how can scientific method solve real world problems examples

Answers

Answer 1
The scientific method is nothing more than a process for discovering answers. While the name refers to “science,” this method of problem solving can be used for any type of problem

Related Questions

Four equal-value resistors are in series with a 5 V battery, and 2.23 mA are measured. What isthe value of each resistor

Answers

Answer:

560.54 Ω

Explanation:

Applying,

V = IR'............... Equation 1

Where V = Voltage of the battery, I = currrent, R' = Total resistance of the resistors

make R' the subject of the equation

R' = V/I............ Equation 2

From the question,

Given: V = 5 V, I = 2.23 mA = 2.23×10⁻³ A

Substitute these values into equation 2

R' = 5/(2.23×10⁻³ )

R' = 2242.15 Ω

Since the fours resistor are connected in series and they are equal,

Therefore the values of each resistor is

R = R'/4

R = 2242.15/4

R = 560.54 Ω

The north pole of magnet A will __?____ the south pole of magnet B

Answers

Answer:

A will attract

B will repare

Warm air rises because faster moving molecules tend to move to regions of less

A) density.
B) pressure.
C) both of these
D) none of the above

Answers

Answer:

76rsfy7zfyuutfzufyztudzutdT7dFy9y8fr6s

Explanation:

rshyyjfshfsgfshfsyhrsyhuydtufhr6ra6yris7toe7r9w7rr6w996ryrowosotusuogsuoufsutot

An ice skater with a mass of 50 kg is gliding acrossthe ice at a speed of 8 m/s when herfriend comes up from behind and gives her a push,causing her speed to increase to 12m/s. How much work did the friend do on the skater

Answers

Answer:

[tex]W=2KJ[/tex]

Explanation:

From the question we are told that:

Mass [tex]M=50kg[/tex]

Initial Velocity [tex]v_1=8m/s[/tex]

Final Velocity [tex]v_2=12m/s[/tex]

Generally the equation for Work-done is mathematically given by

W=\triangle K.E

Therefore

 [tex]W=0.5M(v_2^2-v_1^2)[/tex]

 [tex]W=0.5*50(12^2-8^2)[/tex]

 [tex]W=2KJ[/tex]

Click Stop Using the slider set the following: coeff of restitution to 1.00 A velocity (m/s) to 6.0 A mass (kg) to 6.0 B velocity (m/s) to 0.0 Calculate what range can the mass of B be to cause mass A to bounce off after the collision. Calculate what range can the mass of B be to cause mass A to continue forward after the collision. Check your calculations with the simulation. What are the ranges of B mass (kg)

Answers

Answer:

[tex]M_b=6kg[/tex]

Explanation:

From the question we are told that:

Coefficient of restitution [tex]\mu=1.00[/tex]

Mass A [tex]M_a=6kg[/tex]

Initial Velocity of A [tex]U_a=6m/s[/tex]

Initial Velocity of B [tex]U_b=0m/s[/tex]

Generally the equation for Coefficient of restitution is mathematically given by

 [tex]\mu=\frac{V_b-V_a}{U_a-U_b}[/tex]

 [tex]1=\frac{v_B}{6}[/tex]

 [tex]V_b=6*1[/tex]

 [tex]V_b=6m/s[/tex]

Generally the equation for conservation of linear momentum  is mathematically given by

 [tex]M_aU_a+M_bU_b=M_aV_a+M_bV_b[/tex]

 [tex]6*6+=M_b*6[/tex]

 [tex]M_b=6kg[/tex]

need help pleaseee,question is in the pic​

Answers

Explanation:

For engine 1,

Energy removed = 239 J

Energy added = 567 J

[tex]\eta_1=\dfrac{239}{567}\cdot100=42.15\%[/tex]

For engine 2,

Energy removed = 457 J

Energy added = 789 J

[tex]\eta_2=\dfrac{457}{789}\cdot100=57.92\%[/tex]

For engine 3,

Energy removed = 422 J

Energy added = 1038 J

[tex]\eta_3=\dfrac{422}{1038}\cdot100=40.65\%[/tex]

So, the engine 2 has the highest thermal efficiency.

