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

What will he pay in interest?

How much is his total payment?

What will be his monthly payment?

The next thing Kevin should do is to write the result in standard form

The standard form of scientific notation is expressed as [tex]A \times 10^n[/tex] where:

Given the product of 3.2 × 10^4 and 3.6 × 10^2 as 11.52 × 10^6, the next thing Kevin should do is to write the result in standard form as shown:

[tex]11.52 \times 10^6 = 1.152 \times 10 \times 10^6\\ 11.52 \times 10^6 = 1.152 \times 10^7[/tex]

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Final answer:

To find the position vector r(t) given its derivative and an initial condition, we integrate each component of the derivative and apply the initial condition to solve for constants of integration. The final position function is r(t) = t³i + t⁴j + (0.5t² - 0.5)k.

Explanation:

To find r(t) given that r'(t) = 3t²i + 4t³j + tk and r(1) = i + j, we integrate each component of r'(t) with respect to t. The integral of a derivative returns the original function plus a constant of integration, which we can solve using the initial condition provided.

For the i component: ∑ 3t² dt = t³+ C1

For the j component: ∑ 4t³ dt = t⁴ + C2

For the k component: ∑ t dt = 0.5t² + C3

Applying the initial condition r(1) = i + j, we substitute t = 1 into r(t) to solve for the constants of integration:

(1) + C1 = 1, so C1 = 0

(1) + C2 = 1, so C2 = 0

0.5(1) + C3 = 0, so C3 = -0.5

Therefore, the position function r(t) is given by t³i + t⁴j + (0.5t² - 0.5)k.

r(t) = t^3i + t^4j + ((1/2)t^2 - 1/2)k.

To find r(t) given r'(t) = 3t^2i + 4t^3j + tk and r(1) = i + j, we need to integrate r'(t) with respect to t.

Step 1: Integrate each component of r'(t) separately:

∫3t^2 dt = t^3 + C1 (integration with respect to t)

∫4t^3 dt = t^4 + C2 (integration with respect to t)

∫tk dt = (1/2)t^2k + C3 (integration with respect to t)

Step 2: Combine the results to get r(t):

r(t) = (t^3 + C1)i + (t^4 + C2)j + ((1/2)t^2k + C3)

Step 3: Use the given initial condition r(1) = i + j to find the values of the constants C1, C2, and C3:

r(1) = (1^3 + C1)i + (1^4 + C2)j + ((1/2)(1)^2k + C3)

i + j = i + j + (1/2)k + C3

Comparing the coefficients of k, we get:

(1/2) + C3 = 0

C3 = -1/2

Therefore, r(t) = (t^3 + C1)i + (t^4 + C2)j + ((1/2)t^2 - 1/2)k

The cost function is a linear function

The function is given as:

[tex]\mathbf{C(x) = 17.5x - 10}[/tex]

(a) The cost of 10 tickets

This means that: x= 10

So, we have:

[tex]\mathbf{C(x) = 17.5x - 10}[/tex]

[tex]\mathbf{C(10) = 17.5 \times 10 - 10}[/tex]

[tex]\mathbf{C(10) = 175- 10}[/tex]

[tex]\mathbf{C(10) = 165}[/tex]

Hence, the cost of 10 tickets is $165

(b) The number of tickets that costs $130

This means that C(x) = 130

So, we have:

[tex]\mathbf{C(x) = 17.5x - 10}[/tex]

[tex]\mathbf{130 = 17.5x - 10}[/tex]

Add 10 to both sides

[tex]\mathbf{140 = 17.5x}[/tex]

Divide both sides by 17.5

[tex]\mathbf{8 = x}[/tex]

Rewrite as:

[tex]\mathbf{x = 8}[/tex]

Hence, you can buy 8 tickets with $130

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Final answer:

Paul needs to buy 1 carton of eggs, as there are 6 teams needing 1 egg each, and one carton contains 6 eggs.

Explanation:

Paul needs to calculate how many cartons of eggs to buy for the egg toss at the world egg day fair.

With 48 people participating in teams of 8, we have a total of 48/8 = 6 teams, and each team needs 1 egg.

Since each carton contains 6 eggs, Paul needs to buy 6/6 = 1 carton to provide one egg for each team.

