You start at (-3,0). You move down 2 units and left 1 unit. Where do you end?

Answer:

Step-by-step explanation:

width:length =1:3

width:15=1:3

width =5

The width of the actual car is 45 feet.

Given that,

A automobile model is 3 feet long and 1 foot wide. Considering that the automobile is 15 feet long

To find : The width of the actual car

Model Car's Width (W) = 1 foot

Model Car's Length (L) = 3 feet

Actual Car's Length (l) = 15 feet

Actual Car's Width (w) = x feet

To solve,

We know that,

Product of Means = Product of Extremes which means ,

1 : 3 :: 15 : x

1 * x = 15 *3

x = 45.

Therefore, the Actual Car's Width (w) is 45 feet.

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Answer:

\(\fbox {True}\)

Step-by-step explanation:

Trigonometric ratios apply to only right triangles, and not to oblique triangles. There various ratios of angles less than or equal to 90 degrees is taken.

Step-by-step explanation:

hope you can understand

The final solution to the given differential equation with the given initial conditions is:

\(\( y(t) = \frac{1}{21} e^{-6t} + \frac{2}{7} e^{t} - \frac{1}{3} \)\)

To find the solution y(t)  for the given second-order ordinary differential equation with initial conditions, we can follow these steps:

Find the characteristic equation:

The characteristic equation for the given differential equation is obtained by substituting y(t) = \(e^{rt}\) into the equation, where ( r) is an unknown constant:

r² + 3r - 6 = 0

Solve the characteristic equation:

We can solve the characteristic equation by factoring or using the quadratic formula. In this case, factoring is convenient:

(r + 6)(r - 1) = 0

So we have two possible values for  r :

\(\( r_1 = -6 \) and \( r_2 = 1 \)\)

Step 3: Find the homogeneous solution:

The homogeneous solution is given by:

\(\( y_h(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} \)\)

where \(\( C_1 \) and \( C_2 \)\) are arbitrary constants.

Step 4: Find the particular solution:

To find the particular solution, we assume that y(t) can be expressed as a linear combination of t and a constant term. Let's assume:

\(\( y_p(t) = A t + B \)\)

where \( A \) and \( B \) are constants to be determined.

Taking the derivatives of\(\( y_p(t) \)\):

\(\( y_p'(t) = A \)\)(derivative of  t  is 1, derivative of B is 0)

\(\( y_p''(t) = 0 \)\)(derivative of a constant is 0)

Substituting these derivatives into the original differential equation:

\(\( y_p''(t) + 3t y_p(t) - 6y_p(t) - 2 = 0 \)\( 0 + 3t(A t + B) - 6(A t + B) - 2 = 0 \)\)

Simplifying the equation:

\(\( 3A t² + (3B - 6A)t - 6B - 2 = 0 \)\)

Comparing the coefficients of the powers of \( t \), we get the following equations:

3A = 0  (coefficient of t² term)

3B - 6A = 0 (coefficient of t term)

-6B - 2 = 0 (constant term)

From the first equation, we find that A = 0 .

From the third equation, we find that \(\( B = -\frac{1}{3} \).\)

Therefore, the particular solution is:

\(\( y_p(t) = -\frac{1}{3} \)\)

Step 5: Find the complete solution:

The complete solution is given by the sum of the homogeneous and particular solutions:

\(\( y(t) = y_h(t) + y_p(t) \)\( y(t) = C_1 e^{-6t} + C_2 e^{t} - \frac{1}{3} \)\)

Step 6: Apply the initial conditions:

Using the initial conditions \(\( y(0) = 0 \) and \( y'(0) = 0 \),\) we can solve for the constants \(\( C_1 \) and \( C_2 \).\)

\(\( y(0) = C_1 e^{-6(0)} + C_2 e^{0} - \frac{1}{3} = 0 \)\)

\(\( C_1 + C_2 - \frac{1}{3} = 0 \)     (equation 1)\( y'(t) = -6C_1 e^{-6t} + C_2 e^{t} \)\( y'(0) = -6C_1 e^{-6(0)} + C_2 e^{0} = 0 \)\( -6C_1 + C_2 = 0 \)\)     (equation 2)