1.a machine gun fires a ball with an initial velocity of 600m/s with an elevation of 30° with respect to the ground neglecting air resistance calculate:
a.the maximum height that can be reached?
b.the time of flight of the bullet?
c.the maximum horizontal displacement of the ired bullet?​

Answers

Answer:

See explanation

Explanation:

a) maximum height of a projectile = u sin^2θ/2g

H= 600 × (sin 30)^2/2 × 10

H= 7.5 m

b) Time of flight

t= 2u sinθ/g

t= 2 × 600 sin 30/10

t= 60 seconds

Range

R= u^2sin2θ/g

R= (600)^2 × sin2(30)/10

R= 31.2 m

find the weight of a body of mass 200kg on the earth at a latitude 30°.(R=6400 km ,g=9.8m/s²,ω=7.27×10⁻⁵ rad/sec)

Answers

Answer:

................ftf6x

7. The gravitational potential energy of a body depends on its A speed and position B. mass and volume. C. weight and position D.speed and mass​

Answers

Answer:

Option "D" is the correct answer to the following question.

Explanation:

The gravitational potential energy of an item is determined by its mass, elevation, and gravitational acceleration. As a result, angular momentum and energy are preserved. The gravitational potential energy, on the other hand, varies with distance. When a consequence, kinetic energy varies during each orbit, resulting in a faster speed as a planet approaches the Sun.

Answer:

SPEED AND MASS

Explanation:

TOOK THE TEST

How do you find the product of gamma decay?

Answers

Answer:

The mass and atomic numbers don't change

Explanation:

An excited atom relaxes to the ground state emitting a photon...called a gamma ray.

The answer is that the mass and atomic numbers don't change.

In gamma decay, the product refers to the nucleus resulting from the emission of a gamma ray. Gamma decay occurs when an excited atomic nucleus releases excess energy in the form of a high-energy photon called a gamma ray.

To find the product of gamma decay, you need to identify the nucleus before and after the decay process. The product nucleus is determined by the parent nucleus that undergoes gamma decay.

During gamma decay, the number of protons and neutrons in the nucleus remains unchanged. Therefore, the identity of the element remains the same, but the energy state of the nucleus is altered.

The product nucleus is typically represented by the same chemical symbol as the parent nucleus, followed by a superscript indicating the mass number (total number of protons and neutrons) and a subscript indicating the atomic number (number of protons).

For example, if a parent nucleus with an atomic number of Z and a mass number of A undergoes gamma decay, the product nucleus will have the same atomic number Z and mass number A.

It's important to note that gamma decay does not involve the emission or absorption of any particles, only the release of electromagnetic radiation (gamma ray).

Thus, the product nucleus remains unchanged in terms of atomic number and mass number.

Know more about gamma decay:

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A 1.40-kg block is on a frictionless, 30 ∘ inclined plane. The block is attached to a spring (k = 40.0 N/m ) that is fixed to a wall at the bottom of the incline. A light string attached to the block runs over a frictionless pulley to a 60.0-g suspended mass. The suspended mass is given an initial downward speed of 1.60 m/s .
How far does it drop before coming to rest? (Assume the spring is unlimited in how far it can stretch.)
Express your answer using two significant figures.

Answers

Answer:

0.5

Explanation:

because the block is attached to the pulley of the string

Two forces act on the screw eye. If F = 600 N, determine the magnitude of the resultant force and the angle θ if the resultant force is directed vertically upward.

Answers

Answer:

how to solve this problem ???????

The magnitude of the resultant force is 919.6 N and the value of angle θ is 36.87⁰.

Resultant of the two forces

The resultant of the two forces is determined by resolving the force into x and y component as shown below;

[tex]F_1_x + F_2x_x = F_R_x \ --- (1) \\\\F_1_y + F_2_y = F_R_y\ ---(2)[/tex]

where;

F1 = 500 NF2 = 600 NValue of Angle θ

The value of Angle θ is determined from equation (1)

-500sinθ + 600sin(30) = 0

500sinθ = 600sin(30)

500sinθ = 300

sinθ = 3/5

θ = 36.87⁰

Resultant of the two forces

The resultant of the forces is determined using the second equation;

500cosθ + 600cos(30) = R

500 x cos(36.87) + 600 x cos(30) = R

919.6 N = R

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A horse gallops a distance of 10 kilometers in a time of 30 minutes its average speed is?