Answer:

30.28

Step-by-step explanation:

just did the k12 test so i thought i would help out:)

Answer:

640 m

Step-by-step explanation:

We can consider 4 seconds to be 1 time unit. Then 8 more seconds is 2 more time units, for a total of 3 time units.

The distance is proportional to the square of the number of time units. After 1 time unit, the distance is 1² × 80 m. Then after 3 time units, the distance will be 3² × 80 m = 720 m.

In the additional 2 time units (8 seconds), the ball dropped an additional

... (720 -80) m = 640 m

_____

Alternate solution

You can write the equation for the proportionality and find the constant that goes into it. If we use seconds (not 4-second intervals) as the time unit, then we can say ...

... d = kt²

Filling in the information related to the first 4 seconds, we have ...

... 80 = k(4)²

... 80/16 = k = 5

Then the distance equation becomes ...

... d = 5t²

After 12 seconds (the first 4 plus the next 8), the distance will be ...

... d = 5×12² = 5×144 = 720 . . . meters

That is, the ball dropped an additional 720 -80 = 640 meters in the 12 -4 = 8 seconds after the first data point.

Answer:

Let A = (5, -1) (x1, y1)

Let B = (10, -1) (x2, y2)

Length AB =

[tex] = \sqrt{ {(x2 - x1)}^{2} + {(y2 - y1)}^{2} } [/tex]

[tex] = \sqrt{ {(10 - 5)}^{2} + {( - 1 + ( - 1))}^{2} } [/tex]

[tex] \\ = \sqrt{ {(5)}^{2} + {( - 2)}^{2} } [/tex]

[tex] = √25 - 4[/tex]

[tex] = √21[/tex]

Answer:

[tex]6000x=135000[/tex]

23 fire extinguisher.

Step-by-step explanation:

Let x be the number of fire extinguishers.

We have been given that one fire extinguisher be available for every 6,000 square feet in a building. So the total area covered by x extinguisher will be 6,000x.

Since the building has an area of 135,000 square feet. This means that that total area covered by x extinguisher is 135,000 square feet. We can represent this information in an equation as:

[tex]6000x=135000[/tex]

Therefore, the equation [tex]6000x=135000[/tex] represents the number of fire extinguishers needed for a building that has 135,000 square feet.

Now we will solve for by dividing both sides of our equation by 6000.

[tex]\frac{6000x}{6000}=\frac{135000}{6000}[/tex]

[tex]x=\frac{135}{6}[/tex]

[tex]x=22.5[/tex]

Since we can not have 0.5 of a fire extinguisher, so we will round up our answer as:

[tex]x\approx 23[/tex]

Therefore, 23 fire extinguisher are needed for the building.

Answer:

21.8

Step-by-step explanation:

Leo painted for 2/3 of an hour each day on Monday, Tuesday, Thurday, and friday. Thus, the total time of the week for which he painted is 1.58% of the total time.

A fraction represents a part of a whole number or, more commonly, any number of equal parts. A fraction can be defined as the parts of a certain size, for example, one-half, eight-fifths, three-quarters, etc.

The time for which Leo painted in this week can be calculated as:

Leo painted 2/3 of an hour each day for monday, tuesday, thursday, and friday

Therefore, the time of painting in an hour = 40 minutes

2/3 × 60 = 120/ 3 = 40 minutes = 40 × 4 = 160 minutes

The number of minutes in a day,

1 day = 60 × 24 = 1,440 minutes

The number of minutes in seven days,

7 days = 1440 × 7 = 10,080 minutes

Therefore, the total time for which Leo painted = time for which he painted in minute divided by total time in minutes

Total time for which Leo painted = 160/ 1440 = 0.0158 or 1.58%

Therefore, total time for which he painted in this week is 0.0158 of total time

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If line AB is parallel to line CD and line CD is perpendicular to line EF, we can conclude that line AB is also perpendicular to line EF.

If line AB is parallel to line CD and line CD is perpendicular to line EF, we can conclude that line AB is also perpendicular to line EF.

The reason for this is that if two lines are parallel to the same line, they are parallel to each other. In this case, we have line AB and line CD both parallel to line EF. Then, because line CD is perpendicular to line EF, line AB must also be perpendicular to line EF.

Answer:

1

Step-by-step explanation:

Answer:

2:5!

Step-by-step explanation:

Have a great day!

In an isometric transformation, the preimage and image must not change size. Therefore, the correct option is A.