Solving equations 1 and 2 simultaneously, we can find the values of\(\( C_1 \) and \( C_2 \).\)

From equation 2, we have \(\( C_2 = 6C_1 \).\)

Substituting this into equation 1, we get:

\(\( C_1 + 6C_1 - \frac{1}{3} = 0 \)\( 7C_1 = \frac{1}{3} \)\( C_1 = \frac{1}{21} \)\)

Substituting \(\( C_1 = \frac{1}{21} \)\) into equation 2, we get:

\(\( C_2 = 6 \left( \frac{1}{21} \right) = \frac{2}{7} \)\)

Therefore, the final solution to the given differential equation with the given initial conditions is:

\(\( y(t) = \frac{1}{21} e^{-6t} + \frac{2}{7} e^{t} - \frac{1}{3} \)\)

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Answer:

a) 15 b) 2.8 c) 900 d) 125

Answer:

13r-13n

Step-by-step explanation:

just add 9r+6r get 15r then subtract it by 2r then you get 13r and you can't subtract unlike terms

Answer:

13r-13n

Step-by-step explanation:

6r+9r = 15r

15r-2r = 13r

13r-13n

Answer:

The item that attaches to the bowstring to improve accuracy by giving a consistent anchor point each time the bowstring is drawn is called a "release aid" or "bow release".

It is correct to say that any point on the perpendicular bisector of a segment is equidistant from the endpoints of the segment.

Any point on the perpendicular bisector of a segment is equidistant from the endpoints of the segment. What is a perpendicular bisector? A perpendicular bisector is a line that intersects a line segment and forms a 90-degree angle. It divides the line segment into two equal halves.Each point on the perpendicular bisector of a line segment is equidistant from the endpoints of the segment. This means that the distance from any point on the line to one endpoint of the segment is the same as the distance to the other endpoint of the segment.

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Answer: y = -(1/2)x - 1

Step-by-step explanation:

y=2x+3

This is written in the format of y = mx + b, where m is the slope and b the y-intercept.  The slope of this reference line is 2.  Perpendicular lines will have a slope that is the negative inverse of the slope of the reference line.  That means out new line will have a slope of -(1/2):

y = -(1/2)x + b

We can find b by entering the x and y for the given point (-4,1) and solve for b:

y = -(1/2)x + b

1 = -(1/2)(-4) + b

1 = 2 + b

b = -1

y = -(1/2)x - 1

This equation is perpendicular to y=2x+3 and goes through point (-4,1).  See attachment.

Answer:

its B

Step-by-step explanation:

Answer:

y= x+75

Step-by-step explanation:

We already have 75 and add x money

Answer:

homework hbu?

Step-by-step explanation:

have a nice day and stay safe:)

Answer:

1, 2 and 3  (I, II and III)

Step-by-step explanation:

Since the slope is positive (3) in the equation y = 3x + 1, it means that the graph has a positive slope, meaning the line slopes up from left to right.

The y intercept is 1

the x intercept is:  

0 = 3x + 1

x = 1/3

Therefore, the graph of y = 3x + 1 lies in quadrants I, II and III.  Graph the equation to prove this.  

Answer:

0.38

Step-by-step explanation:

0×x=0.3 bexuse the angle need works

Iterative is used when the equations cannot be solved simply.

The given value is put in the equation to solve.

Walter estimation is incorrect.

No, both 3. 61^2 and 3. 62^2 are greater than 13.

First he checks that is 13 between 3.61 and 3.62.

But suppose he cannot because he wants to use iteration.

Suppose he squares both the numbers 3.61 and 3.62 to determine if 13 is between the two numbers.

3.61² = 13.0321  and

3.62² =13.1044

We see that when both the numbers are squared then the answer in each case is greater than 13.

Therefore 13 cannot lie as both numbers are greater than 13.

Hence 13 does not lie between 3.61 and 3.62

From the given choices the best choice is  No, both 3. 61^2 and 3. 62^2 are greater than 13.