Answers

Answer:

20 km/hr

Explanation:

Distance = 10km

Time = 30 minutes = 1/2 hour

Average Speed = Total distance / Total Time Taken

                           = 10 ÷  1/2

                           = 10 x 2

                           = 20 km/hr

Average speed = (distance covered) / (time to cover the distance)

Average speed = (10 km) / (30 minutes)

Average speed =  1/3 km/min

Most people would probably want to see it in a more convenient, more familiar unit, such as km/hour or m/second.

(10 km / 30 min) x (60 min / hour) = (10 x 60 / 30) (km-min / min-hour)

Average speed = 20 km/hour

AvgSpd = (10 km / 30 min) x (1,000 m / km) x (min / 60 sec)

AvgSpd = (10x1,000 / 30x60) (km-m-min / min-km-sec)

Averge Speed =  5.56 m/s

Of the following, which have the highest frequency in the electromagnetic
spectrum?
A. Visible light
B. Infrared waves
C. Ultraviolet rays
D. X-rays

Answers

d. X rays ..........

A satellite is launched to orbit the Earth at an altitude of 2.90 x10^7 m for use in the Global Positioning System (GPS). Take the mass of the Earth to be 5.97 x 10^24 kg and its radius 6.38 x10^6 m.

Required:
What is the orbital period of this GPS satellite?

Answers

Answer:

[tex]T=66262.4s[/tex]

Explanation:

From the question we are told that:

Altitude [tex]A=2.90 *10^7[/tex]

Mass [tex]m=5.97 * 10^{24} kg[/tex]

Radius [tex]r=6.38 *10^6 m.[/tex]

Generally the equation for Satellite Speed is mathematically given by

[tex]V=(\frac{GM}{d} )^{0.5}[/tex]

[tex]V=(\frac{6.67*10^{-11}*5.97 * 10^{24}}{6.38 *10^6+2.90 *10^7} )^{0.5}[/tex]

[tex]V=3354.83m/s[/tex]

Therefore

Period T is Given as

[tex]T=\frac{2 \pi *a}{V}[/tex]

[tex]T=\frac{2 \pi *(6.38 *10^6+2.90 *10^7}{3354.83}[/tex]

[tex]T=66262.4s[/tex]

Puck B has twice the mass of puck A. Starting from rest, both pucks are pulled the same distance across frictionless ice by strings with the same tension.a. Compare the final kinetic energies of pucks A and B. b. Compare the final speeds of pucks A and B.

Answers

Answer:

(a) 1 : 2

(b) same

Explanation:

Let the mass of puck A is m and the mass of puck B is 2 m.

initial speed for both the pucks is same as u and the distance is same for both is s.

let the tension is T for same.

The kinetic energy is given by

[tex]K = 0.5 mv^2[/tex]

(a) As the speed is same, so the kinetic energy depends on the mass.

So, kinetic energy of A : Kinetic energy of B = m : 2m  = 1 : 2

(b) A the distance s same so the final velocities are also same.

(a)  The kinetic energy of puck B is 2 times the kinetic energy of puck A.

(b)  The final speed of both the puck A and B are same.

Let the mass of puck A is m and the mass of puck B is 2 m.

Initial speed for both the pucks is same as u and the distance is same for both is s.

Let the tension is T for same.

Then, the kinetic energy is given as,

[tex]KE = \dfrac{1}{2}mv^{2}[/tex]

(a)

As the speed is same, so the kinetic energy depends on the mass.

Then,

[tex]\dfrac{KE_{A}}{KE_{B}} = \dfrac{1/2 \times mv^{2}}{1/2 \times (2m)v^{2}}\\\\\\\dfrac{KE_{A}}{KE_{B}} =\dfrac{1}{2}[/tex]

So, kinetic energy of A : Kinetic energy of B = 1 : 2.

Thus, we can conclude that the kinetic energy of puck B is 2 times the kinetic energy of puck A.

