Express 12/7 as a mixed number

This acquisition of used buses reflects Sheffield Transportation's commitment to providing reliable and cost-effective transportation options for their customers.

In recent years, Sheffield Transportation has acquired three used buses for their transportation fleet. The decision to purchase these buses was likely made to meet the growing demand for transportation services in the area, while also keeping costs down by opting for used vehicles instead of new ones.

It is important to note that when purchasing used buses, Sheffield Transportation would have had to ensure that the vehicles were in good condition and met safety standards before putting them into service. Overall, this acquisition of used buses reflects Sheffield Transportation's commitment to providing reliable and cost-effective transportation options for their customers.

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The angles of a right angled triangle with acute angles of ratio 4:5 are as follows:

40 degrees = 0.698132 radian

50 degrees = 0.872665 radians

90 degrees = 1.5708 radians

Right angle triangle has one of its angle as 90 degrees. Therefore, the sum of the other two angles are less than 90 degrees.

The acute angle are in the ratio 4:5. Therefore,

The other angle = 90 - 40 = 50 degrees.

The angles in radian are as follows:

40 degrees = 0.698132 radian

50 degrees = 0.872665 radians

90 degrees = 1.5708 radians

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Both angle K and angle L must be acute angles, measuring less than 90 degrees, in order to satisfy the conditions of the given triangle.

In triangle JKL, if angle J is less than 90 degrees, then angle K and angle L are both acute angles.

An acute angle is defined as an angle that measures less than 90 degrees. Since angle J is given to be less than 90 degrees, it is an acute angle.

In a triangle, the sum of the interior angles is always 180 degrees. Therefore, if angle J is less than 90 degrees, the sum of angles K and L must be greater than 90 degrees in order to satisfy the condition that the angles of a triangle add up to 180 degrees.

Hence, both angle K and angle L must be acute angles, measuring less than 90 degrees, in order to satisfy the conditions of the given triangle.

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Answer:

2 solutions

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Algebra I

Standard Form: ax² + bx + c = 0

Discriminant: b² - 4ac

Step-by-step explanation:

Step 1: Define

-2x² + 3 = 0

Step 2: Identify

a = -2

b = 0

c = 3

Step 3: Find

Answer:

NO REAL SOLUTIONS

Step-by-step explanation:

a=2 b=0 c=3

0-4(2)(3)=0-24=-24. Since the discriminant is negative, there are NO REAL SOLUTIONS.

Answer:

D.

Step-by-step explanation:

It’s the only option that makes sense. Option A. Is invalid because they need to make $350. Option B. Is just purely invalid and has no context to support the theory. C. Is possible, but we can have students wash more than one car, so we don’t need fifty four kids.

D. Is your answer.

Hope this helped!!

LunarRose3

7x + 3 = 9x - 3 - 2x

7x - 9x + 2x = -3 - 3

0 = -6

equation doesn't have any solution

Anwser:

2  X=2

Step-by-step explanation:

Because I am right!

Recall that variation of parameters is used to solve second-order ODEs of the form

y''(t) + p(t) y'(t) + q(t) y(t) = f(t)

so the first thing you need to do is divide both sides of your equation by t :

y'' + (2t - 1)/t y' - 2/t y = 7t

You're looking for a solution of the form

\(y=y_1u_1+y_2u_2\)

where

\(u_1(t)=\displaystyle-\int\frac{y_2(t)f(t)}{W(y_1,y_2)}\,\mathrm dt\)

\(u_2(t)=\displaystyle\int\frac{y_1(t)f(t)}{W(y_1,y_2)}\,\mathrm dt\)

and W denotes the Wronskian determinant.

Compute the Wronskian:

\(W(y_1,y_2) = W\left(2t-1,e^{-2t}\right) = \begin{vmatrix}2t-1&e^{-2t}\\2&-2e^{-2t}\end{vmatrix} = -4te^{-2t}\)

Then

\(u_1=\displaystyle-\int\frac{7te^{-2t}}{-4te^{-2t}}\,\mathrm dt=\frac74\int\mathrm dt = \frac74t\)

\(u_2=\displaystyle\int\frac{7t(2t-1)}{-4te^{-2t}}\,\mathrm dt=-\frac74\int(2t-1)e^{2t}\,\mathrm dt=-\frac74(t-1)e^{2t}\)

The general solution to the ODE is

\(y = C_1(2t-1) + C_2e^{-2t} + \dfrac74t(2t-1) - \dfrac74(t-1)e^{2t}e^{-2t}\)

which simplifies somewhat to

\(\boxed{y = C_1(2t-1) + C_2e^{-2t} + \dfrac74(2t^2-2t+1)}\)