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Answer: Situation:

Step-by-step explanation:

You are walking in a grocery store and you see a nice deal saying that if you buy 4 batches of 2 bananas, you will get 5 extra bananas free of charge. However you see in another store that for the same price, you can buy 12 batches of 2 bananas but 11 of them are always bad and are needed to be thrown out. You calculate which deal is better and you decide on the first one because you hate stinking bananas. Hope this helps!

:)

Answer:

AS = 46

Step-by-step explanation:

The circumcentre is equally distant from the triangle' s 3 vertices , then

AS = BS = 46

The current value of Certificate A is approximately $4322.41, and the current value of Certificate B is approximately $4319.03.

For Certificate A the first payment of $2200 is received in three months. The second payment of $2200 is received in six months.

PV = CF / (1 + r)ⁿ

PV = Present Value

CF = Cash Flow

r = Rate of return

n = Number of periods

n = 3/12

n = 0.25

PV₁ = \(\frac{2200}{(1+0.455)^{0.25}}\)

PV₁= 2200 / 1.011349

PV₁ = 2174.24

n = 6/12

n = 0.5

PV₂ =  \(\frac{2200}{(1+0.455)^{0.5}}\)

PV₂ = 2200 / 1.022595

PV₂ = 2148.17

Therefore, the current value of Certificate A is approximately $2174.24 + $2148.17 = $4322.41.

For Certificate B the first payment of $2200 is received in four months. The second payment of $2200 is received in seven months.

PV₁ =  \(\frac{2200}{(1+0.455)^{\frac{6}{12} } }\)

PV₁ = 2200 / 1.014674

PV₁ = 2161.40

PV₂ = \(\frac{2200}{(1+0.455)^{\frac{7}{12} } }\)

PV₂ = 2200 / 1.018202

PV₂ = 2158.63

Therefore, the current value of Certificate B is approximately $2161.40 + $2158.63 = $4319.03.

The current value of Certificate A is approximately $4322.41, and the current value of Certificate B is approximately $4319.03.

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A congruent rectangle is a rectangle that has the same size and shape as another rectangle. In other words, two rectangles are congruent if they have the same length and width.

Since the rectangle is 4 inches by 7 inches, one of the dimensions represents the length and the other represents the width. The length and the width can be interchanged to form a different rectangle with the same area.

Since the rectangle is congruent to the 4-inch by 7-inch rectangle, the length and width of the rectangle are either 4 inches or 7 inches. Therefore, the length of segment JM can be either 4 inches or 7 inches.

So, there are two possible values for the length of segment JM: 4 inches and 7 inches.

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In the end, Moe played 51 games, while  Meeny played 17 games and Miny played 35 games.

We can start solving the problem by using algebra. Let x be the number of games Moe played. Then, since Meeny and Miny played 17 and 35 games respectively, the total number of games played is 17 + 35 + x = x + 52.

We know that in the second game and on, the one who sat out the preceding game would replace the loser of that game. Therefore, the number of games played by Moe is one less than the number of games played by Meeny and Miny combined.

x = (17 + 35) - 1 = 51.

So, Moe played 51 games.

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The complete question is -

Meeny, Miny and Moe were playing tennis. From the second game on, the one who sat out the preceding game would replace the loser of that game. In the end, Meeny played 17 games and Miny played 35 games. How many games did Moe play?

Answer:

I believe that the error is that sqrt(-2) should be equal to sqrt(2) * i

Step-by-step explanation:

The rest of the steps all use proper algebraic rules, but the square root of a*b = sqrt(a)*sqrt(b)

The difference in the dollar value of Assistantship (Stipend) between art and science fields with  standard errors is equal to 1.214.

Mean      24041.62203                 25952.36501

Sample mean difference between Assistantship (stipend) in arts and science is equal to

=  $25952.36501 - $24041.62203

= $1910.74298.

Hypothesized mean difference is 0 (there is no difference in stipend between the two fields).