(b)

The final speed for the puck is given as,

v = s/t

here, s is the distance covered.

Since, both pucks are pulled the same distance across frictionless ice. Then, the final speed of each puck is also same.

Thus, we can conclude that the final speed of both the puck A and B are same.

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What is the length of the x-component of the vector shown below?
у
6
28°

Answers

Answer:

Explanation:

6cos28

=5.3 N

The thrust F of a screw propeller is known to depend upon the diameter d, Speed of advance v, fluid density e, revolution per second N, and the coefficient of viscosity M, of the fluid. Find the expression for F, in terms of the quantities

Answers

Answer:

[tex]{ \bf{F = { \tt{ \frac{4}{3} \pi {r}^{3}v gM}}}}[/tex]

b) Two skaters collide and grab on to each other on a frictionless ice. One of them, of mass 80 kg, is moving to the right at 5.0 m/s, while the other of mass 70 kg is moving to the left at 2.0 m/s. What are the magnitude and direction of the two skaters just after they collide

Answers

Answer:

The two skaters move with a speed of 1.73 m/s after the collision in the right direction.

Explanation:

Given that,

The mas of skater 1, m₁ = 80 kg

The speed of skater 1, u₁ = 5 m/s (right)

The mass of skater 2, m₂ = 70 kg

The speed of skater 2, u₂ = -2 m/s (left)

Let v is the magnitude of the two skaters just after they collide. They must have a common speed. So, using the conservation of momentum as follows :

[tex]m_1u_1+m_2u_2=(m_1+m_2)v\\\\v=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}[/tex]

Put all the values,

[tex]v=\dfrac{80(5)+70(-2)}{(80+70)}\\\\=1.73m /s[/tex]

So, the two skaters move with a speed of 1.73 m/s after the collision in the right direction.

A 190 g glider on a horizontal, frictionless air track is attached to a fixed ideal spring with force constant 160 N/m. At the instant you make measurements on the glider, it is moving at 0.835 m/sm/s and is 4.00 cmcm from its equilibrium point.

Required:
a. Use energy conservation to find the amplitude of the motion.
b. Use energy conservation to find the maximum speed of the glider.
c. What is the angular frequency of the oscillations?

Answers

(a) Let x be the maximum elongation of the spring. At this point, the glider would have zero velocity and thus zero kinetic energy. The total work W done by the spring on the glider to get it from the given point (4.00 cm from equilibrium) to x is

W = - (1/2 kx ² - 1/2 k (0.0400 m)²)

(note that x > 4.00 cm, and the restoring force of the spring opposes its elongation, so the total work is negative)

By the work-energy theorem, the total work is equal to the change in the glider's kinetic energy as it moves from 4.00 cm from equilibrium to x, so

W = ∆K = 0 - 1/2 m (0.835 m/s)²

Solve for x :

- (1/2 (160 N/m) x ² - 1/2 (160 N/m) (0.0400 m)²) = -1/2 (0.190 kg) (0.835 m/s)²

==>   x ≈ 0.0493 m ≈ 4.93 cm

(b) The glider attains its maximum speed at the equilibrium point. The work done by the spring as it is stretched away from equilibrium to the 4.00 cm position is

W = - 1/2 k (0.0400 m)²

If v is the glider's maximum speed, then by the work-energy theorem,

W = ∆K = 1/2 m (0.835 m/s)² - 1/2 mv ²

Solve for v :

- 1/2 (160 N/m) (0.0400 m)² = 1/2 (0.190 kg) (0.835 m/s)² - 1/2 (0.190 kg) v ²

==>   v1.43 m/s

(c) The angular frequency of the glider's oscillation is

√(k/m) = √((160 N/m) / (0.190 kg)) ≈ 29.0 Hz

The amplitude of the motion is 0.049 cm. The maximum speed of the glider is 1.429 m/s. The angular frequency of the oscillation is 29.02 rad/s

From the given information;

the mass of the glider = 190 gForce constant k = 160 N/mthe horizontal speed of the glider [tex]v_x[/tex] = 0.835 m/sthe distance away from the equilibrium = 4.0 cm = 0.04 m

