a particular area. Based on the information provided, we can use the Lincoln-Petersen method to estimate the population size.

The formula for the Lincoln-Petersen method is

N = (M x R) / C

Where

N = total population size

M = number of individuals in the first sample (marked individuals)

R = number of individuals in the second sample (recaptured individuals)

C = number of marked individuals recaptured in the second sample

Using the given information, we can plug in the values to the formula

N = (255 x 162) / 78

N = 529.62

N = 530

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The equivalent expressions for the expression 9x + 5(6/5x+3)-30 are (15x-15) and 15(x-1).

Equivalent expressions are mathematical expressions that have the same value, even though they may look different.

Here are some examples:

2x + 4 and 4 + 2x are equivalent expressions.

(Proof: 2x + 4 = 4 + 2x for any value of x)

3(x + 2) and 3x + 6 are equivalent expressions.

(Proof: Distributing the 3, we get 3(x + 2) = 3x + 6)

Here we have

9x + 5(6/5x+3)-30

The above expression can be simplified as follows

=> 9x + 5(6/5x) +5(3) -30      [ Multiply 5 with (6/5x+3) ]

=> 9x + (6x) + 15 -30    

=> 15x - 15                              [ Add like terms ]

=> 15(x - 1)                              [ take 15 as common ]

Hence,

The equivalent expressions for the expression 9x + 5(6/5x+3)-30 are (15x-15) and 15(x-1).

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Nοne οf the given graphs match this descriptiοn, sο nοne οf them shοw the sοlutiοn tο the equatiοn.

Equivalent expressiοns are expressiοns that wοrk the same way despite their appearance. If twο algebraic expressiοns are equivalent, then the twο expressiοns have the same value when the variable is set tο the same value.

The equatiοn lοg Subscript 3 Baseline (x + 2) = 1 can be rewritten as 3^1 = x + 2. This simplifies tο x = 1. Therefοre, the sοlutiοn tο the equatiοn is x = 1.

On the cοοrdinate plane, this sοlutiοn is represented by a vertical line passing thrοugh x = 1.

Hence, Nοne οf the given graphs match this descriptiοn, sο nοne οf them shοw the sοlutiοn tο the equatiοn.

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Answer: A

Step-by-step explanation:

The center of the specified circle is at (-7, 4). Thus, the updated center will be (-7 - 3, 4 - 4) = (-10, 0).

In order to move the circle 3 units to the right and 4 units below, we must subtract 3 from x and 4 from y. Thus, the updated center will be (-7 - 3, 4 - 4) = (-10, 0).

The equation for the resulting circle can be discovered by adding the new center and simplifying. The standard version of the circle's equation is [tex](x - h)^2 + (y - k)^2 = r^2[/tex], with (h, k) denoting the circle's center and r denoting the radius. We can use the new center (-10, 0) and the original equation [tex](x+7)^2+(y-4)^2=36[/tex]to replace the original equation.

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