Variance        621483.0801            615193.5853

Sample size    521                             479

Standard error of the difference is,

=√[(variance in arts / sample size in arts) + (variance in science / sample size in science)]

= √[(615193.5853 / 479) + (621483.0801 / 521)]

= $49.77

t-statistic

= (sample mean difference - hypothesized mean difference) / standard error of the difference

Substitute the values into the t-statistic formula,

⇒ t-statistic = ($1910.74298 - 0) / $49.77

⇒ t-statistic = 38.39

t-critical value for a two-tailed test with 992 degrees of freedom at a significance level of 0.05 is 1.9624.

t-statistic (38.39) > t-critical value (1.9624)

⇒Reject the null hypothesis that there is no difference in stipend between the two fields.

standard errors

= t-statistic / √(sample size)

= 38.39 / √(479 + 521)

= 38.39 /31.62

=1.214

Therefore,  the difference in the dollar value of Assistantship (Stipend) between these two fields is 1.214 standard errors away from the hypothesized difference.

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Answer:

Step-by-step explanation:

Area of the square = 6^2 = 36 cm^2

radius = 6 : 2 = 3 cm

Area of the four semicircles = 3^2 * π * 2 = 18 π cm^2

Total area in terms of π= 36 + 18π = 18(2+π) cm^2

Total area = 18(5.141593) = 92.548668 = 92.5 cm^2

The side resistance of the 20 ft long open-ended 27-inch diameter smooth steel pipe pile driven in sand with a friction angle of 35 degrees, using the beta method, is X pounds.

To calculate the side resistance of the steel pipe pile, we can use the beta method, which considers the soil properties and geometry of the pile. In this case, we have a 20 ft long pile with an open end and a diameter of 27 inches, driven into sand with a friction angle of 35 degrees. We are assuming that the water table is at the ground surface, and the unit weight of the soil is 126 pounds per cubic foot.

The beta method involves calculating the skin friction along the pile shaft based on the effective stress and the soil properties. In sandy soils, the side resistance is typically estimated using the formula:

Rs = beta * N * σ'v * Ap

Where:

Rs = Side resistance

beta = Empirical coefficient (dependent on soil type and pile geometry)

N = Number of times the pile diameter

σ'v = Effective vertical stress

Ap = Perimeter of the pile shaft

The value of beta can vary depending on the soil conditions and is typically determined from empirical correlations. For this calculation, we'll assume a beta value based on previous studies or available literature.

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Answer:

40.9

Step-by-step explanation:

In ΔRST, the measure of ∠T=90°, ST = 71 feet, and RS = 94 feet. Find the measure of ∠S to the nearest tenth of a degree.

The rate of constant for the above-given first-order reaction is b.) 8.77 × 10–3 h–1.

The half-life of a first-order reaction can be calculated using the equation t1/2 = ln(2)/k, where t1/2 is the half-life and k is the rate constant.

In this case, we know that the reaction goes to half completion in 79 hours. Therefore, the half-life is also 79 hours.

Plugging this into the equation, we get:

79 = ln(2)/k

Solving for k, we get:

k = ln(2)/79

Using a calculator, we can evaluate this expression to get:

k = 8.77 × 10–3 h–1

Therefore, the correct answer is b) 8.77 × 10–3 h–1.

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Answer:

55 days

Step-by-step explanation:

Given

Jim ran 15 miles in 5 days

no. of miles ran in 5 days = 15 miles

dividing LHS and RHS by 5

no. of miles ran in 5/5(=1) days = 15/5 miles = 3 miles

no. of miles ran in 1 day =  3 miles

let the no. of days taken to run 165 miles be x   ----A

No of miles ran in x days = x*no. of miles ran in 1 day = 3x miles

thus,  From A

3x  = 165

x = 165/3 = 55

Thus, it took 55 days for JIM to run 165 miles

Answer:

B: the accuracy rate of most people is barely above the 50% guessing level

Step-by-step explanation:

Bond and DePaul carried out a research and analysis on the accuracy of deception judgments in 2016.

This research involved synthesizing of results gotten from 206 research documents as well as a total of 24483 judges.

In this research and analysis, there were a couple of discoveries when people tried to discriminate lies from the truth and these discoveries include; average percentage of correct lie-truth judgments, percentage of those who correctly classified lies as deception and also percentage of those who classified truths as non-deceptive e.t.c. In conclusion, part of their discoveries was that the accuracy rate of most people is barely above the 50% guessing level.

This corresponds to option B.