Using energy conservation E, the amplitude of the motion can be calculated by using the formula:

[tex]\mathbf{E = \dfrac{1}{2}mv^2 + \dfrac{1}{2}kx^2}[/tex]

[tex]\mathbf{E = \dfrac{1}{2}(0.19 \ kg )\times (0.835)^2 + \dfrac{1}{2}(160) (0.04)^2}[/tex]

[tex]\mathbf{E =0.194 \ J}[/tex]

Similarly, we know that:

[tex]\mathbf{E = \dfrac{1}{2}kA^2}[/tex]

Making amplitude A the subject, we have:

[tex]\mathbf{A = \sqrt{\dfrac{2E}{k}}}[/tex]

[tex]\mathbf{A = \sqrt{\dfrac{2(0.194)}{160}}}[/tex]

[tex]\mathbf{A =0.049 \ cm}[/tex]

Again, using the energy conservation, the maximum speed of the glider can be calculated by using the formula:

[tex]\mathbf{E =\dfrac{1}{2} mv^2 _{max}}[/tex]

[tex]\mathbf{v _{max} = \sqrt{\dfrac{2E}{m}}}[/tex]

[tex]\mathbf{v _{max} = \sqrt{\dfrac{2\times 0.194}{0.19}}}[/tex]

[tex]\mathbf{v _{max} = 1.429 \ m/s}[/tex]

The angular frequency of the oscillation can be computed by using the expression:

[tex]\mathbf{\omega = \sqrt{\dfrac{k}{m}}}[/tex]

[tex]\mathbf{\omega = \sqrt{\dfrac{160}{0.19}}}[/tex]

ω = 29.02 rad/s

Learn more about energy conservation here:

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A
Fluids in which the shear stress must reach
certain minimum value(yield stress)
before flow commences are called

Answers

Answer:

Plastic

Explanation:

Shear Modulus can be defined as the ratio of shear stress to shear strain with respect to a physical object.

This ultimately implies that, Shear Modulus arises as a result of the application of a shear force on an object or body which eventually leads to its deformation. Thus, this phenomenon is simply used by scientists to measure or determine the rigidity of an object or body.

Fluids in which the shear stress must reach certain minimum value (yield stress) before flow commences are called plastic. Thus, a plastic would only begin to flow when its shear stress attain a certain minimum value (yield stress). The unit of measurement of yield stress is usually mega pascal (MPa).

A grade 12 Physics student shoots a basketball
from the ground at a hoop which is 2.0 m above
her release. The shot was at a velocity of 10 m/s
and at an angle of 80° to the ground.
a. Determine the vertical velocity of the ball
when it is at the level of the net. You
should get two answers.
Please show ALL steps

Answers

Answer:

7.84 m/s

Explanation:

Height, h = 2 m

Initial velocity, u = 10 m/s

Angle, A = 80°

(a) Let the time taken to go to the net is t.

Use second equation of motion

[tex]h = u t + 0.5 at^2\\\\- 2 = - 10 sin 80 t - 4.9 t^2\\\\4.9 t^2 + 9.8 t - 2 = 0 \\\\t= \frac{- 9.8\pm\sqrt{9.8^2 + 4\times 4.9\times 2}}{9.8}\\\\t = \frac{- 9.8 \pm 11.6}{9.8}\\\\t = - 2.2 s , 0.2 s[/tex]

Time cannot be negative.

So, t = 0.2 s

The vertical velocity at t = 0.2 s is

v = u + at

v = 10 sin 80 - 9.8 x0.2

v = 9.8 - 1.96 = 7.84 m/s

(a) What is the maximum frictional force (in N) in the knee joint of a person who supports 45.0 kg of her mass on that knee if the coefficient of static friction is 0.016

Answers

Answer:

f = 7.06 N

Explanation:

The maximum frictional force on the knee joint of the person can be given by the following formula:

[tex]f = \mu R = \mu W \\[/tex]

where,

f = maximum frictional force = ?

