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3 years ago

A bond payable is similar to which of the following?

2 answers:

4 0

A Bond payable is are likely similar to note payable. They are similar because they have both written premises to pay the interest and the principal amount on a specific futures dates. They are both liability and also the interest is accrued in current liability.

Trava [24]3 years ago

3 0

The answer to your question is "Notes Payable."

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HELP WILL GIVE BRAINLIEST IF CORRECT

marysya [2.9K]

C. 1 only because csc x is the only one that has asymptotes at npi

4 0

3 years ago

Some body can help me with a geometric mean maze

Mars2501 [29]

Answer:

See explanation

Step-by-step explanation:

Theorem 1: The length of the altitude to the hypotenuse of a right triangle is the geometric mean of the lengths of the segments of the hypotenuse.

Theorem 2: The length of each leg of a right triangle is the geometric mean of the length of the hypotenuse and the length of the segment of the hypotenuse adjacent to that leg.

1. Start point: By the 1st theorem,

$x^2=25\cdot (49-25)=25\cdot 24=5^2\cdot 2^2\cdot 6\Rightarrow x=5\cdot 2\cdot \sqrt{6}=10\sqrt{6}.$

2. South-East point from the Start: By the 2nd theorem,

$x^2=40\cdot (40+5)=4\cdot 5\cdot 2\cdot 9\cdot 5\Rightarrow x=2\cdot 5\cdot 3\cdot \sqrt{2}=30\sqrt{2}.$

3. West point from the previous: By the 2nd theorem,

$x^2=(32-20)\cdot 32=4\cdot 3\cdot 16\cdot 2\Rightarrow x=2\cdot 4\cdot \sqrt{6}=8\sqrt{6}.$

4. West point from the previous: By the 1st theorem,

$9^2=x\cdot 15\Rightarrow x=\dfrac{81}{15}=\dfrac{27}{5}=5.4.$

5. West point from the previous: By the 2nd theorem,

$10^2=8\cdot (8+x)\Rightarrow 8+x=12.5,\ x=4.5.$

6. North point from the previous: By the 1st theorem,

$x^2=48\cdot 6=6\cdot 4\cdot 2\cdot 6\Rightarrow x=6\cdot 2\cdot \sqrt{2}=12\sqrt{2}.$

7. East point from the previous: By the 2nd theorem,

$x^2=22.5\cdot 30=225\cdot 3\Rightarrow x=15\sqrt{3}.$

8. North point from the previous: By the 1st theorem,

$x^2=7.5\cdot 36=270\Rightarrow x=3\sqrt{30}.$

8. West point from the previous: By the 2nd theorem,

$x^2=12.5\cdot (12.5+13.5)=12.5\cdot 26=25\cdot 13\Rightarrow x=5\sqrt{13}.$

9. North point from the previous: By the 1st theorem,

$12^2=x\cdot 30\Rightarrow x=\dfrac{144}{30}=4.8.$

101. East point from the previous: By the 1st theorem,

$6^2=1.6\cdot (x-1.6)\Rightarrow x-1.6=22.5,\ x=24.1.$

11. East point from the previous: By the 2nd theorem,

$20^2=32\cdot (32-x)\Rightarrow 32-x=12.5,\ x=19.5.$

12. South-east point from the previous: By the 2nd theorem,

$18^2=x\cdot 21.6\Rightarrow x=15.$

13. North point=The end.

6 0

3 years ago

The heights in cm of some pupils are

tatuchka [14]

The median is a halfway point in the set of data.

List the numbers in order from smallest to largest:

111, 129, 144, 149, 152, 162, 166, 171

Because there are an even number of values in the data set, find the two middle numbers, add them together and then divide by 2:

Median = ( 149 + 152 ) /2

Median = 150.5

4 0

3 years ago

A castle is surrounded by a circular moat which is 5 m wide

Elanso [62]

Answer:

1414 kL

Step-by-step explanation:

The volume of the donut-shaped moat is the product of its surface area and depth. The area is the product of its centerline length and its width.

The diameter of the centerline of the moat is (50 m -5 m) = 45 m. The length of that centerline is ...

C = πd = π(45 m) = 45π m

The area is this length times the width of the moat:

moat area = (45π m)(5 m) = 225π m²

<h3>Moat volume</h3>

The volume is the product of the area and the depth of the moat:

V = Ah = (225π m²)(2 m) = 450π m³ ≈ 1413.72 m³

1 cubic meter is 1000 liters, 1 kiloliter.

The volume of the moat is about 1414 kL.

5 0

1 year ago

Help me solve this problem

oee [108]

Answer:

The answer is True. Just trust me on this one.

8 0

3 years ago