Answer:

0.75

Step-by-step explanation:

Answer:

$1.33 per ticket cost

Step-by-step explanation:

5 / $3.75 = $1.33 per 1 ticket cost

Answer:

15/2

Step-by-step explanation:

Answer:

c? I think

Step-by-step explanation:

sorry if i'm wrong

Answer:

Its D

Step-by-step explanation:

Answer: m<1 = 163°

Step-by-step explanation: -73° + 163° = 90°

Answer:91.875

Step-by-step explanation:

A pound contains 16 ounces

1 gram contains 1000mg

1.47*1000=1470

Which means 1 pound contains 1470 mg

1 ounce contains 1470/16 =91.875 mg

1/5

Step-by-step explanation:

(2,6) = 4/20 = 1/5

answer is 1/5

The weight of the 0.8-meter piece is 19.6 kilograms.

We can use the ratio of length to weight to determine the weight of the 0.8-meter piece.

Let's call the weight of the 2.8-meter piece "W₁" and the weight of the 0.8-meter piece "W₂". Then we have:

W₁/2.8m = 24.5kg/1m

Solving for W₁, we get:

W₁ = (24.5kg/1m) x 2.8m = 68.6kg

Now we can use the same ratio to find W₂:

W₂/0.8m = 24.5kg/1m

Solving for W₂, we get:

W₂ = (24.5kg/1m) x 0.8m = 19.6kg

Therefore, the weight of the 0.8-meter piece is 19.6 kilograms.

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Answer:

Points: (1,-1) (4,-1) (4,-3)

Step-by-step explanation:

Hope this helps

Answer:

(1,-1) (4,-1) (4,-3)

Step-by-step explanation:

bye have a good day

Answer:

45∘

Step-by-step explanation:

∠2 = 180∘ - 135∘ = 45∘

∠2 = ∠1 = 45∘

∠1 = ∠4 = 45∘

\(\\ \rm\Rrightarrow \sqrt{36x^{21}y^{64}}=6y^8x^{10.5}\)

\(\\ \rm\Rrightarrow \sqrt[15]{7^{10}a^{25}b^{-35}}=7^{2/3}a^5b^{-7}{3}\)

\(\\ \rm\Rrightarrow \sqrt[3]{648(3m-2n)^{64}}=648^{1/3}(3m-2n)^4\)

\(\\ \rm\Rrightarrow \sqrt[4]{(a^2-2a+1)^{29}(a^2-1)^7}=(a^2-a+1)^{29/4}(a^2-1)^{7/4}\)

\(\\ \rm\Rrightarrow \dfrac{3a^2b}{\sqrt[3]{18a^{21}b^{17}}}=\dfrac{3a^2b}{18^{1/3}a^7b^{17/3}}=\dfrac{3a^{-5}b^{-14/3}}{18^{1/3}}=\dfrac{3}{18^{1/3}a^5b^{14/3}}\)

The result of adding -4 (row 1) to row 3 is determined as (0  - 9   2)|12.

The resultant of the row multiplication in the Matrice is calculated by applying the following method;

row 1 in the given matrices = [1  2  1] | -5

To multiply row by -4, we will multiply each entity by 4 as shown below;

= -4(1 2  1) | -5

= (-4  -8  - 4) |20

To add the result to 3;

(-4  -8  - 4)|20 + (4  - 1   6) | -8

= (0  - 9   2)|12

Thus, the result of the row multiplication is determined by multiplying each entry in row 1, by - 4.

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The binomial probability of x successes is 0.00928

Since we are dealing with a binomial probability distribution. We are going to use the binomial distribution formula for determining the probability of x successes:

P(x = r) = nCr . p^r  . q^n-r

Given: n = 30, p = 1/5 = 0.2, x = 1 (single trial)

The failures can be calculated using: q = 1 - p = 1 - 0.2 = 0.8

P(x = 1) =  30C1 x 0.2^1  x 0.8^30-1

              =  30C1 x 0.2^1  x 0.8^29

              = 30!/(30-1)! 1!  x 0.2 x 0.8^29

              =  30x29!/29!  x  0.2 x 0.8^29

             =  30 x  0.2 x 0.8^29

             = 0.00928

Therefore, the probability of x successes on a single trial is 0.00928

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(4x³- 4 + 7x) - (2x³ - x - 8) is equivalent to expression 2x³ + 8x + 4. Option B

(-3x² + x⁴ + x) + (2x⁴ - 7+ 4x)  is equivalent to expression 3x⁴ - 3x² + 5x - 7. Option D

(x² - 2x) (2x + 3) is equivalent to expression 2x³ - x² - 6x. Option A

From the information given, we have that the expressions are;