μ = static friction coefficient = 0.016

W = Weight load on knee = mg

m = mass supported by knee = 45 kg

g = acceleration due to gravity = 9.81 m/s²

Therefore,

[tex]f = \mu mg\\f = (0.016)(45\ kg)(9.81\ m/s^2)\\[/tex]

f = 7.06 N

The 1 kg box is sliding along a frictionless surface. It collides with and sticks to the 2 kg box. Afterward, the speed of the two boxes is:__________.
A) 0 m/s
B) 1 m/s
C) 2 m/s
D) 3 m/s
E) Not enough info

Answers

Answer:

The correct option is (E).

Explanation:

Given that,

Mass of object 1, m₁ = 1 kg

Mass of object 2, m₂ = 2 kg

They collides after the collision. We need to find the speed of the two boxes after the collision.

The initial speeds of both boxes is not given. So, we can't put the values of their speeds in the momentum conservation equation.

So, the information is not enough.

water contracts on freezing is it incorrect or conrrect

Answers

Answer:

hope it helps

much as you can

Drag the titles to the correct boxes to complete the pairs.

Answers

Can you input a picture??

A cylindrical container with a cross sectional area of 65.2 cm^2 holds a fluid of density 806 kg/m^3. At the bottom of the container the pressure is 116 kPa.
(a) What is the depth of the fluid?
(b) Find the pressure at the bottom of the container after an additional 2.05 X 10^-3 m^3 of this fluid is added to the container. Assume that no fluid spills out of the container.

Answers

The right answer is (b)

Three forces of magnitude 10N, 5N and 4N act on an object in the directions North, West and East respectively. Find the magnitude and directions of their resultant

Answers

Answer:

19N to the south

Explanation:

F =10N + 5N + 4N

The bulk modulus of water is B = 2.2 x 109 N/m2. What change in pressure ΔP (in atmospheres) is required to keep water from expanding when it is heated from 10.9 °C to 40.0 °C?

Answers

Answer:

A change of 160.819 atmospheres is required to keep water from expanding when it is heated from 10.9 °C to 40.0 °C.

Explanation:

The bulk modulus of water ([tex]B[/tex]), in newtons per square meters, can be estimated by means of the following model:

[tex]B = \rho_{o}\cdot \frac{\Delta P}{\rho_{f} - \rho_{o}}[/tex] (1)

Where:

[tex]\rho_{o}[/tex] - Water density at 10.9 °C, in kilograms per cubic meter.

[tex]\rho_{f}[/tex] - Water density at 40 °C, in kilograms per cubic meter.

[tex]\Delta P[/tex] - Pressure change, in pascals.

If we know that [tex]\rho_{o} = 999.623\,\frac{kg}{m^{3}}[/tex], [tex]\rho_{f} = 992.219\,\frac{kg}{m^{3}}[/tex] and [tex]B = 2.2\times 10^{9}\,\frac{N}{m^{2}}[/tex], then the bulk modulus of water is:

[tex]\Delta P = B\cdot \left(\frac{\rho_{f}}{\rho_{o}}-1 \right)[/tex]

[tex]\Delta P = \left(2.2\times 10^{9}\,\frac{N}{m^{3}} \right)\cdot \left(\frac{992.219\,\frac{kg}{m^{3}} }{999.623\,\frac{kg}{m^{3}} }-1 \right)[/tex]

[tex]\Delta P = -16294943.19\,Pa \,(-160.819\,atm)[/tex]

A change of 160.819 atmospheres is required to keep water from expanding when it is heated from 10.9 °C to 40.0 °C.

A 1,760 W toaster, a 1,420 W electric frying pan, and an 85 W lamp are plugged into the same outlet in a 15 A, 120 V circuit. (The three devices are in parallel when plugged into the same socket.) (a) What current (in A) is drawn by each device

Answers

Answer:

Toaster = I = 14.67 A

Frying Pan = 11.83 A

Lamp = 0.71 A

Explanation:

The electric power is given as:

[tex]P = VI\\\\I = \frac{P}{V}[/tex]

where,

I = current

P = Power

V = Voltage = 120 V

FOR TOASTER:

P = 1760 W

Therefore,

[tex]I = \frac{1760\ W}{120\ V}[/tex]

I = 14.67 A

FOR FRYING PAN:

