is the sum of two polynomials of degree 5 always a polynomial of degree 5? Give an example to explain your answer

Ben and Josh went to the roof of their 40-foot tall high school to throw their math books offthe edge.The initial velocity of Be

Taya2010 [7]

Answer

Josh's textbook reached the ground first

Josh's textbook reached the ground first by a difference of $t=0.6482$

Step-by-step explanation:

Before we proceed let us re write correctly the height equation which in correct form reads:

$h(t)=-16t^2 +v_{o}t+s$ Eqn(1).

Where:

$h(t)$ : is the height range as a function of time

$v_{o}$ : is the initial velocity

$s$ : is the initial heightin feet and is given as 40 feet, thus Eqn(1). becomes:

$h(t)=-16t^2 + v_{o}t + 40$ Eqn(2).

Now let us use the given information and set up our equations for Ben and Josh.

Ben:

We know that $v_{o}=60ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+60t+40$ Eqn.(3)

Josh:

We know that $v_{o}=48ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+48t+40$ Eqn. (4).

Now since we want to find whose textbook reaches the ground first and by how many seconds we need to solve each equation (i.e. Eqns. (3) and (4)) at $h(t)=0$ . Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one.

Thus we have for Ben, Eqn. (3) gives:

$h(t)=0-16t^2+60t+40=0$

Using the quadratic expression to find the roots of the quadratic we have:

$t_{1,2}=\frac{-b+/-\sqrt{b^2-4ac} }{2a} \\t_{1,2}=\frac{-60+/-\sqrt{60^2-4(-16)(40)} }{2(-16)} \\t_{1,2}=\frac{-60+/-\sqrt{6160} }{-32} \\t_{1,2}=\frac{15+/-\sqrt{385} }{8}\\\\t_{1}=4.3276 sec\\t_{2}=-0.5776 sec$

Since time can only be positive we reject the $t_{2}$ solution and we keep that Ben's book took $t=4.3276$ seconds to reach the ground.

Similarly solving for Josh we obtain

$t_{1}=3.6794sec\\t_{2}=-0.6794sec$

Thus again we reject the negative and keep the positive solution, so Josh's book took $t=3.6794$ seconds to reach the ground.

Then we can find the difference between Ben and Josh times as

$t_{Ben}-t_{Josh}= 4.3276 - 3.6794 = 0.6482$

So to answer the original question:

Whose textbook reaches the ground first and by how many seconds?

Josh's textbook reached the ground first
Josh's textbook reached the ground first by a difference of $t=0.6482$

3 0

1 year ago

A sphere has a diameter of 14 ft. Which equation finds the volume of the sphere?

aev [14]

Answer:

The volume of the sphere is 1436 in³

The equation is

V = ⁴⁄₃ * 3.14 * (7ft)³

Step-by-step explanation:

radius = half of diameter

d = 14ft

r = 14ft / 2 = 7ft

To calculate the volume of a sphere we have to use the following formula:

V = volume

r = radius = 7ft

π = 3.14

V = ⁴⁄₃πr³

we replace with the known values

V = ⁴⁄₃ * 3.14 * (7ft)³

V = 4.187 * 343 in³

V = 1436 in³

The volume of the sphere is 1436 in³

6 0

1 year ago

There are two approaches to solving the equation -3x + 10 = 4x - 20 Subtract 4x from both sides. OR Add 3x to both sides. –7x +

Scrat [10]

Answer:

There are two approaches to solving the equation -3x + 10 = 4x - 20. The value of x after solving in both methods is 4.2857.

Solution:

Given, equation is -3x + 10 = 4x – 20.

We have to solve the given equation in two methods, now let us see the first method.

1st method ⇒ by subtracting 4x from both sides.

Then, -3x + 10 = 4x – 20 ⇒ -3x + 10 – 4x = 4x – 20 – 4x ⇒ -3x – 4x = -20 – 10 ⇒ -7x = -30 ⇒ 7x = 30

Then, x = $\frac{30}{7}$

Now, let us use the 2nd method.

2nd method ⇒ by adding 3x on both sides.

Then, -3x + 10 = 4x – 20 ⇒ -3x + 10 + 3x = 4x – 20 + 3x

⇒ 10 + 20 = 4x + 3x

⇒ 7x = 30

⇒ x = $\frac{30}{7}$

Hence, the value of x after solving in both methods is $\frac{30}{7}$ = 4.2857.

6 0

1 year ago

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Find the smallest relation containing the relation {(1, 2), (1, 4), (3, 3), (4, 1)} that is:

professor190 [17]

Answer:

Remember, if B is a set, R is a relation in B and a is related with b (aRb or (a,b))

1. R is reflexive if for each element a∈B, aRa.

2. R is symmetric if satisfies that if aRb then bRa.

3. R is transitive if satisfies that if aRb and bRc then aRc.

Then, our set B is $\{1,2,3,4\}$ .

a) We need to find a relation R reflexive and transitive that contain the relation $R1=\{(1, 2), (1, 4), (3, 3), (4, 1)\}$

Then, we need:

1. That 1R1, 2R2, 3R3, 4R4 to the relation be reflexive and,

2. Observe that

1R4 and 4R1, then 1 must be related with itself.
4R1 and 1R4, then 4 must be related with itself.
4R1 and 1R2, then 4 must be related with 2.

Therefore $\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(4,1),(4,2)\}$ is the smallest relation containing the relation R1.

b) We need a new relation symmetric and transitive, then

since 1R2, then 2 must be related with 1.
since 1R4, 4 must be related with 1.

and the analysis for be transitive is the same that we did in a).

Observe that

1R2 and 2R1, then 1 must be related with itself.
4R1 and 1R4, then 4 must be related with itself.
2R1 and 1R4, then 2 must be related with 4.
4R1 and 1R2, then 4 must be related with 2.
2R4 and 4R2, then 2 must be related with itself

Therefore, the smallest relation containing R1 that is symmetric and transitive is

$\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}$

c) We need a new relation reflexive, symmetric and transitive containing R1.

For be reflexive

1 must be related with 1,

2 must be related with 2,

3 must be related with 3,

4 must be related with 4

For be symmetric

since 1R2, 2 must be related with 1,
since 1R4, 4 must be related with 1.

For be transitive

Since 4R1 and 1R2, 4 must be related with 2,
since 2R1 and 1R4, 2 must be related with 4.

Then, the smallest relation reflexive, symmetric and transitive containing R1 is

$\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}$

5 0

1 year ago

Devin borrowed $1,058 at 13 percent for nine months. What will he pay in interest?

Viktor [21]

Devin borrowed $1,058 at 13 percent for nine months.

We have to calculate the interest paid.

Interest = $\frac{P \times R \times T}{100}$

Substituting the values of

Principal = $1058

Rate = 13%

Time = 9 months = $\frac{9}{12}$ year

Interest = $\frac{1058 \times 13 \times 9}{12 \times 100}$

Interest = 103.155

= 103.16

So, Devin will pay 103.16 as the interest.

Therefore, Option A is the correct answer.

8 0

2 years ago

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