(4x³- 4 + 7x) - (2x³ - x - 8)

expand the bracket, we have;

4x³ - 4 + 7x - 2x³ + x + 8

collect the like terms, we have;

2x³ + 8x + 4

(-3x² + x⁴ + x) + (2x⁴ - 7+ 4x)

expand the bracket, we have;

-3x² + x⁴ + x + 2x⁴ - 7+ 4x

collect the like terms

3x⁴ - 3x² + 5x - 7

(x² - 2x) (2x + 3)

expand

2x³ + 3x² - 4x² - 6x

2x³ - x² - 6x

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A linear function rule to model the distance in feet {d} the car has traveled any number of seconds {t} after passing the timing device is

d = 92t.

Given is that a car moving at a constant speed passed a timing device at t = 0 s. After 9s, the car has traveled 828 ft.

We can write the linear function as -

y = mx

Now, we can write the slope as -

{m} = (828/9) = 92

So -

d = 92t

Therefore, a linear function rule to model the distance in feet {d} the car has traveled any number of seconds {t} after passing the timing device is

d = 92t.

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Answer:

9 hours > 450 minutes

Step-by-step explanation:

An hour is a period of 60 minutes. 9 times 60 (9 hours) is 540 minutes.

540 > 450 minutes.

9 hours is more than 450 minutes.

9 hours > 450 minutes

Answer:

9 hours > 450 minutes

Step-by-step explanation:

We want to compare 9 hours with 450 minutes.

Let's first convert them to the same units of time: minutes.

There are 60 minutes in an hour, so there are 60 * 9 = 540 minutes in 9 hours.

Clearly, 540 is greater than 450, so we would use > to show that:

9 hours > 450 minutes

~ an aesthetics lover

The point where the line 2x + 3y = 6 intersects the x-axis is (3, 0). This means that when x is equal to 3, y is equal to 0.

To graph the equation 2x + 3y = 6, we can rewrite it in the slope-intercept form, y = mx + b, where m represents the slope and b represents the y-intercept.

Starting with the given equation, we isolate y to one side:

3y = -2x + 6

y = (-2/3)x + 2

Now, we have the equation in slope-intercept form, y = (-2/3)x + 2. The slope is -2/3, and the y-intercept is (0, 2).

To find the point where the graph intersects the x-axis, we need to determine the coordinates where y is equal to zero. This occurs when the line crosses the x-axis.

Setting y = 0 in the equation, we have:

0 = (-2/3)x + 2

(-2/3)x = -2

x = (-2)(-3/2) = 3

Therefore, the point where the line 2x + 3y = 6 intersects the x-axis is (3, 0). This means that when x is equal to 3, y is equal to 0, indicating the point of intersection with the x-axis on the graph.

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Suppose H (x)=(2x+4)⁶. Find two functions ƒ and g such that (f°g)(x) = H (x). One of the possible solution is

f(x) = (2x + 4)^3

g(x) = x^2

(f°g)(x) = f(g(x)) = f(x^2) = (2x^2 + 4)^3

One possible solution is to set f(x) = (2x + 4)^3 and g(x) = x^2.

f(x) = (2x + 4)^3

g(x) = x^2

Then, (f°g)(x) = f(g(x)) = f(x^2) = (2x^2 + 4)^3

And, (2x^2 + 4)^3 = (2(x^2) + 4)^3 = (2x^2 + 4)^3 = H(x).

Note that this is just one of the possible solutions. There could be other functions ƒ and g that would satisfy the equation (f°g)(x) = H (x).

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Answer:

Answer:

1. B. (x^2 + 4x)(x)

2. A. (x^3+4x^2)

3. 9

4. 0

5. 1

Step-by-step explanation:

Answer and Explanation:

Please find full question attached

From the bar chart, we can see that the the interval 130-150 has two bars in it:

Interval 130-140 with the highest frequency of 5

Interval 140-150 with frequency of 2

Therefore to find the percentage between 130-150 we add up the frequencies 2+5=7 and divide by the total frequency of 15 people all multiplied by 100

=7/15×100= 46.67%

Percentage of participants with heart rate between 130 and 150 =46.67%

The y-coordinate of -3f(x-1) when x = 0 is -63

From the question, we have the following parameters that can be used in our computation:

The table of values

So, we have

-3f(x - 1)

When x = 0, we have

y = -3f(0 - 1)

This gives

y = -3f(-1)

From the table, we have

y = -3 * 21

Evaluate

y = -63

Hence, the y-coordinate is -63

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Answer:

(5, 8)

Step-by-step explanation:

(5. – 5) down 3 units => (5, -8)

reflect x-axis => (5 , 8)