P = 1420 W

Therefore,

[tex]I = \frac{1420\ W}{120\ V}[/tex]

I = 11.83 A

FOR LAMP:

P = 85 W

Therefore,

[tex]I = \frac{85\ W}{120\ V}[/tex]

I = 0.71 A

Other Questions
Answer EIGHT questions.1(a) Whai do you mean by generation of computer? Describe briely5about third and fourth generations of computer.Bmoga dele Can someone help please Find two power series solutions of the given differential equation about the ordinary point x = 0. Compare the series solutions with the solutions of the differential equation obtained using the method of Section 4.3. Try to explain any differences between the two forms of the solution. y'' y' = 0 y1 = 1 x2 2! + x4 4! x6 6! + and y2 = x x3 3! + x5 5! x7 7! + y1 = x and y2 = 1 + x + x2 2! + x3 3! + y1 = 1 + x2 2! + x4 4! + x6 6! + and y2 = x + x3 3! + x5 5! + x7 7! + y1 = 1 + x and y2 = x2 2! + x3 3! + x4 4! + x5 5! + y1 = 1 and y2 = x + x2 2! + x3 3! + x4 4! + Food Deliveries Should Be Schedule A boat goes 24 miles at a constant speed in 4 hours against a river current. It takes the boat 3 hours to go the same distance at the same speed with the current of the river. What is the speed of the boat and what is the current of the river?Boat speed: Speed of Current: Identify each of the following sentences as simple, compound, complex, or compound-complex.1. He didnt want any vegetables or rice with dinner.2. Do you want the pasta, or would you prefer the steak?3. In Paris last year we saw many attractions, including the Eiffel Tower.4. After the game on Thursday, we are going to the movies.5. After you go to the game on Thursday, come to dinner with us.6. The book that is on the shelf is yours, and you can take it whenever you want it.7. Although the cookies were burned, they tasted good.8. Jamie and Ralph called me last night and then came over for a visit. Karen is listening to a colleague's idea for reducing customer wait time at the stora Which behavior can Karen exhibit to best demonstrate that she agrees with her colleague's idea? O a) Nod her head Cross her arms in front of her chest b) C) Rest her chin in one hand O d) Rub her hands together The weekly amount of money spent on maintenance and repairs by a company was observed, over a long period of time, to be approximately normally distributed with mean $440 and standard deviation $20. How much should be budgeted for weekly repairs and maintenance so that the probability the budgeted amount will be exceeded in a given week is only 0.1 Fill in the blank: Data-inspired decision-making can discover _____ when exploring different data sources. 1 point if a decision was properly made where the largest amount of data is which experts can give advice what the data has in common what is the value of the smallest of five consecutive integers if the least minus twice the greates equals -3A. -9B. -5C. -3D. 5 identify words with positive and negative connotationinformationnewsdetailsgossip (please help me ) find the missing side lengths answered in the simplest radical form with the denominator rationalized Write a balanced nuclear reaction for one complete cycle What was the purpose of the Bill of Rights?to grant more power to the federal governmentto grant more power to the statesto guarantee and protect individual libertiesto declare that all men were created equal If Dedra has a mass of 50 kg, where would she weigh the most?SunGravity = 274 m/s2MoonGravity = 1.6 m/s2EarthGravity = 9.8 m/sa In the fall, Professor Cole remembered the names of his 44 students. Now, in the spring, he has learned the names of his 42 new students. When Professor Cole ran into one of the students from the fall class, he could not remember the student's name. He could think only of the names of students in his spring class. This is most likely due to ________. Group of answer choices Which planet is (close) . to the Sun? Find the area of the triangle. What is the most likely reason the Florida Declaration of Rights has twenty-seven sections, while the Bill of Rightshas only ten amendments?O The Florida Declaration of Rights is more up-to-date than the US Constitution, so it includes more.The Florida Declaration of Rights includes many more rights that are only necessary for FloridaO The Florida Declaration of Rights also includes things that are in other parts of the US ConstitutionO The Florida Declaration of Rights lists specific rights, while the Bill of Rights deals with bigger ideas. Michael has a project due in exactly 83 hours. It is currently 8:30 on a Monday morning. What time is his project due